This question already has answers here:
Why does shell ignore quoting characters in arguments passed to it through variables? [duplicate]
(3 answers)
Closed 3 years ago.
Explaining the question through examples...
Demonstrates that the single-quotes after --chapters is gets escaped when the variable is expanded (I didn't expect this):
prompt#ubuntu:/my/scripts$ cat test1.sh
#!/bin/bash
actions="--tags all:"
actions+=" --chapters ''"
mkvpropedit "$1" $actions
prompt#ubuntu:/my/scripts$ ./test1.sh some.mkv
Error: Could not open '''' for reading.
And now for some reason mkvpropedit receives the double quotes as part of the filename (I didn't expect this either):
prompt#ubuntu:/my/scripts$ cat test1x.sh
#!/bin/bash
command="mkvpropedit \"$1\""
command+=" --tags all:"
command+=" --chapters ''"
echo "$command"
$command
prompt#ubuntu:/my/scripts$ ./test1x.sh some.mkv
mkvpropedit "some.mkv" --tags all: --chapters ''
Error: Could not open '''' for reading.
The above echo'd command seems to be correct. Putting the same text in another script gives the expected result:
prompt#ubuntu:/my/scripts$ cat test2.sh
#!/bin/bash
mkvpropedit "$1" --tags all: --chapters ''
prompt#ubuntu:/my/scripts$ ./test2.sh some.mkv
The file is being analyzed.
The changes are written to the file.
Done.
Could anyone please explain why the quotes are not behaving as expected. I found searching on this issue difficult as there are so many other quoting discussions on the web. I wouldn't even know how to explain the question without examples.
I am afraid that some day the file name in the argument contains some character that breaks everything, hence the maybe excessive quoting. I do not understand why the same command executes differently when typed directly in the script or when provided via a variable. Please enlighten me.
Thanks for reading.
The important thing to keep in mind is that quotes are only removed once, when the command line is originally parsed. A quote which is inserted into the command line as a result of parameter substitution ($foo) or command substitution ($(cmd args)) is not treated as a special character. [Note 1]
That seems different from whitespace and glob metacharacters. Word splitting and pathname expansion happen after parameter/command substitution (unless the substitution occurs inside quotes). [Note 2]
The consequence is that it is almost impossible to create a bash variable $args such that
cmd $args
If $args contains quotes, they are not removed. Words inside $args are delimited by sequences of whitespace, not single whitespace characters.
The only way to do it is to set $IFS to include some non-whitespace character; that character can then be used inside $args as a single-character delimiter. However, there is no way to quote a character inside a value, so once you do that, the character you chose cannot be used other than as a delimiter. This is not usually very satisfactory.
There is a solution, though: bash arrays.
If you make $args into an array variable, then you can expand it with the repeated-quote syntax:
cmd "${args[#]}"
which produces exactly one word per element of $args, and suppresses word-splitting and pathname expansion on those words, so they end up as literals.
So, for example:
actions=(--tags all:)
actions+=(--chapters '')
mkvpropedit "$1" "${actions[#]}"
will probably do what you want. So would:
args=("$1")
args+=(--tags)
args+=(all:)
args+=(--chapters)
args+=('')
mkvpropedit "${args[#]}"
and so would
command=(mkvpropedit "$1" --tags all: --chapters '')
"${command[#]}"
I hope that's semi-clear.
man bash (or the online version) contains a blow-by-blow account of how bash assembles commands, starting at the section "EXPANSION". It's worth reading for a full explanation.
Notes:
This doesn't apply to eval or commands like bash -c which evaluate their argument again after command line processing. But that's because command-line processing happens twice.
Word splitting is not the same as "dividing the command into words", which happens when the command is parsed. For one thing, word-splitting uses as separator characters the value of $IFS, whereas command-line parsing uses whitespace. But neither of these are done inside quotes, so they are similar in that respect. In any case, words are split in one way or another both before and after parameter substitution.
Related
This question already has answers here:
echo "#!" fails -- "event not found"
(5 answers)
Closed 7 years ago.
