From the bash man page:
"$*" is equivalent to "$1c$2c...", where c is the first character of the value of the IFS variable.
"$#" is equivalent to "$1" "$2" ...
Any example where "$#" cannot work in place of "$*"?
My favorite use is replacing the field separator.
$ set -- 'My word' but this is a bad 'example!'
$ IFS=,
$ echo "$*"
My word,but,this,is,a,bad,example!
There are other ways to replace delimiters, but IFS and "$*" are often one of the simplest.
These are completely different tools and should be used in completely different situations. There's no reasonable question about one substituting for the other, because in any given situation, only one or the other will be correct.
"$*" is most applicable when you're trying to form a single string argument from an argument list -- mostly for logging (but not cases where division between arguments is important; then, "$#" is appropriate with something like print '%q '). "$#" is useful in... well, any other case.
Examples:
die() {
local stat=$1; shift
log "ERROR: $*"
exit $stat
}
Using "$*" when formatting a string to be passed to log uses only a single argv entry, allowing other optional positional arguments to be added to log's usage in the future.
$* expands to all parameters as just one word with IFS between parameters.
$# expands to all parameters as a list.
Try next code in a file named list.sh:
#!/bin/bash
echo "using '$*'"
for i in "$*"
do
echo $i
done
echo "using '$#'"
for i in "$#"
do
echo $i
done
use it:
./list.sh apple pear kiwi
Related
I don't seem to be able to assign a string that begins with "-e" or "-E" to a bash shell variable:
$ options="-e stuff"
$ echo $options
stuff
Other letters work fine:
$ options="-g stuff"
$ echo $options
-g stuff
What is the reason for this?
You should quote your variable:
echo "${options}"
otherwise it's being expanded to
echo -g stuff
which is being interpreter by echo as its -e option, which actually exists (see man echo), and that's why -e "did not work" while other letters you tried "did".
First: To reliably determine the value of a variable in bash, use declare -p, not echo. Thus:
declare -p options
will emit something like:
declare -- options="-e stuff"
This tells you much more than echo does:
Because it's declare -- rather than declare -x, you know that the variable is not exported.
Because it's not declare -a, you know it's not giving you an array (echo "$array" will print only the first element of a shell array and ignore the rest).
Because it's not declare -i, you know the value wasn't declared to be an integer... etc.
If you're only worried about the string case, but want to ensure that you get a printable value no matter which version of bash is in use (as some historical releases will not always guarantee printable escaping for values printed with declare -p), consider instead:
printf '%q=%q\n' options "$options"
...which will emit unambiguous output even if there are cursor control characters, newlines, or other non-textual contents in your string.
Follow the advice of the POSIX specification for echo, and use printf instead. To quote the APPLICATION USAGE section in full, emphasis added, noting that in bash, -e enables XSI-style interpretation of escape sequences:
It is not possible to use echo portably across all POSIX systems
unless both -n (as the first argument) and escape sequences are
omitted.
The printf utility can be used portably to emulate any of the
traditional behaviors of the echo utility as follows (assuming that
IFS has its standard value or is unset):
The historic System V echo and the requirements on XSI implementations
in this volume of POSIX.1-2008 are equivalent to:
printf "%b\n" "$*"
The BSD echo is equivalent to:
if [ "X$1" = "X-n" ]
then
shift
printf "%s" "$*"
else
printf "%s\n" "$*"
fi
New applications are encouraged to use printf instead of echo.
So, how does this apply to you? Since you want -e to be treated as data, not part of echo's setup, the BSD, non--n branch of that applies:
options="-e stuff"
printf '%s\n' "$options"
Example:
bash script.sh "hello world"
(in script echo "$1")
hello world
Question:
bash script.sh "good" "morning" "everybody"
What do I have to write in my script to output directly:
goodmorningeverybody
So, in general, I want $1, $2, $3, ... (can be 100 but I don't know) to be saved in one variable for example VAR1.
You can refer to all the positional arguments with $* and $#.
From 3.4.2 Special Parameters
*
Expands to the positional parameters, starting from one. When the expansion occurs within double quotes, it expands to a single word with the value of each parameter separated by the first character of the IFS special variable. That is, "$*" is equivalent to "$1c$2c…", where c is the first character of the value of the IFS variable. If IFS is unset, the parameters are separated by spaces. If IFS is null, the parameters are joined without intervening separators.
