I have the following code:
script_list=$'1 a\n2\n3'
old_IFS="${IFS}"
IFS=$'\n'
for line in "${script_list}"; do
echo "__${line}__";
done
IFS="${old_IFS}"
which shall produce:
__1 a__
__2__
__3__
However, instead it gives me:
__1 a
2
3__
I double quoted "${script_list}" because it can contain spaces (see: https://stackoverflow.com/a/10067297). But I believe this is also where the problem lies, because when I remove the double quotes around it, it works as expected.
What am I missing?
edit:
As Cyrus suggested, I ran the code at ShellCheck and it tells me:
$ shellcheck myscript
Line 5:
for line in "${script_list}"; do
^-- SC2066: Since you double quoted this, it will not word split, and the loop will only run once.
Is it safe to simply remove the double quotes or do I need to be careful with that?
I double quoted "${script_list}" because it can contain spaces
This is only done to prevent the shell from splitting the string at spaces. However you explicitly tell the shell (by setting IFS) that your IFS is now a newline, not a space. The shell would split by default here on newlines, not on spaces, unless you quote it. Hence, remove the quotes.
Related
I am trying to execute script with commands:
sed -i "USER/c\export USER=${signumid}" .bashrc
sed -i "DEVENVHOME=$/c\export DEVENVHOME=${DEVENVHOME:-/home/${signumid}/CPM_WORKAREA/devenv.x}" .bashrc
I want to replace the line with string "USER" in .bashrc with export USER=${signumid} where $signumid variable is being provided through Cygwin prompt. Similarly I want to replace line with string DEVENVHOME=$ with export DEVENVHOME=${DEVENVHOME:-/home/${signumid}/CPM_WORKAREA/devenv.x} in .bashrc file, where $signumid variable is provided through Cygwin prompt.
But I am getting following errors on Cygwin termminal.:
sed: -e expression #1, char 1: unknown command: `U'
sed: -e expression #1, char 3: extra characters after command
The general syntax of a sed script is a sequence of address command arguments statements (separated by newline or semicolon). The most common command is the s substitution command, with an empty address, so we can perhaps assume that that is what you want to use here. You seem to be attempting to interpolate a shell variable $signumid which adds a bit of a complication to this exposition.
If your strings were simply static text, it would make sense to use single quotes; then, the shell does not change the text within the quotes at all. The general syntax of the s command is s/regex/replacement/ where the slash as the argument separator is just a placeholder, as we shall soon see.
sed -i 's/.*USER.*/export USER=you/
s% DEVENVHOME=\$%export DEVENVHOME=${DEVENVHOME:-/home/you/CPM_WORKAREA/devenv.x}%' .bashrc
This will find any line with USER and substitute the entire line with export USER=you; then find any line which contains DEVENVHOME=$ (with a space before, and a literal dollar character) and replace the matched expression with the long string. Because the substitution string uses slashes internally, we use a different regex separator % -- alternatively, we could backslash-escape the slashes which are not separators, but as we shall see, that quickly becomes untenable when we add the following twist. Because the dollar sign has significance as the "end of line" metacharacter in regular expressions, we backslash-escape it.
I have ignored the c\ in your attempt on the assumption that it is simply a misunderstanding of sed syntax. If it is significant, what do you hope to accomplish with it? c\export is not a valid Bash command, so you probably mean something else, but I cannot guess what.
Now, to interpolate the value of the shell variable signumid into the replacement, we cannot use single quotes, because those inhibit interpolation. You have correctly attempted to use double quotes instead (in your edited question), but that means we have to make some additional changes. Inside double quotes, backslashes are processed by the shell, so we need to double all backslashes, or find alternative constructs. Fortunately for us, the only backslash is in \$ which can equivalently be expressed as [$], so let's switch to that notation instead. Also, where a literal dollar sign is wanted in the replacement string, we backslash-escape it in order to prevent the shell from processing it.
sed -i "s/.*USER.*/export USER=$signumid/
s% DEVENVHOME=[$]%export DEVENVHOME=\${DEVENVHOME:-/home/$signumid/CPM_WORKAREA/devenv.x}%" .bashrc
Equivalenty, you could use single quotes around the parts of the script which are meant to be untouched by the shell, and then put an adjacent double-quoted string around the parts which need interpolation, like
'un$touched*by$(the!shell)'"$signumid"'more$[complex]!stuff'
This final script still rests on a number of lucky or perhaps rather unlucky guesses about what you actually want. On the first line, I have changed just USER to a regular expression which matches the entire line -- maybe that's not what you want? On the other hand, the second line makes the opposite assumption, just so you can see the variations -- it only replaces the actual text we matched. Probably one or the other needs to be changed.
