bash array passing args

[uuperl]

Hi, I have question about the following script. If I define the variable in the script, then the output came out correct. My question is if I passing FOO and BAR as args through script, then the output is wrong.

How can I passing correct through args through script?

thanks!

#!/bin/sh
#FOO=( bar string 'some text' )
#BAR=( foo string 'other text' )

FOO=( $1 )
BAR=( $2 )
foonum=${#FOO}

for ((i=0;i<$foonum;i++)); do
echo ${FOO[${i}]} ${BAR[${i}]}
done

 

[SamBones]

How about this?

#!/bin/bash
#FOO=( bar string 'some text' )
#BAR=( foo string 'other text' )

FOO="$1"
BAR="$2"
foonum=${#FOO}

for ((i=0;i<$foonum;i++)); do
        echo ${FOO[${i}]} ${BAR[${i}]}
done

If not, show us some sample input and the wrong output. It's not clear what you are expecting it to do.

 

[uuperl]

Here is my issue:

When I define the FOO and BAR variable inside the script, then the output will be correct like the following:

bar foo
string string
some text other text

When I passing the variable FOO and BAR through script, then the output will not work

$ ./test "bar string 'some text'" "foo string 'other text'"

output like following:
bar bar string 'some text' foo string 'other text'

which is not what I want like above first output.

thanks!

 

[SamBones]

Then my example should work. The quotes around $1 and $2 should protect the variable. It has spaces in the value you are passing, so they need to be quoted to keep it looking like one value to the shell.

 

[Annihilannic]

The var=(one two three) syntax you are using is a Bash-specific feature, so you should make it clear by using #!/bin/bash at the beginning of your script.

This seems to fix your problem:

eval FOO=($1)
eval BAR=($2)

Annihilannic.