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bashshellquotes

Nested quotes in Bash

Please let me explain my scenario with some sample script which is much simpler than my actual application but shows the same behavior:

For the demo I use two bash shell scripts. The first one is called argtest.sh simply outputs the command line parameters:

#/bin/bash

echo "Argument 1: $1"
echo "Argument 2: $2"

Test:

root@test:~# ./argtest.sh 1 2
Argument 1: 1
Argument 2: 2

and

root@test:~# ./argtest.sh "1 2" 3
Argument 1: 1 2
Argument 2: 3

This works as expected

I've created another command like this:

./argtest.sh $(for i in ` seq 1 2 ` ; do echo Number $i; done)

This is the result

Argument 1: Number
Argument 2: 1

I've tried many different combinations of quotation marks (") and escaped quotation marks (\") around Number, but no combination produced the desired output of:

Argument 1: Number 1
Argument 2: Number 2

How can I get to that output?

Solution

  • because arguments are subject to word splitting, you may change $IFS and put the the new separator into your echo

    IFS=$'_\n'
./argtest.sh $(for i in {1..2} ; do echo "Number ${i}_"; done )

    don't forget to restore the old $IFS

    oIFS=$IFS
...
IFS=$oIFS

    btw {1..2} is equivalent to seq 1 2 but you don't need an external command