Hello @ilya-levadny and All,
Actually, you want to delete all characters located after the second colon character, included, of any line
Indeed, not difficult with regular expressions !
Open the Replace dialog ( Ctrl + H )
SEARCH (?-s)^(.+?:.+?):.+
REPLACE \1
Select the Regular expression search mode
Check, preferably, the Wrap around option
Click, once on the Replace All button or several times on the Replace button
An other formulation would be :
SEACRH (?-s)^.+?:.+?\K:.+
REPLACE Leave EMPTY
This time, due to the \K syntax you must use the Replace All button, exclusively
Notes :
The (?-s) in-line modifier means that dot ( . ) matches any single standard character, only
The .+?: part looks for the shortest non-null range of characters till a colon ( : )
The part between parentheses ( .+?:.+? ) is stored as group 1, which must be rewritten, during the replacement phase
At the end, the .+ syntax represents the rest of the line, after the second colon char, included, of the current line
In the second version, after matching the implicit group 1, the \K syntax resets the regex engine, so the final overall match is the part :.+ which corresponds to the second colon char with all the subsequent chars of current line, which are deleted as the replacement zone is empty
Best Regards,
guy038