for example I have the following code
Module MPI
Use MPI
!
! MPI info
If (true) Then
Print *, \'\'
! empty line 1
! empty line 2
End If
Since the engine is already passed on a comment line to indent it, I think it is impossible to use the same entire edited line for the next match to get the number of spaces. So I think you have to repeat the same replacement more times. Try with:
^(\s*)([^!\s].*?\r\n(\1\!.*?\r\n)*)\s*\!
always replacing it with $1$2$1!
.
Like I said in the comment, if you have at most N consecutive comment lines, you will click on the "replace all" button N times
Using the search text ^( +)(.*\R)(!)
and the replace text \1\2\1\3
then clicking on "Replace all" twice does what is wanted on the sample text. I cannot see a way of doing this in one pass.
The expression looks for a line with leading spaces followed by a line starting with a !
. The capture groups are the leading spaces in \1
, the rest of that line including the newline in \2
and the leading !
in \3
. The replacement just assembles the captures in the right order. Note that you could omit the capture group around the !
and just have an explicit `! in the replacement, but I like to use captures in such contexts as they often allow for shorter replacements (although not in this case) and easier enhancements.