How to handle quoted arguments in $@?

Question

For example here is test.sh

#!/bin/bash for i in "${@}"; do echo "${i}" done

If you run ./test.sh one 'two' "three", you will get

one two three

But what I want to get is:

one 'two' "three"

How can I get this?

Thanks.

./test.sh one "'two'" '"three"' :P Just to inform you, the problem is not in your script. bash calling this script would strip out the quotes... — anishsane
– anishsane, Commented Jun 25, 2015 at 7:49
Thanks for the info. But a user(of the script) usually don't know to do that. So there is no way to pass quoted arguments transparently in bash. — Haishan
– Haishan, Commented Jun 25, 2015 at 8:25

patapouf_ai · Accepted Answer · 2015-06-25 07:55:08Z

Bash strips the quotes from strings. Thus if you want quotes, you need to put them inside other quotes.

Doing:

./test.sh one "'two'" '"three"'

should give you the result you want.

Another possibility is to escape your quotes so that bash knows that it is part of the string:

./test.sh one \'two\' \"three\"

will work, so will:

./test.sh "one" "\'two\'" "\"three\""

Otherwise you can always add the quotes again in your script by doing:

echo "\"${i}\""

for example.

vlp · Accepted Answer · 2015-06-25 07:58:05Z

you need to adapt your input. For example like this:

# ./test.sh one \'two\' \"three\" one 'two' "three" #

2 Answers 2