Regex how to insert object found in replacement

[cutcopypaste]

How to replace the
example:

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_e("Full Implementation")

find:

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_e\("(?:[^\\"]+|\\.)*"\)

replace with:
_e\("$1", "my_theme"\)

As is, the $1 is just being replaced with nothing, but I would like it to be the content between the quotes

[agibsonsw]

You are not capturing a group-1. Try this

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_e\("([^\\"]+|\\..*)"\)

but I'm not sure what you are trying to achieve with the backslashes, dots, etc.

[highend]

I don't know if it meets your requirement, but this regex will find the contents inside the brackets:

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_e\("(.*?)"\)

replace with:

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_e\("$1", "my_theme"\)

Leads to:

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_e("Full Implementation", "my_theme")

[cutcopypaste]

thanks both! I just copied this from somewhere as an example of grabbing a text string.

What needs to be done exactly to 'capture' a group?

[highend]

You have to surround them by round brackets.
E.g.:

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(.*?)\.(.*)

Two capture groups.

Let's say you have a Windows file named "file.xml"

Replacing by $1$2 would lead to:
"filexml"

[adzenith]

To clarify, using the ?: after the ( tells the parser that you want this set of parentheses to be non-capturing. Parentheses capture the matched text by default.