#
Expands to the positional parameters, starting from one. When the expansion occurs within double quotes, each parameter expands to a separate word. That is, "$#" is equivalent to "$1" "$2" …. If the double-quoted expansion occurs within a word, the expansion of the first parameter is joined with the beginning part of the original word, and the expansion of the last parameter is joined with the last part of the original word. When there are no positional parameters, "$#" and $# expand to nothing (i.e., they are removed).
So to get good morning everybody as output you could just use echo "$*" or echo "$#".
In general # is the more useful of the two variables.
However, if you really want the worlds all smushed together the way you indicate then you have a few options.
The most straightforward of which is a simple loop:
for word; do
s+=$word
done
(for without the in <list> part operates on the positional arguments).
However, you can also do this with * by controlling IFS.
So you could also do s=$(IFS=; echo "$*"). You want/need the sub-shell to avoid setting IFS for the current shell.
Try doing this:
#!/bin/sh
var=$(printf '%s' "$#")
echo "$var"
or even better, credits to chepner :
printf -v var '%s' "$#"
or
#!/bin/bash
for arg; do
str+="$arg"
done
echo "$str"
Output :
goodmorningeverybody
Note :
"$#" expands to each positional parameter as its own argument: "$1" "$2" "$3"...
You could do it with a loop:
for x in "$#"; do
input="$input$x"
done
You can do it using Shell-Parameter-Expansion in this way:
#!/bin/bash
VAR="$*"
VAR=${VAR// /}
echo $VAR
Example
$ script.sh "good" "morning" "everybody"
goodmorningeverybody
How can I traverse through 2nd argument to last argument
like:
for arg in $2-$#
do
echo $i
done
Please help
Actually, don't trust the arguments... they may contain spaces, or other special meta characters to the shell. Double quotes are your friends in shell scripts.
shift; #eat $1
for arg in "$#"
do
echo "$arg"
done
When put in double quotes the "$#" takes on special magic to assure words are retained. Much better then $*. "$*" would process all arguments at once.
Double quotes are not a perfect solution, just the best easy one.
Use shift in bash.
shift
for arg in $#
do
echo $arg
done
args=("$#")
for arg in $(seq 2 `expr $# - 1`)
do
echo ${args[$arg]}
done
VAR="-e xyz"
echo $VAR
This prints xyz, for some reason. I don't seem to be able to find a way to get a string to start with -e.
What is going on here?
The answers that say to put $VAR in quotes are only correct by side effect. That is, when put in quotes, echo(1) receives a single argument of -e xyz, and since that is not a valid option string, echo just prints it out. It is a side effect as echo could just as easily print an error regarding malformed options. Most programs will do this, but it seems GNU echo (from coreutils) and the version built into bash simply echo strings that start with a hyphen but are not valid argument strings. This behaviour is not documented so it should not be relied upon.
Further, if $VAR contains a valid echo option argument, then quoting $VAR will not help:
$ VAR="-e"
$ echo "$VAR"
$
Most GNU programs take -- as an argument to mean no more option processing — all the arguments after -- are to be processed as non-option arguments. bash echo does not support this so you cannot use it. Even if it did, it would not be portable. echo has other portability issues (-n vs \c, no -e).
The correct and portable solution is to use printf(1).
printf "%s\n" "$VAR"
The variable VAR contains -e xyz, if you access the variable via $ the -e is interpreted as a command-line option for echo. Note that the content of $VAR is not wrapped into "" automatically.
Use echo "$VAR" to fix your problem.
In zsh, you can use a single dash (-) before your arguments. This ensures that no following arguments are interpreted as options.
% VAR="-e xyz"
% echo - $VAR
-e xyz
From the zsh docs:
echo [ -neE ] [ arg ... ]
...
Note that for standards compliance a double dash does not
terminate option processing; instead, it is printed directly.
However, a single dash does terminate option processing, so the
first dash, possibly following options, is not printed, but
everything following it is printed as an argument.
The single dash behaviour is different from other shells.
Keep in mind this behavior is specific to zsh.
Try:
echo "$VAR"
instead.
(-e is a valid option for echo - this is what causes this phenomenon).
This question already has answers here:
How can I preserve quotes in printing a bash script's arguments
(7 answers)
Closed 3 years ago.
I have a Bash script where I want to keep quotes in the arguments passed.
Example:
./test.sh this is "some test"
then I want to use those arguments, and re-use them, including quotes and quotes around the whole argument list.
I tried using \"$#\", but that removes the quotes inside the list.