I am attempting to parse the output of a VNC server startup event and have run into a problem in parsing using sed in a command substitution. Specifically, the remote VNC server is started in a manner such as the following:
address1="user1#lxplus.cern.ch"
VNCServerResponse="$(ssh "${address1}" 'vncserver' 2>&1)"
The standard error output produced in this startup event is then to be parsed in order to extract the server and display information. At this point the content of the variable VNCServerResponse is something such as the following:
New 'lxplus0186.cern.ch:1 (user1)' desktop is lxplus0186.cern.ch:1
Starting applications specified in /afs/cern.ch/user/u/user1/.vnc/xstartup
Log file is /afs/cern.ch/user/u/user1/.vnc/lxplus0186.cern.ch:1.log
This output can be parsed in the following way in order to extract the server and display information:
echo "${VNCServerResponse}" | sed '/New.*desktop.*is/!d' \
| awk -F" desktop is " '{print $2}'
The result is something such as the following:
lxplus0186.cern.ch:1
What I want to do is use this parsing in a command substitution something like the following:
VNCServerAndDisplayNumber="$(echo "${VNCServerResponse}" \
| sed '/New.*desktop.*is/!d' | awk -F" desktop is " '{print $2}')"
On attempting to do this, I am presented with the following error:
bash: !d': event not found
I am not sure how to address this. It appears to be a problem in the way sed is being used in the command substitution. I would appreciate guidance.
Bash history expansion is a very odd corner in the bash command line parser, and you are clearly running into an unexpected history expansion, which is explained below. However, any sort of history expansion in a script is unexpected, because normally history expansion is not enabled in scripts; not even scripts run with the source (or .) builtin.
How history expansion is enabled (or disabled)
There are two shell options which control history expansion:
set -o history: Required for the history to be recorded.
set -H (or set -o histexpand): Additionally required for history expansion to be enabled.
Both of these options must be set for history expansion to be recognized. (I found the manual unclear on this interaction, but it's logical enough.)
According to the bash manual, these options are unset for non-interactive shells, so if you want to enable history expansion in a script (and I cannot imagine a reason you would want this), you would need to set both of them:
set -o history -o histexpand
The situation for scripts run with source is more complicated (and what I'm about to say only applies to bash v4, and since it's undocumented in might change in the future). [Note 3]
History recording (and consequently expansion) is turned off in source'd scripts, but through an internal flag which, as far as I know, is not made visible. It certainly does not appear in $SHELLOPTS. Since a sourced script runs in the current bash context, it shares the current execution environment, including shell options. So in the execution of a sourced script initiated from an interactive session, you'll see both history and histexpand in $SHELLOPTS, but no history expansion will take place. In order to enable it, you need to:
set -o history
which is not a no-op because it has the side-effect of resetting the internal flag which suppresses history recording. Setting the histexpand shell option does not have this side-effect.
In short, I'm not sure how you managed to enable history expansion in a script (if, indeed, the misbehaving command was in a script and not in an interactive shell), but you might want to consider not doing so, unless you have a really good reason.
How history expansion is parsed
The bash implementation of history expansion is designed to work with readline, so that it can be performed during command input. (By default this function is bound to Meta-^; generally Meta is ESC, but you can customize that as well.) However, it is also performed immediately after each line is input, before any bash parsing is performed.
By default, the history expansion character is !, and -- as mostly documented -- that will trigger history expansion except:
when it is followed by whitespace or =
if the shell option extglob is set, and it is followed by ( [Note 1]
if it appears in a single-quoted string
if it is preceded by a \ [Note 2 and see below]
if it is preceded by $ or ${ [Note 1]
if it is preceded by [ [Note 1]
(As of bash v4.3) if it is the last character in a double-quoted string.
The immediate issue here is the precise interpretation of the third case, an ! appearing inside of a single-quoted string. Normally, bash starts a new quoting context for a command substitution ($(...) or the deprecated backtick notation). For example:
$ s=SUBSTITUTED
$ # The interior single quotes are just characters
$ echo "'Echoing $s'"
'Echoing SUBSTITUTED'
$ # The interior single quotes are single quotes
$ echo "$(echo 'Echoing $s')"
Echoing $s
However, the history expansion scanner isn't that intelligent. It keeps track of quotes, but not of command substitution. So as far as it is concerned, both of the single quotes in the above example are double-quoted single quotes, which is to say ordinary characters. So history expansion occurs in both of them:
# A no-op to indicated history expansion
$ HIST() { :; }
# Single-quoted strings inhibit history expansion
$ HIST
$ echo '!!'
