Store a bash array with spaces into variable / file-CodePudding

How can I pass a variable that contains spaces to a script so that the script processes them correctly?

My minimal example:

test.sh

#!/bin/bash
COUNTER=1
for i in "$@"; do echo "$COUNTER '$i'"; COUNTER=`expr $COUNTER   1`; done

If I call the script directly with the arguments it works:

./test.sh 1 2 3\ 4 5

Output (as expected):

1 '1'
2 '2'
3 '3 4'
4 '5'

If I store the arguments into a variable before, the backslash is not interpreted correctly anymore as a escape char.

TEST_ARGS="1 2 3\ 4 5"
./test.sh $TEST_ARGS

Output:

1 '1'
2 '2'
3 '3\'
4 '4'
5 '5'

I would like to get the same output like before. How can I reach it?

TEST_ARGS="1 2 3\ 4 5"
# how to call ./test.sh here?

CodePudding user response：

Use an array, not a plain variable.

(Also, there is no reason to use ancient external tools like expr with Bash; (( COUNTER)) would work just fine. (Also, it is better to avoid uppercase variable names, because those are (by convention) used for environment variables.))

test.sh:

for ((i = 1; i <= $#;   i)); do echo "${i} '${!i}'"; done

Passing the arguments:

test_args=(1 2 '3 4' 5)
./test.sh "${test_args[@]}"

CodePudding user response：

I just found some solution which seems to work so far:

TEST_ARGS="1 2 3\ 4 5"
bash -c "./test.sh $TEST_ARGS"

Output:

1 '1'
2 '2'
3 '3 4'
4 '5'