Hello, @josé-luis-montero-castellanos, @aln-kilborn and all,
@josé-luis-montero-castellanos, regarding your regex S/R :
SEARCH (?s)(<St)(.)(/S)(.)(<Xt)(.)(/X)|(<Xt)(.)(/X)(.)(<St)(.)(/S)
REPLACE ?1\5\2\7\4\1\6\3:${12}${9}${14}${11}${8}${13}${10}
Why do you need the leading in-line modifier (?s), in the search part ? Do you mean that the single char . in group 2, 4, 6, 9, 11 and 13 may also represent a line-break char ( \r or \n ) ?
I suppose not ! Then, your correct search regex is rather :
(?-s)(<St)(.)(/S)(.)(<Xt)(.)(/X)|(<Xt)(.)(/X)(.)(<St)(.)(/S)
where any . single char will represent a non-break char, only !
Now, I noticed that the groups used, in replacement, for each alternative of the search regex, are in the same order ! Then, you can use, preferably, a specific feature of the Boost Regex engine, called Branch Reset
For instance, in your regex, the numbers of each group are :
(?-s)(<St)(.)(/S)(.)(<Xt)(.)(/X)|(<Xt)(.)(/X)(.)(<St)(.)(/S)
Gr : 1 2 3 4 5 6 7 |8 9 10 11 12 13 14
If we use the Branch Reset feature the search regex becomes :
(?-s)(?|(<St)(.)(/S)(.)(<Xt)(.)(/X)|(<Xt)(.)(/X)(.)(<St)(.)(/S))
Gr: 1 2 3 4 5 6 7 |1 2 3 4 5 6 7
Thus, your entire regex S/R can be simplified as below :
SEARCH (?-s)(?|(<St)(.)(/S)(.)(<Xt)(.)(/X)|(<Xt)(.)(/X)(.)(<St)(.)(/S))
REPLACE \5\2\7\4\1\6\3
For instance, given this INPUT text :
<St2/S4<Xt6/X
<Xt9/XB<StD/S
You would get the OUTPUT text, below :
<Xt2/X4<St6/S
<St9/SB<XtD/X
Notes :
The Regular expression option must be selected
Preferably, tick the [Wrap around](https://www.ahijoy.com/) option
Best Regards,
guy038