!!
# Double-quoted strings allow history expansion
$ HIST
$ echo "'!!'"
echo "'HIST'"
'HIST'
# ... and it applies also to interior command substitution.
$ HIST
$ echo "$(echo '!!')"
echo "$(echo 'HIST')"
HIST
So if you have a perfectly normal command like sed '/foo/!d' file, where you would expect the single-quotes to protect you from history-expansion, and you put it inside a double-quoted command substitution:
result="$(sed '/foo/!d' file)"
you suddenly find that the ! is a history expansion character. Worse, you can't fix this by backslash escaping the exclamation point, because although "\!" inhibits history expansion, it doesn't remove the backslash:
$ echo "\!"
\!
In this particular example -- and the one in the OP -- the double quotes are completely unnecessary, because the right-hand side of a variable assignment does not undergo either filename expansion nor word splitting. However, there are other contexts in which removing the double quotes would change the semantics:
# Undesired history expansion
printf "The answer is '%s'\n" "$(sed '/foo/!d' file)"
# Undesired word splitting
printf "The answer is '%s'\n" $(sed '/foo/!d' file)
In this case, the best solution is probably to put the sed argument in a variable
# Works
sed_prog='/foo/!d'
printf "The answer is '%s'\n" "$(sed "$sed_prog" file)"
(The quotes around $sed_prog were not necessary in this case but usually they would be, and they do no harm.)
Notes:
The inhibition of history expansion when the following character is some form of open parenthesis only works if there is a corresponding close parenthesis in the rest of the string. However, it doesn't have to really match the open parenthesis. For example:
# No matching close parenthesis
$ echo "!("
bash: !: event not found
# The matching close parenthesis has nothing to do with the open
$ echo "!(" ")"
!( )
# An actual extended glob: files whose names don't start with a
$ echo "!(a*)"
b
As indicated in the bash manual, a history-expansion character is treated as an ordinary character if immediately preceded by a backslash. This is literally true; it doesn't matter whether the backslash will later be considered an escape character or not:
$ echo \!
!
$ echo \\!
\!
$ echo \\\!
\!
\ also inhibits history expansion inside double quotes, but \! is not a valid escape sequence inside the double quoted string, so the backslash is not removed:
$ echo "\!"
\!
$ echo "\\!"
\!
$ echo "\\\!"
\\!
I'm referring to the source code for bash v4.2 as I write this, so any undocumented behaviour may be completely different as of v4.3.
The problem is that within double quotes, bash is trying to expand !d before passing it to the subshell. You can get around this problem by removing the double quotes but I would also propose a simplification to your script:
VNCServerAndDisplayNumber=$(echo "$VNCServerResponse" | awk '/desktop/ {print $NF}')
This simply prints the last field on the line containing the word "desktop".
On a newer bash, you can use a herestring rather than piping an echo:
VNCServerAndDisplayNumber=$(awk '/desktop/ {print $NF}' <<<"$VNCServerResponse")
Don't wrap the $(...) command substitution in double quotes. You are asking the shell to perform evaluation on the contents of the quotes and are hitting the history substitution expansion feature. Drop the quotes and you stop telling the shell to do that and you won't hit that problem.
And yes, dropping those quotes is safe on that assignment line even if the output may contain spaces or newlines or whatever. Assignments of that sort are not going to split on those the way command substitution or variable evaluation will on a normal shell execution line.
Alternatively, disable history expansion in your shell/script before you run that. (It should be off when running a script by default I believe anyway.)
This only happens when history expansion is enabled, which it normally isn't and definitely shouldn't be for scripts.
Rather than trying to work around it, figure out why history expansion is enabled and what to do so it isn't.
If you're executing your script with . foo or source foo, use ./foo instead.