Finally, notice how the two separate sed commands have been conflated into a single script. Many newcomers do not realize that sed is a scripting language which accepts an arbitrary number of commands in a script, and simply treat it as a "replace" program with a funny syntax.
Another common source of confusion is the evaluation order. The shell processes the double-quoted string even before sed starts to execute, so if you have mistakes in the quoting, you can easily produce syntax errors in the sed script which lead to rather uninformative error messages (because what sed tells you in the error message is based on what the script looks like after the shell's substutions). For example, if signumid contains slashes, it will produce syntax errors, because sed will see those as terminating separators for the s/// command. An easy workaround is to switch to a separator which does not occur in the value of signumid.
I want to run a command from a bash script which has single quotes and some other commands inside the single quotes and a variable.
e.g. repo forall -c '....$variable'
In this format, $ is escaped and the variable is not expanded.
I tried the following variations but they were rejected:
repo forall -c '...."$variable" '
repo forall -c " '....$variable' "
" repo forall -c '....$variable' "
repo forall -c "'" ....$variable "'"
If I substitute the value in place of the variable the command is executed just fine.
Please tell me where am I going wrong.
Inside single quotes everything is preserved literally, without exception.
That means you have to close the quotes, insert something, and then re-enter again.
'before'"$variable"'after'
'before'"'"'after'
'before'\''after'
Word concatenation is simply done by juxtaposition. As you can verify, each of the above lines is a single word to the shell. Quotes (single or double quotes, depending on the situation) don't isolate words. They are only used to disable interpretation of various special characters, like whitespace, $, ;... For a good tutorial on quoting see Mark Reed's answer. Also relevant: Which characters need to be escaped in bash?
Do not concatenate strings interpreted by a shell
You should absolutely avoid building shell commands by concatenating variables. This is a bad idea similar to concatenation of SQL fragments (SQL injection!).
Usually it is possible to have placeholders in the command, and to supply the command together with variables so that the callee can receive them from the invocation arguments list.
For example, the following is very unsafe. DON'T DO THIS
script="echo \"Argument 1 is: $myvar\""
/bin/sh -c "$script"
If the contents of $myvar is untrusted, here is an exploit:
myvar='foo"; echo "you were hacked'
Instead of the above invocation, use positional arguments. The following invocation is better -- it's not exploitable:
script='echo "arg 1 is: $1"'
/bin/sh -c "$script" -- "$myvar"
Note the use of single ticks in the assignment to script, which means that it's taken literally, without variable expansion or any other form of interpretation.
The repo command can't care what kind of quotes it gets. If you need parameter expansion, use double quotes. If that means you wind up having to backslash a lot of stuff, use single quotes for most of it, and then break out of them and go into doubles for the part where you need the expansion to happen.
repo forall -c 'literal stuff goes here; '"stuff with $parameters here"' more literal stuff'
Explanation follows, if you're interested.
When you run a command from the shell, what that command receives as arguments is an array of null-terminated strings. Those strings may contain absolutely any non-null character.
But when the shell is building that array of strings from a command line, it interprets some characters specially; this is designed to make commands easier (indeed, possible) to type. For instance, spaces normally indicate the boundary between strings in the array; for that reason, the individual arguments are sometimes called "words". But an argument may nonetheless have spaces in it; you just need some way to tell the shell that's what you want.
You can use a backslash in front of any character (including space, or another backslash) to tell the shell to treat that character literally. But while you can do something like this:
reply=\”That\'ll\ be\ \$4.96,\ please,\"\ said\ the\ cashier
...it can get tiresome. So the shell offers an alternative: quotation marks. These come in two main varieties.
