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CodeCode
Posts: 1366 Joined: September 7th, 2020, 2:24 pm
Location: QLD, Australia
Post
by CodeCode » December 22nd, 2021, 6:10 am
I am trying this but it seems i am not doing it right:
Code: Select all
RegExpSubstitute=1
Substitute="^(.*)\\":"\1"
I think I am getting close, but I am struggling with ignoring a string of any length that has a \ at the end.
Examples are mostly in perl or java so the methods of substituting are not obviously transferrable.
Last edited by Brian on December 22nd, 2021, 6:24 am, edited 1 time in total.
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ƈǟռ'ȶ ʄɨӼ ɨȶ ɨʄ ɨȶ ǟɨռ'ȶ ɮʀօӄɛ - ʊռʟɛֆֆ ɨȶ ɨֆ ɨռ ƈօɖɛ.
CodeCode
Posts: 1366 Joined: September 7th, 2020, 2:24 pm
Location: QLD, Australia
Post
by CodeCode » December 22nd, 2021, 2:14 pm
CodeCode wrote: ↑ December 22nd, 2021, 6:10 am
I am trying this but it seems i am not doing it right:
Code: Select all
RegExpSubstitute=1
Substitute="^(.*)\\":"\1"
I think I am getting close, but I am struggling with ignoring a string of any length that has a \ at the end.
Examples are mostly in perl or java so the methods of substituting are not obviously transferrable.
Well. Once again, I both overcomplicated things as well as trying the the substitute in the wrong measure.
The substitute is simply this "\\$":""
Ok I have necroposted here too much.
I have another generally related question, but I will start a new post for it.
ƈǟռ'ȶ ʄɨӼ ɨȶ ɨʄ ɨȶ ǟɨռ'ȶ ɮʀօӄɛ - ʊռʟɛֆֆ ɨȶ ɨֆ ɨռ ƈօɖɛ.