If you're writing this as a function in .bashrc or similar, consider making it a separate script.
If your script (or BASH_ENV) explicitly does set -H, don't.
Quote it with '' or \ or disable history expansion with set +H or shopt -u -o histexpand. See History Expansion.
I'm writing a shell, and am getting unexpected parsing from both bash, dash, and busybox's ash:
echo "`echo a #`"
prints a, however
echo "$(echo a #)"
gives an error about missing a closing ).
How is a comment in a command-substitution parsed according to POSIX?
So, for the commands:
echo "`echo a #`"
and
echo "$(echo a #)"
Will the shell parse the comment as extending to the end of the command substitution, or to the end of the line?
Also, will the shell parse it differently if the command substitutions are not in double quotes?
Finally, are there any other constructs (either in POSIX or bash) where a comment can start inside quotes like this?
According to Posix (Shell&Utilities, §2.6.3), "`echo a #`" is undefined (implying that it should not be used):
The search for the matching backquote shall be satisfied by the first unquoted non-escaped backquote; during this search, if a non-escaped backquote is encountered within a shell comment, … undefined results occur. (emphasis added)
However, the $( command substitution marker is terminated by the "first matching )"; the implication (made explicit by examples in the Rationale, Note 1) is that the matching ) cannot be inside of a shell comment, here-doc or quoted string.
The quotes surrounding the command substitution are not relevant in either case (although, of course, "undefined results" could be different in the quoted case, since they are undefined.)
In bash and certain other shells, comments could also be present inside process substitution (eg., <(…)); however, process substitution cannot be quoted.
Notes:
Thanks to #mklement0, who included this link in a comment.
This question already has answers here:
echo "#!" fails -- "event not found"
(5 answers)
Closed 7 years ago.
I am attempting to parse the output of a VNC server startup event and have run into a problem in parsing using sed in a command substitution. Specifically, the remote VNC server is started in a manner such as the following:
address1="user1#lxplus.cern.ch"
VNCServerResponse="$(ssh "${address1}" 'vncserver' 2>&1)"
The standard error output produced in this startup event is then to be parsed in order to extract the server and display information. At this point the content of the variable VNCServerResponse is something such as the following:
New 'lxplus0186.cern.ch:1 (user1)' desktop is lxplus0186.cern.ch:1
Starting applications specified in /afs/cern.ch/user/u/user1/.vnc/xstartup
Log file is /afs/cern.ch/user/u/user1/.vnc/lxplus0186.cern.ch:1.log
This output can be parsed in the following way in order to extract the server and display information:
echo "${VNCServerResponse}" | sed '/New.*desktop.*is/!d' \
| awk -F" desktop is " '{print $2}'
The result is something such as the following:
lxplus0186.cern.ch:1
What I want to do is use this parsing in a command substitution something like the following:
VNCServerAndDisplayNumber="$(echo "${VNCServerResponse}" \
| sed '/New.*desktop.*is/!d' | awk -F" desktop is " '{print $2}')"
On attempting to do this, I am presented with the following error:
bash: !d': event not found
I am not sure how to address this. It appears to be a problem in the way sed is being used in the command substitution. I would appreciate guidance.
Bash history expansion is a very odd corner in the bash command line parser, and you are clearly running into an unexpected history expansion, which is explained below. However, any sort of history expansion in a script is unexpected, because normally history expansion is not enabled in scripts; not even scripts run with the source (or .) builtin.
How history expansion is enabled (or disabled)
There are two shell options which control history expansion:
set -o history: Required for the history to be recorded.
set -H (or set -o histexpand): Additionally required for history expansion to be enabled.
Both of these options must be set for history expansion to be recognized. (I found the manual unclear on this interaction, but it's logical enough.)
