Hello!

There is a text of the form:

$var = .\`[qwerty`] qwerty.txt.

It is necessary to find the escape characters "``" (backquotes) and replace them.
How to do it with regulars? I do this:

 $var -replace "``[^.]+``"

Finds, replaces, but with the text inside these return quotes.
Those. it turns out:

 .\] qwerty.txt

How to exclude everything inside?

    1 answer 1

    If you need backquotes with a backslash, then like this:

     \\?`

    Finds a backslash with a quotation mark, and just a quote,

    Otherwise the regular expression is out of place here.