Double-quotation marks are called "grouping quotes". They prevent wildcards and aliases from being expanded, but mostly they're for including spaces in a word. Other things like parameter and command expansion (the sorts of thing signaled by a $) still happen. And of course if you want a literal double-quote inside double-quotes, you have to backslash it:
reply="\"That'll be \$4.96, please,\" said the cashier"
Single-quotation marks are more draconian. Everything between them is taken completely literally, including backslashes. There is absolutely no way to get a literal single quote inside single quotes.
Fortunately, quotation marks in the shell are not word delimiters; by themselves, they don't terminate a word. You can go in and out of quotes, including between different types of quotes, within the same word to get the desired result:
reply='"That'\''ll be $4.96, please," said the cashier'
So that's easier - a lot fewer backslashes, although the close-single-quote, backslashed-literal-single-quote, open-single-quote sequence takes some getting used to.
Modern shells have added another quoting style not specified by the POSIX standard, in which the leading single quotation mark is prefixed with a dollar sign. Strings so quoted follow similar conventions to string literals in the ANSI standard version of the C programming language, and are therefore sometimes called "ANSI strings" and the $'...' pair "ANSI quotes". Within such strings, the above advice about backslashes being taken literally no longer applies. Instead, they become special again - not only can you include a literal single quotation mark or backslash by prepending a backslash to it, but the shell also expands the ANSI C character escapes (like \n for a newline, \t for tab, and \xHH for the character with hexadecimal code HH). Otherwise, however, they behave as single-quoted strings: no parameter or command substitution takes place:
reply=$'"That\'ll be $4.96, please," said the cashier'
The important thing to note is that the single string that gets stored in the reply variable is exactly the same in all of these examples. Similarly, after the shell is done parsing a command line, there is no way for the command being run to tell exactly how each argument string was actually typed – or even if it was typed, rather than being created programmatically somehow.
Below is what worked for me -
QUOTE="'"
hive -e "alter table TBL_NAME set location $QUOTE$TBL_HDFS_DIR_PATH$QUOTE"
EDIT: (As per the comments in question:)
I've been looking into this since then. I was lucky enough that I had repo laying around. Still it's not clear to me whether you need to enclose your commands between single quotes by force. I looked into the repo syntax and I don't think you need to. You could used double quotes around your command, and then use whatever single and double quotes you need inside provided you escape double ones.
just use printf
instead of
repo forall -c '....$variable'
use printf to replace the variable token with the expanded variable.
For example:
template='.... %s'
repo forall -c $(printf "${template}" "${variable}")
Variables can contain single quotes.
myvar=\'....$variable\'
repo forall -c $myvar
I was wondering why I could never get my awk statement to print from an ssh session so I found this forum. Nothing here helped me directly but if anyone is having an issue similar to below, then give me an up vote. It seems any sort of single or double quotes were just not helping, but then I didn't try everything.
check_var="df -h / | awk 'FNR==2{print $3}'"
getckvar=$(ssh user#host "$check_var")
echo $getckvar
What do you get? A load of nothing.
Fix: escape \$3 in your print function.
Does this work for you?
eval repo forall -c '....$variable'
I want to run a command from a bash script which has single quotes and some other commands inside the single quotes and a variable.
e.g. repo forall -c '....$variable'
In this format, $ is escaped and the variable is not expanded.
I tried the following variations but they were rejected:
repo forall -c '...."$variable" '
repo forall -c " '....$variable' "
" repo forall -c '....$variable' "
repo forall -c "'" ....$variable "'"
If I substitute the value in place of the variable the command is executed just fine.
Please tell me where am I going wrong.
Inside single quotes everything is preserved literally, without exception.
That means you have to close the quotes, insert something, and then re-enter again.
'before'"$variable"'after'
'before'"'"'after'
'before'\''after'
Word concatenation is simply done by juxtaposition. As you can verify, each of the above lines is a single word to the shell. Quotes (single or double quotes, depending on the situation) don't isolate words. They are only used to disable interpretation of various special characters, like whitespace, $, ;... For a good tutorial on quoting see Mark Reed's answer. Also relevant: Which characters need to be escaped in bash?
Do not concatenate strings interpreted by a shell
You should absolutely avoid building shell commands by concatenating variables. This is a bad idea similar to concatenation of SQL fragments (SQL injection!).