According to the bash manual, these options are unset for non-interactive shells, so if you want to enable history expansion in a script (and I cannot imagine a reason you would want this), you would need to set both of them:
set -o history -o histexpand
The situation for scripts run with source is more complicated (and what I'm about to say only applies to bash v4, and since it's undocumented in might change in the future). [Note 3]
History recording (and consequently expansion) is turned off in source'd scripts, but through an internal flag which, as far as I know, is not made visible. It certainly does not appear in $SHELLOPTS. Since a sourced script runs in the current bash context, it shares the current execution environment, including shell options. So in the execution of a sourced script initiated from an interactive session, you'll see both history and histexpand in $SHELLOPTS, but no history expansion will take place. In order to enable it, you need to:
set -o history
which is not a no-op because it has the side-effect of resetting the internal flag which suppresses history recording. Setting the histexpand shell option does not have this side-effect.
In short, I'm not sure how you managed to enable history expansion in a script (if, indeed, the misbehaving command was in a script and not in an interactive shell), but you might want to consider not doing so, unless you have a really good reason.
How history expansion is parsed
The bash implementation of history expansion is designed to work with readline, so that it can be performed during command input. (By default this function is bound to Meta-^; generally Meta is ESC, but you can customize that as well.) However, it is also performed immediately after each line is input, before any bash parsing is performed.
By default, the history expansion character is !, and -- as mostly documented -- that will trigger history expansion except:
when it is followed by whitespace or =
if the shell option extglob is set, and it is followed by ( [Note 1]
if it appears in a single-quoted string
if it is preceded by a \ [Note 2 and see below]
if it is preceded by $ or ${ [Note 1]
if it is preceded by [ [Note 1]
(As of bash v4.3) if it is the last character in a double-quoted string.
The immediate issue here is the precise interpretation of the third case, an ! appearing inside of a single-quoted string. Normally, bash starts a new quoting context for a command substitution ($(...) or the deprecated backtick notation). For example:
$ s=SUBSTITUTED
$ # The interior single quotes are just characters
$ echo "'Echoing $s'"
'Echoing SUBSTITUTED'
$ # The interior single quotes are single quotes
$ echo "$(echo 'Echoing $s')"
Echoing $s
However, the history expansion scanner isn't that intelligent. It keeps track of quotes, but not of command substitution. So as far as it is concerned, both of the single quotes in the above example are double-quoted single quotes, which is to say ordinary characters. So history expansion occurs in both of them:
# A no-op to indicated history expansion
$ HIST() { :; }
# Single-quoted strings inhibit history expansion
$ HIST
$ echo '!!'
!!
# Double-quoted strings allow history expansion
$ HIST
$ echo "'!!'"
echo "'HIST'"
'HIST'
# ... and it applies also to interior command substitution.
$ HIST
$ echo "$(echo '!!')"
echo "$(echo 'HIST')"
HIST
So if you have a perfectly normal command like sed '/foo/!d' file, where you would expect the single-quotes to protect you from history-expansion, and you put it inside a double-quoted command substitution:
result="$(sed '/foo/!d' file)"
you suddenly find that the ! is a history expansion character. Worse, you can't fix this by backslash escaping the exclamation point, because although "\!" inhibits history expansion, it doesn't remove the backslash:
$ echo "\!"
\!
In this particular example -- and the one in the OP -- the double quotes are completely unnecessary, because the right-hand side of a variable assignment does not undergo either filename expansion nor word splitting. However, there are other contexts in which removing the double quotes would change the semantics:
# Undesired history expansion
printf "The answer is '%s'\n" "$(sed '/foo/!d' file)"
# Undesired word splitting
printf "The answer is '%s'\n" $(sed '/foo/!d' file)
In this case, the best solution is probably to put the sed argument in a variable
# Works
sed_prog='/foo/!d'
printf "The answer is '%s'\n" "$(sed "$sed_prog" file)"
(The quotes around $sed_prog were not necessary in this case but usually they would be, and they do no harm.)
Notes:
The inhibition of history expansion when the following character is some form of open parenthesis only works if there is a corresponding close parenthesis in the rest of the string. However, it doesn't have to really match the open parenthesis. For example:
# No matching close parenthesis
$ echo "!("
bash: !: event not found
# The matching close parenthesis has nothing to do with the open
$ echo "!(" ")"
!( )
# An actual extended glob: files whose names don't start with a
$ echo "!(a*)"
b
As indicated in the bash manual, a history-expansion character is treated as an ordinary character if immediately preceded by a backslash. This is literally true; it doesn't matter whether the backslash will later be considered an escape character or not:
$ echo \!
