For example, using this script:
#!/bin/bash
for a in $@
do
echo $a
done
And running:./script "x y" z returns:
x
y
z
and not:
x y
z
Why is that?
And how would I pass string arguments with spaces to bash?
I use Bash 4.3.33.
-
See the documentation about this.– TotorMar 24, 2015 at 9:06
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2 Answers
Quote $@:
#!/bin/bash
for a in "$@"
do
echo "$a"
done
Output:
x y z
-
7For some versions (at least) of
bash, it's enough to sayfor a. It will assume that you meanin "$@". Mar 23, 2015 at 19:25 -
On a side note, how does one do this with introducing a variable? For example
T=$@ for a in $T do echo "$a" donedoes not work. Mar 23, 2015 at 23:01 -
2@orangeorange You can't safely store multiple arguments in a plain variable. But you can store them in an array with
t=("$@"), and then reference it as"${t[@]}"to recover the original argument list intact. Note that all of the double-quotes I used are necessary. Mar 24, 2015 at 1:16
Note that in when dealing with $@ you can simplify it to
for a
do
echo "$a"
done
If `in WORDS ...;' is not present, then `in "$@"' is assumed.