How do I accomplish this?
using "$#" will substitute the arguments as a list, without re-splitting them on whitespace (they were split once when the shell script was invoked), which is generally exactly what you want if you just want to re-pass the arguments to another program.
Note that this is a special form and is only recognized as such if it appears exactly this way. If you add anything else in the quotes the result will get combined into a single argument.
What are you trying to do and in what way is it not working?
There are two safe ways to do this:
1. Shell parameter expansion: ${variable#Q}:
When expanding a variable via ${variable#Q}:
The expansion is a string that is the value of parameter quoted in a format that can be reused as input.
Example:
$ expand-q() { for i; do echo ${i#Q}; done; } # Same as for `i in "$#"`...
$ expand-q word "two words" 'new
> line' "single'quote" 'double"quote'
word
'two words'
$'new\nline'
'single'\''quote'
'double"quote'
2. printf %q "$quote-me"
printf supports quoting internally. The manual's entry for printf says:
%q Causes printf to output the corresponding argument in a format that can be reused as shell input.
Example:
$ cat test.sh
#!/bin/bash
printf "%q\n" "$#"
$
$ ./test.sh this is "some test" 'new
>line' "single'quote" 'double"quote'
this
is
some\ test
$'new\nline'
single\'quote
double\"quote
$
Note the 2nd way is a bit cleaner if displaying the quoted text to a human.
Related: For bash, POSIX sh and zsh: Quote string with single quotes rather than backslashes
Yuku's answer only works if you're the only user of your script, while Dennis Williamson's is great if you're mainly interested in printing the strings, and expect them to have no quotes-in-quotes.
Here's a version that can be used if you want to pass all arguments as one big quoted-string argument to the -c parameter of bash or su:
#!/bin/bash
C=''
for i in "$#"; do
i="${i//\\/\\\\}"
C="$C \"${i//\"/\\\"}\""
done
bash -c "$C"
So, all the arguments get a quote around them (harmless if it wasn't there before, for this purpose), but we also escape any escapes and then escape any quotes that were already in an argument (the syntax ${var//from/to} does global substring substitution).
You could of course only quote stuff which already had whitespace in it, but it won't matter here. One utility of a script like this is to be able to have a certain predefined set of environment variables (or, with su, to run stuff as a certain user, without that mess of double-quoting everything).
Update: I recently had reason to do this in a POSIX way with minimal forking, which lead to this script (the last printf there outputs the command line used to invoke the script, which you should be able to copy-paste in order to invoke it with equivalent arguments):
#!/bin/sh
C=''
for i in "$#"; do
case "$i" in
*\'*)
i=`printf "%s" "$i" | sed "s/'/'\"'\"'/g"`
;;
*) : ;;
esac
C="$C '$i'"
done
printf "$0%s\n" "$C"
I switched to '' since shells also interpret things like $ and !! in ""-quotes.
If it's safe to make the assumption that an argument that contains white space must have been (and should be) quoted, then you can add them like this:
#!/bin/bash
whitespace="[[:space:]]"
for i in "$#"
do
if [[ $i =~ $whitespace ]]
then
i=\"$i\"
fi
echo "$i"
done
Here is a sample run:
$ ./argtest abc def "ghi jkl" $'mno\tpqr' $'stu\nvwx'
abc
def
"ghi jkl"
"mno pqr"
"stu
vwx"
You can also insert literal tabs and newlines using Ctrl-V Tab and Ctrl-V Ctrl-J within double or single quotes instead of using escapes within $'...'.
A note on inserting characters in Bash: If you're using Vi key bindings (set -o vi) in Bash (Emacs is the default - set -o emacs), you'll need to be in insert mode in order to insert characters. In Emacs mode, you're always in insert mode.
I needed this for forwarding all arguments to another interpreter.
What ended up right for me is:
bash -c "$(printf ' %q' "$#")"
Example (when named as forward.sh):
$ ./forward.sh echo "3 4"
3 4
$ ./forward.sh bash -c "bash -c 'echo 3'"
3
(Of course the actual script I use is more complex, involving in my case nohup and redirections etc., but this is the key part.)
Like Tom Hale said, one way to do this is with printf using %q to quote-escape.