!
$ echo \\!
\!
$ echo \\\!
\!
\ also inhibits history expansion inside double quotes, but \! is not a valid escape sequence inside the double quoted string, so the backslash is not removed:
$ echo "\!"
\!
$ echo "\\!"
\!
$ echo "\\\!"
\\!
I'm referring to the source code for bash v4.2 as I write this, so any undocumented behaviour may be completely different as of v4.3.
The problem is that within double quotes, bash is trying to expand !d before passing it to the subshell. You can get around this problem by removing the double quotes but I would also propose a simplification to your script:
VNCServerAndDisplayNumber=$(echo "$VNCServerResponse" | awk '/desktop/ {print $NF}')
This simply prints the last field on the line containing the word "desktop".
On a newer bash, you can use a herestring rather than piping an echo:
VNCServerAndDisplayNumber=$(awk '/desktop/ {print $NF}' <<<"$VNCServerResponse")
Don't wrap the $(...) command substitution in double quotes. You are asking the shell to perform evaluation on the contents of the quotes and are hitting the history substitution expansion feature. Drop the quotes and you stop telling the shell to do that and you won't hit that problem.
And yes, dropping those quotes is safe on that assignment line even if the output may contain spaces or newlines or whatever. Assignments of that sort are not going to split on those the way command substitution or variable evaluation will on a normal shell execution line.
Alternatively, disable history expansion in your shell/script before you run that. (It should be off when running a script by default I believe anyway.)
This only happens when history expansion is enabled, which it normally isn't and definitely shouldn't be for scripts.
Rather than trying to work around it, figure out why history expansion is enabled and what to do so it isn't.
If you're executing your script with . foo or source foo, use ./foo instead.
If you're writing this as a function in .bashrc or similar, consider making it a separate script.
If your script (or BASH_ENV) explicitly does set -H, don't.
Quote it with '' or \ or disable history expansion with set +H or shopt -u -o histexpand. See History Expansion.
Thank you very much in advance for helping.
The title says everything: what's the difference between using:
echo `basename $HOME`
and
echo $(basename $HOME)
Please notice that I know what the basename command does, that both syntax are valid and both commands give the same output.
I was just wondering if there is any difference between both and if it's possible, why there are two syntaxes for this.
Cheers
Rafael
The second form has different escaping rules making it much easier to nest. e.g.
echo $(echo $(basename $HOME))
I'll leave working out how to do that with ` as an exercise for the reader, it should prove enlightening.
They are one of the same.
please read this.
EDIT (from the link):
Command substitution
Command substitution allows the output of a command to replace the command itself. Command substitution occurs when a command is enclosed like this:
$(command)
or like this using backticks:
`command`
Bash performs the expansion by executing COMMAND and replacing the command substitution with the standard output of the command, with any trailing newlines deleted. Embedded newlines are not deleted, but they may be removed during word splitting.
$ franky ~> echo `date`
Thu Feb 6 10:06:20 CET 2003
When the old-style backquoted form of substitution is used, backslash retains its literal meaning except when followed by "$", "`", or "\". The first backticks not preceded by a backslash terminates the command substitution. When using the $(COMMAND) form, all characters between the parentheses make up the command; none are treated specially.
Command substitutions may be nested. To nest when using the backquoted form, escape the inner backticks with backslashes.
If the substitution appears within double quotes, word splitting and file name expansion are not performed on the results.
They are alternative syntaxes for command substitution. as #Steve mentions they have different quoting rules and th backticks are harder to nest with. On the other hand they are more portable with older version of bash, and other shells eg csh.
This question already has answers here:
What is the difference between $(command) and `command` in shell programming?
(6 answers)
Closed last year.
There are two ways to capture the output of command line in bash:
Legacy Bourne shell backticks ``:
var=`command`
$() syntax (which as far as I know is Bash specific, or at least not supported by non-POSIX old shells like original Bourne)
var=$(command)
Is there any benefit to using the second syntax compared to backticks? Or are the two fully 100% equivalent?