Usually it is possible to have placeholders in the command, and to supply the command together with variables so that the callee can receive them from the invocation arguments list.
For example, the following is very unsafe. DON'T DO THIS
script="echo \"Argument 1 is: $myvar\""
/bin/sh -c "$script"
If the contents of $myvar is untrusted, here is an exploit:
myvar='foo"; echo "you were hacked'
Instead of the above invocation, use positional arguments. The following invocation is better -- it's not exploitable:
script='echo "arg 1 is: $1"'
/bin/sh -c "$script" -- "$myvar"
Note the use of single ticks in the assignment to script, which means that it's taken literally, without variable expansion or any other form of interpretation.
The repo command can't care what kind of quotes it gets. If you need parameter expansion, use double quotes. If that means you wind up having to backslash a lot of stuff, use single quotes for most of it, and then break out of them and go into doubles for the part where you need the expansion to happen.
repo forall -c 'literal stuff goes here; '"stuff with $parameters here"' more literal stuff'
Explanation follows, if you're interested.
When you run a command from the shell, what that command receives as arguments is an array of null-terminated strings. Those strings may contain absolutely any non-null character.
But when the shell is building that array of strings from a command line, it interprets some characters specially; this is designed to make commands easier (indeed, possible) to type. For instance, spaces normally indicate the boundary between strings in the array; for that reason, the individual arguments are sometimes called "words". But an argument may nonetheless have spaces in it; you just need some way to tell the shell that's what you want.
You can use a backslash in front of any character (including space, or another backslash) to tell the shell to treat that character literally. But while you can do something like this:
reply=\”That\'ll\ be\ \$4.96,\ please,\"\ said\ the\ cashier
...it can get tiresome. So the shell offers an alternative: quotation marks. These come in two main varieties.
Double-quotation marks are called "grouping quotes". They prevent wildcards and aliases from being expanded, but mostly they're for including spaces in a word. Other things like parameter and command expansion (the sorts of thing signaled by a $) still happen. And of course if you want a literal double-quote inside double-quotes, you have to backslash it:
reply="\"That'll be \$4.96, please,\" said the cashier"
Single-quotation marks are more draconian. Everything between them is taken completely literally, including backslashes. There is absolutely no way to get a literal single quote inside single quotes.
Fortunately, quotation marks in the shell are not word delimiters; by themselves, they don't terminate a word. You can go in and out of quotes, including between different types of quotes, within the same word to get the desired result:
reply='"That'\''ll be $4.96, please," said the cashier'
So that's easier - a lot fewer backslashes, although the close-single-quote, backslashed-literal-single-quote, open-single-quote sequence takes some getting used to.
Modern shells have added another quoting style not specified by the POSIX standard, in which the leading single quotation mark is prefixed with a dollar sign. Strings so quoted follow similar conventions to string literals in the ANSI standard version of the C programming language, and are therefore sometimes called "ANSI strings" and the $'...' pair "ANSI quotes". Within such strings, the above advice about backslashes being taken literally no longer applies. Instead, they become special again - not only can you include a literal single quotation mark or backslash by prepending a backslash to it, but the shell also expands the ANSI C character escapes (like \n for a newline, \t for tab, and \xHH for the character with hexadecimal code HH). Otherwise, however, they behave as single-quoted strings: no parameter or command substitution takes place:
reply=$'"That\'ll be $4.96, please," said the cashier'
The important thing to note is that the single string that gets stored in the reply variable is exactly the same in all of these examples. Similarly, after the shell is done parsing a command line, there is no way for the command being run to tell exactly how each argument string was actually typed – or even if it was typed, rather than being created programmatically somehow.
Below is what worked for me -
QUOTE="'"
hive -e "alter table TBL_NAME set location $QUOTE$TBL_HDFS_DIR_PATH$QUOTE"
EDIT: (As per the comments in question:)
I've been looking into this since then. I was lucky enough that I had repo laying around. Still it's not clear to me whether you need to enclose your commands between single quotes by force. I looked into the repo syntax and I don't think you need to. You could used double quotes around your command, and then use whatever single and double quotes you need inside provided you escape double ones.
just use printf
instead of
repo forall -c '....$variable'
use printf to replace the variable token with the expanded variable.