For example:
send_all_args.sh
#!/bin/bash
if [ "$#" -lt 1 ]; then
quoted_args=""
else
quoted_args="$(printf " %q" "${#}")"
fi
bash -c "$( dirname "${BASH_SOURCE[0]}" )/receiver.sh${quoted_args}"
send_fewer_args.sh
#!/bin/bash
if [ "$#" -lt 2 ]; then
quoted_last_args=""
else
quoted_last_args="$(printf " %q" "${#:2}")"
fi
bash -c "$( dirname "${BASH_SOURCE[0]}" )/receiver.sh${quoted_last_args}"
receiver.sh
#!/bin/bash
for arg in "$#"; do
echo "$arg"
done
Example usage:
$ ./send_all_args.sh
$ ./send_all_args.sh a b
a
b
$ ./send_all_args.sh "a' b" 'c "e '
a' b
c "e
$ ./send_fewer_args.sh
$ ./send_fewer_args.sh a
$ ./send_fewer_args.sh a b
b
$ ./send_fewer_args.sh "a' b" 'c "e '
c "e
$ ./send_fewer_args.sh "a' b" 'c "e ' 'f " g'
c "e
f " g
Just use:
"${#}"
For example:
# cat t2.sh
for I in "${#}"
do
echo "Param: $I"
done
# cat t1.sh
./t2.sh "${#}"
# ./t1.sh "This is a test" "This is another line" a b "and also c"
Param: This is a test
Param: This is another line
Param: a
Param: b
Param: and also c
Changed unhammer's example to use array.
printargs() { printf "'%s' " "$#"; echo; }; # http://superuser.com/a/361133/126847
C=()
for i in "$#"; do
C+=("$i") # Need quotes here to append as a single array element.
done
printargs "${C[#]}" # Pass array to a program as a list of arguments.
My problem was similar and I used mixed ideas posted here.
We have a server with a PHP script that sends e-mails. And then we have a second server that connects to the 1st server via SSH and executes it.
The script name is the same on both servers and both are actually executed via a bash script.
On server 1 (local) bash script we have just:
/usr/bin/php /usr/local/myscript/myscript.php "$#"
This resides on /usr/local/bin/myscript and is called by the remote server. It works fine even for arguments with spaces.
But then at the remote server we can't use the same logic because the 1st server will not receive the quotes from "$#". I used the ideas from JohnMudd and Dennis Williamson to recreate the options and parameters array with the quotations. I like the idea of adding escaped quotations only when the item has spaces in it.
So the remote script runs with:
CSMOPTS=()
whitespace="[[:space:]]"
for i in "$#"
do
if [[ $i =~ $whitespace ]]
then
CSMOPTS+=(\"$i\")
else
CSMOPTS+=($i)
fi
done
/usr/bin/ssh "$USER#$SERVER" "/usr/local/bin/myscript ${CSMOPTS[#]}"
Note that I use "${CSMOPTS[#]}" to pass the options array to the remote server.
Thanks for eveyone that posted here! It really helped me! :)
Quotes are interpreted by bash and are not stored in command line arguments or variable values.
If you want to use quoted arguments, you have to quote them each time you use them:
val="$3"
echo "Hello World" > "$val"
As Gary S. Weaver shown in his source code tips, the trick is to call bash with parameter '-c' and then quote the next.
e.g.
bash -c "<your program> <parameters>"
or
docker exec -it <my docker> bash -c "$SCRIPT $quoted_args"
If you need to pass all arguments to bash from another programming language (for example, if you'd want to execute bash -c or emit_bash_code | bash), use this:
escape all single quote characters you have with '\''.
then, surround the result with singular quotes
The argument of abc'def will thus be converted to 'abc'\''def'. The characters '\'' are interpreted as following: the already existing quoting is terminated with the first first quote, then the escaped singular single quote \' comes, then the new quoting starts.
Yes, seems that it is not possible to ever preserve the quotes, but for the issue I was dealing with it wasn't necessary.
I have a bash function that will search down folder recursively and grep for a string, the problem is passing a string that has spaces, such as "find this string". Passing this to the bash script will then take the base argument $n and pass it to grep, this has grep believing these are different arguments. The way I solved this by using the fact that when you quote bash to call the function it groups the items in the quotes into a single argument. I just needed to decorate that argument with quotes and pass it to the grep command.
If you know what argument you are receiving in bash that needs quotes for its next step you can just decorate with with quotes.
Just use single quotes around the string with the double quotes:
./test.sh this is '"some test"'
So the double quotes of inside the single quotes were also interpreted as string.
But I would recommend to put the whole string between single quotes:
./test.sh 'this is "some test" '
In order to understand what the shell is doing or rather interpreting arguments in scripts, you can write a little script like this:
#!/bin/bash
echo $#
echo "$#"
Then you'll see and test, what's going on when calling a script with different strings