The major one is the ability to nest them, commands within commands, without losing your sanity trying to figure out if some form of escaping will work on the backticks.
An example, though somewhat contrived:
deps=$(find /dir -name $(ls -1tr 201112[0-9][0-9]*.txt | tail -1l) -print)
which will give you a list of all files in the /dir directory tree which have the same name as the earliest dated text file from December 2011 (a).
Another example would be something like getting the name (not the full path) of the parent directory:
pax> cd /home/pax/xyzzy/plugh
pax> parent=$(basename $(dirname $PWD))
pax> echo $parent
xyzzy
(a) Now that specific command may not actually work, I haven't tested the functionality. So, if you vote me down for it, you've lost sight of the intent :-) It's meant just as an illustration as to how you can nest, not as a bug-free production-ready snippet.
Suppose you want to find the lib directory corresponding to where gcc is installed. You have a choice:
libdir=$(dirname $(dirname $(which gcc)))/lib
libdir=`dirname \`dirname \\\`which gcc\\\`\``/lib
The first is easier than the second - use the first.
The backticks (`...`) is the legacy syntax required by only the very oldest of non-POSIX-compatible bourne-shells and $(...) is POSIX and more preferred for several reasons:
Backslashes (\) inside backticks are handled in a non-obvious manner:
$ echo "`echo \\a`" "$(echo \\a)"
a \a
$ echo "`echo \\\\a`" "$(echo \\\\a)"
\a \\a
# Note that this is true for *single quotes* too!
$ foo=`echo '\\'`; bar=$(echo '\\'); echo "foo is $foo, bar is $bar"
foo is \, bar is \\
Nested quoting inside $() is far more convenient:
echo "x is $(sed ... <<<"$y")"
instead of:
echo "x is `sed ... <<<\"$y\"`"
or writing something like:
IPs_inna_string=`awk "/\`cat /etc/myname\`/"'{print $1}' /etc/hosts`
because $() uses an entirely new context for quoting
which is not portable as Bourne and Korn shells would require these backslashes, while Bash and dash don't.
Syntax for nesting command substitutions is easier:
x=$(grep "$(dirname "$path")" file)
than:
x=`grep "\`dirname \"$path\"\`" file`
because $() enforces an entirely new context for quoting, so each command substitution is protected and can be treated on its own without special concern over quoting and escaping. When using backticks, it gets uglier and uglier after two and above levels.
Few more examples:
echo `echo `ls`` # INCORRECT
echo `echo \`ls\`` # CORRECT
echo $(echo $(ls)) # CORRECT
It solves a problem of inconsistent behavior when using backquotes:
echo '\$x' outputs \$x
echo `echo '\$x'` outputs $x
echo $(echo '\$x') outputs \$x
Backticks syntax has historical restrictions on the contents of the embedded command and cannot handle some valid scripts that include backquotes, while the newer $() form can process any kind of valid embedded script.
For example, these otherwise valid embedded scripts do not work in the left column, but do work on the rightIEEE:
echo ` echo $(
cat <<\eof cat <<\eof
a here-doc with ` a here-doc with )
eof eof
` )
echo ` echo $(
echo abc # a comment with ` echo abc # a comment with )
` )
echo ` echo $(
echo '`' echo ')'
` )
Therefore the syntax for $-prefixed command substitution should be the preferred method, because it is visually clear with clean syntax (improves human and machine readability), it is nestable and intuitive, its inner parsing is separate, and it is also more consistent (with all other expansions that are parsed from within double-quotes) where backticks are the only exception and ` character is easily camouflaged when adjacent to " making it even more difficult to read, especially with small or unusual fonts.
Source: Why is $(...) preferred over `...` (backticks)? at BashFAQ
See also:
POSIX standard section "2.6.3 Command Substitution"
POSIX rationale for including the $() syntax
Command Substitution
bash-hackers: command substitution
From man bash:
$(command)
or
`command`
Bash performs the expansion by executing command and replacing the com-
mand substitution with the standard output of the command, with any
trailing newlines deleted. Embedded newlines are not deleted, but they
may be removed during word splitting. The command substitution $(cat
file) can be replaced by the equivalent but faster $(< file).