For example:
template='.... %s'
repo forall -c $(printf "${template}" "${variable}")
Variables can contain single quotes.
myvar=\'....$variable\'
repo forall -c $myvar
I was wondering why I could never get my awk statement to print from an ssh session so I found this forum. Nothing here helped me directly but if anyone is having an issue similar to below, then give me an up vote. It seems any sort of single or double quotes were just not helping, but then I didn't try everything.
check_var="df -h / | awk 'FNR==2{print $3}'"
getckvar=$(ssh user#host "$check_var")
echo $getckvar
What do you get? A load of nothing.
Fix: escape \$3 in your print function.
Does this work for you?
eval repo forall -c '....$variable'
I have a bash script,
echo 'abcd'
in shell, I want to show ab'c'd and I have tried following approach but without success
echo 'ab\'c\'d'
I am asking is it possible to show single quote in single quoted text?
From the bash manual section on Single Quotes:
A single quote may not occur between single quotes, even when preceded by a backslash.
You'll need to use double quotes instead. It's not pretty, but the following gives the output you are looking for:
echo 'ab'"'"'c'"'"'d'
A bash-specific feature, not part of POSIX, is a $'...'-quoted string:
echo $'ab\'c\'d'
Such a string behaves identically to a single-quoted string, but does allow for a selection of \-escaped characters (such as \n, \t, and yes, \').
A="echo 'q'"
$A
I got the result 'q'
but if I type echo 'q' directly, it's q (without single quote)
So, I wonder what rule does bash follow while facing single quotes inside the double quotes.
The original problem is
A="curl http://123.196.125.62/send -H 'Host: poj.org' -e http://poj.org/send"
$A
I got curl: (6) Couldn't resolve host 'poj.org''
it will be all right if I just type the command into the terminal..
P.S. I'd like to use $A for excuting the command inside A
Please see BashFAQ/050: I'm trying to put a command in a variable, but the complex cases always fail!
It's best to avoid putting commands in variables for the reason you've experienced, among others.
You should use a function and pass it arguments.
Why do you want to do this rather than simply executing the command directly?
If you must do it, use an array:
A=(curl http://123.196.125.62/send -H 'Host: poj.org' -e http://poj.org/send)
${A[#]}
Regarding the treatment of single quotes within double quotes, they are treated literally and as part of the rest of the string. Here is the relevant paragraph from man bash:
Enclosing characters in double quotes preserves the literal value of
all characters within the quotes, with the exception of $, `, \, and,
when history expansion is enabled, !. The characters $ and ` retain
their special meaning within double quotes. The backslash retains its
special meaning only when followed by one of the following characters:
$, `, ", \, or <newline>. A double quote may be quoted within double
quotes by preceding it with a backslash. If enabled, history expansion
will be performed unless an ! appearing in double quotes is escaped
using a backslash. The backslash preceding the ! is not removed.
If you want to "save" a command for later execution, you do NOT want a variable. You want a function.
a() { curl http://123.196.125.62/send -H 'Host: poj.org' -e http://poj.org/send; }
Putting code in variables is bad since variables are containers for data not code. Additionally, you are seeing the problem because $A does NOT execute the bash code in A, what it really does is split the value of A into words, then it performs Pathname Expansion on these words, and as a result of those two operations, it executes a program named by the first resulting word and passes the other words as arguments. In your particular case, this is what happens (I use [] to indicate "units"):
A: [echo 'q']
after wordsplitting: [echo] ['q']
after pathname expansion: [echo] ['q']
Now bash looks for a program called echo and passes the argument 'q' to it.
This is NOT executing bash code, because if you execute echo 'q' as bash code, bash removes the single quotes after it's done with them. Similarly, you cannot do pipes, redirection et al. like this, because they too are bash syntax (just like your single quotes).
Recap: never put code in bash variables. never leave parameters unquoted (if you think doing that fixes something, you are wrong, you've just made things worse, go fix the real problem). The solution is to use a function.
You'd better use backquotes in this case:
A=`curl http://123.196.125.62/send -H 'Host: poj.org' -e 'http://poj.org/send'`