Wildcard in replace field
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A big thank you to everyone who replied to my question offering help. @Alan-Kilborn’s suggestion seems to work perfectly.
All the best to this great community!
-Dax. -
Hey all!
I have no idea why, but for some reason, this solution has stopped working…
Seriosuly, I have no idea what’s going on, but (\d\d:\d\d):(?=\d\d) no longer finds a string of text that looks like 0:10:02 using RegEx.Is it possible a recent update to NPP might have broken this or changed syntax??
I’m so confused.Thanks in advance!
-Dax. -
Hi @daxliniere
The quoted regex matches strings with two leading numbers, as “00:12:12”.
In order to match strings with only one leading number, just remove a “\d” from the expression, as(\d:\d\d):(?=\d\d).Have fun!
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@astrosofista Thank you so very much!! It works exactly as before.
I have no clue what could have changed, but this has solved the problem.I really appreciate your time on this. I was in the middle of a job and got stuck on this. Was comtemplating manual correction of ~400 lines!
Have a great weekend. :)
-Dax. -
AHHH!!! I understand it now!
The data I had been receiving previously had trailing zeros so every line had xx:xx:xx formatting. It seems whoever was typing this started dropping the trailing zeros.
Is there a more robust form of(\d:\d\d):(?=\d\d)needed? -
You say “trailing”, but the regex shown (as modified by @astrosofista) has removed the requirement for leading zeros. To make it match both, I’d do
(\d{1,2}:\d\d):(?=\d\d), which will match whether there is one digit or two.But if you give an example of “with” or “without” “trailing zeros”, we can be more confident of what you’re really asking for.
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@PeterJones
instead:
(\d{1,2}:\d\d):(?=\d\d)maybe that’s better?
(:\d{1,2}):the shorter the pattern, the better
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@Pan-Jan ,
Your regex does not provide all the same groups (and thus features) of the regex developed through this thread, and thus likely misses some of the functionality needed
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@PeterJones said in Wildcard in replace field:
and thus likely misses some of the functionality needed
show on the example in this thread that it doesn’t work
(\d{1,2}:\d\d):(?=\d\d)
but this one has disadvantages, and he will also mark it here
126:23:45
126:23:450
12:23:45:67it will be correct
(^| )(\d{1,2}:\d\d):(?=\d\d( |$))
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(^| )(\d{1,2}:\d\d):(?=\d\d([ ,'!\.\?"”\)]|$))