When the old-style backquote form of substitution is used, backslash
retains its literal meaning except when followed by $, `, or \. The
first backquote not preceded by a backslash terminates the command sub-
stitution. When using the $(command) form, all characters between the
parentheses make up the command; none are treated specially.
In addition to the other answers,
$(...)
stands out visually better than
`...`
Backticks look too much like apostrophes; this varies depending on the font you're using.
(And, as I just noticed, backticks are a lot harder to enter in inline code samples.)
$() allows nesting.
out=$(echo today is $(date))
I think backticks does not allow it.
It is the POSIX standard that defines the $(command) form of command substitution. Most shells in use today are POSIX compliant and support this preferred form over the archaic backtick notation. The command substitution section (2.6.3) of the Shell Language document describes this:
Command substitution allows the output of a command to be substituted in place of the command name itself. Command substitution shall occur when the command is enclosed as follows:
$(command)
or (backquoted version):
`command`
The shell shall expand the command substitution by executing command
in a subshell environment (see Shell Execution Environment) and
replacing the command substitution (the text of command plus the
enclosing "$()" or backquotes) with the standard output of the
command, removing sequences of one or more <newline> characters at the
end of the substitution. Embedded <newline> characters before the end
of the output shall not be removed; however, they may be treated as
field delimiters and eliminated during field splitting, depending on
the value of IFS and quoting that is in effect. If the output contains
any null bytes, the behavior is unspecified.
Within the backquoted style of command substitution, <backslash> shall
retain its literal meaning, except when followed by: '$' , '`', or
<backslash>. The search for the matching backquote shall be satisfied
by the first unquoted non-escaped backquote; during this search, if a
non-escaped backquote is encountered within a shell comment, a
here-document, an embedded command substitution of the $(command)
form, or a quoted string, undefined results occur. A single-quoted or
double-quoted string that begins, but does not end, within the "`...`"
sequence produces undefined results.
With the $(command) form, all characters following the open
parenthesis to the matching closing parenthesis constitute the
command. Any valid shell script can be used for command, except a
script consisting solely of redirections which produces unspecified
results.
The results of command substitution shall not be processed for further
tilde expansion, parameter expansion, command substitution, or
arithmetic expansion. If a command substitution occurs inside
double-quotes, field splitting and pathname expansion shall not be
performed on the results of the substitution.
Command substitution can be nested. To specify nesting within the
backquoted version, the application shall precede the inner backquotes
with <backslash> characters; for example:
\`command\`
The syntax of the shell command language has an ambiguity for expansions beginning with "$((",
which can introduce an arithmetic expansion or a command substitution that starts with a subshell.
Arithmetic expansion has precedence; that is, the shell shall first determine
whether it can parse the expansion as an arithmetic expansion
and shall only parse the expansion as a command substitution
if it determines that it cannot parse the expansion as an arithmetic expansion.
The shell need not evaluate nested expansions when performing this determination.
If it encounters the end of input without already having determined
that it cannot parse the expansion as an arithmetic expansion,
the shell shall treat the expansion as an incomplete arithmetic expansion and report a syntax error.
A conforming application shall ensure that it separates the "$(" and '(' into two tokens
(that is, separate them with white space) in a command substitution that starts with a subshell.
For example, a command substitution containing a single subshell could be written as:
$( (command) )
I came up with a perfectly valid example of $(...) over `...`.
I was using a remote desktop to Windows running Cygwin and wanted to iterate over a result of a command. Sadly, the backtick character was impossible to enter, either due to the remote desktop thing or Cygwin itself.
It's sane to assume that a dollar sign and parentheses will be easier to type in such strange setups.
Here in 2021 it is worth mentioning a curious fact as a supplement to the other answers.
The Microsoft DevOps YAML "scripting" for pipelines may include Bash tasks. However, the notation $() is used for referring to variables defined in the YAML context, so in this case backticks should be used for capturing the output of commands.
This is mostly a problem when copying scripting code into a YAML script since the DevOps preprocessor is very forgiving about nonexisting variables, so there will not be any error message.