Bash Variable inside double quotes inside single quotes - Common Expression Language

Question

I need to put variable inside a double quotes inside single quotes.

I want to create common expression language like below.

passbolt list resource --filter 'FolderParentID == "**some-imaginary-id**"'

Just help replace some-imaginary with some bash variable. Thanks in advance.

Answer 1

You need to concatenate separate strings via juxtaposition (i.e., putting them directly next to each other, with no intervening whitespace).

passbolt list resource --filter 'FolderParentID == "'"$some_variable"'"'

Note there are three separate quoted strings:

1. FolderParentID == ", in single quotes
2. $some_variable, in double quotes
3. ", in single quotes

Regardless of the type of quotes used to quote each of the three, the results are concatenated into a single word.

At the expense of a command substitution, might want to use printf to generate the string for you.

passbolt list resource --filter "$(printf 'FolderParentID == "%s"' "$some_variable")"

Here you can "nest" the double quotes, because the command substitution starts a new quoting context: the first " inside the command substitution does not close the initial " that contains the command substitution.

Answer 2

What I would do:

id=12345
echo passbolt list resource --filter "FolderParentID == '$id'"

passbolt list resource --filter FolderParentID == '12345'