how to pass variable value with double quote in shell script [duplicate]

Question

USER_UID=$1
echo 'generate_token("$USER_UID")'

I want output like

generate_token("1234567")

i tried multiple ways but didn't worked. it just print same line without value generate_jwt("$USER_UID")

Answer 1

When you use single quotes, it causes the shell to preserve the literal value of each character within the quotes. This means the $ will be treated as a literal $ character.

You should use double quotes:

USER_UID="$1"
echo "generate_token(\"$USER_UID\")"

From the bash man page, under the Quoting section:

Enclosing characters in double quotes preserves the literal value of all characters within the quotes, with the exception of $, `, \, and, when history expansion is enabled, !.

For POSIX details on quoting, see here.

Example in an interactive shell:

$ USER_UID='foo'
$ echo "generate_token(\"$USER_UID\")"
generate_token("foo")

This will also work if USER_UID contains spaces:

$ USER_UID='var with spaces'
$ echo "generate_token(\"$USER_UID\")"
generate_token("var with spaces")