Say I have this:
foo="bar"
echo "\$foo"
That won't work because it will escape the $.
I just want to concatenate the backslash with the content of variable foo.
One way is mentioned by h0ch5tr4355 in a comment, another way you can try if you just want to print and concern with out put then you can try this.
#!/bin/bash
foo="bar"
echo '\'$foo
As mentioned in comment, I am also expecting that you try print foo instead of bar, as bar is empty.
Single-Quotes
Double-Quotes
Yeah this was a dumb question, one solution is as simple as:
foo='bar'
back_slash='\'
echo "$back_slash$foo"
or if you want to be extra careful:
echo "${back_slash}${foo}"
The question is basically, how does bash deal with double quotes and brackets?
I couldn't come up with a better example, so apologies about that, but imagine something like this:
"$(echo "${foo}")"
What happens there? Is ${foo} technically out of the quotation scope? Is bash smart enough to also consider that it is inside of the parentheses and work inside-out? Would \"${foo}\" work better?
I've tried Googling this, but no luck. Please note that this is different from just nested quoting questions, as it includes expansion.
Thanks!
Each command substitution establishes a new quoting context, so the correct way to avoid word splitting inside nested command substitutions is to use double quotes.
In this example, white space is preserved:
$ echo "$(echo "$(echo 'foo bar')")" # outer echo sees 'foo bar'
foo bar
However, missing any of the double quotes means that the string is split by the shell:
$ echo $(echo "$(echo 'foo bar')") # outer echo sees 'foo' 'bar'
foo bar
$ echo "$(echo $(echo 'foo bar'))" # middle echo sees 'foo' 'bar'
foo bar
echo outputs each argument, separated by a spaces, which is where the single space between "foo" and "bar" comes from.
The example you gave does save you from word splitting completely:
"$(echo "${foo}")"
while,
"$(echo ${foo})"
would save you from word splitting on the output but not on the variable $foo.
$() opens a subshell and you need to quote your vars in this subshell too. Consider the following example:
"$(echo "$(echo "$foo")")"
None of the pair of quotes above is optional, if you miss any of them, you will be subject to word splitting.
Consider another example:
"$foo var in same shell, $(cmd) - this output is also in the same shell"
In above, $foo and the output of cmd are safe, but if the cmd includes another variable inside $(), then that variable will need to be quoted.
What is foo? Is it a value? try
"$(echo "$\{foo\}")"
Is there a way to tell bash/zsh to not parse the quotes at all but give them to a shell function verbatim?
$ argtest abc def "ghi jkl" $'mno\tpqr' $'stu\nvwx'
abc
def
"ghi jkl"
$'mno\tpqr'
$'stu\nvwx'
You might be thinking why I don't just do
argtest abc def '"ghi jkl"' "$'mno\tpqr'" "$'stu\nvwx'"
But the argtest function I'm trying to create tries to wrap around other commands which can have noglob prefixes. So I need a way of being able to tell apart * and '*' .
In zsh, you can use the q parameter expansion flag, but it's messy. One q escapes individual characters as necessary; two expands the text in single quotes; three in double quotes. (The notation ${:-stuff} simply expands to the text following :-; it's a wrapper that allows you to create anonymous parameters.)
$ echo "foo bar"
foo bar
$ echo ${(qq):-"foo bar"}
'foo bar'
$ echo ${(qqq):-"foo bar"}
"foo bar"
$ argtest () {
function> echo "$1"
function> }
$ argtest "foo bar"
foo bar
$ argtest ${(qqq):-"foo bar"}
"foo bar"
No, there is no way to disable the shell's quote handling.
If you want this for your own convenience, reading arguments from a here document or similar might be acceptable.
But if you want your users to be able to write quotes at the shell prompt and have them preserved, there is no way to do that (short of writing your own shell).
In bash, you can access $BASH_COMMAND to see the literal, pre-parsing command being executed. Thus, while you can't prevent the shell from parsing an argument list, you can see its pre-parsed state.
However -- this gives you only the entire argument; you need to do string-splitting yourself, if going this route. As such, I would describe this as an ill-advised course of action.
I have a script I am trying to call that needs to have the $ symbol passed to it. If I run the script as
./script "blah$blah"
it is passed in fine but then the script calls another program I have no control over which then expands the parameter to just "blah". The program is being called by the command program $#. I was wondering if there was a way to prevent the parameter from being expanded when passed to the next script.
Escape the character $ with: \, e.g.: "This will not expand \$hello"
use single quotes: 'This will not expand $hello'
Use a HERE DOC:
<<'EOF'
This will not expand $hello
EOF
In your case I recommend using single quotes for readability: ./script 'blah$blah'.
A couple of options involving changing the quoting:
./script 'blah$blah'
./script "blah\$blah"
I hope this helps.
Call using single quotes:
./script 'blah$blah'
Or escape the $
./script "blah\$blah"
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