Bash: How to escape $@?

Question

I need to write a bash script that, among other things, should pass all its arguments intact to another program.

Minimal example:

 $ cat >proxy.sh #!/bin/bash ./script.sh $@ ^D $ chmod +x proxy.sh $ cat >script.sh #!/bin/bash echo one $1 echo two $2 echo three $3 ^D $ chmod +x script.sh

This naïve approach does not work for arguments with spaces:

 $ ./proxy.sh "a b" c one a two b three c

Expected:

 $ ./proxy.sh "a b" c one a b two c three

What should I write in proxy.sh for this to happen?

Note that I can't use aliases, proxy.sh must be a script — it does some stuff before invoking script.sh.

"...should all its arguments intact to another program." -- is that a typo? What did you mean to say? — user541686
– user541686, Commented Dec 31, 2010 at 9:17

moinudin · Accepted Answer · 2010-12-31 09:11:24Z

13

Quote $@, making it "$@":

$ cat >proxy.sh #!/bin/bash ./script.sh "$@" ^D

Then it retains the original quotes:

one a b two c three

answered Dec 31, 2010 at 9:11

moinudin

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4 Comments

Alexander Gladysh Over a year ago

It works, thanks. I did not expect that, I thought that this would glue all arguments to one... Need to read up on Bash syntax, I guess.

psmears Over a year ago

This is correct. $@ must always be quoted, or else it doesn't do what you want (it does the same as $* instead, which is usually a bug).

Roland Illig Over a year ago

You also have to enclose all other variables in "double quotes" if you want to keep the spaces in them. Except in a few rare cases, so always write "$var" instead of $var.

Andy Over a year ago

does ssh -t some-host -C "some-command $@" quote $@ as desired?

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