Hello, Markus-Ruhsam,
Of course, it’s a bit late, by now, but here is a shorter regex, that works perfectly well !
SEARCH (?-s)^(?=.{10}\.).+\R
REPLACE EMPTY
Et voilà !
Notes :
The first part (?-s) is an in-line modifier, which ensures that the dot meta-character will match only standard characters, even if you previously checked the . matches newline option
From the beginning of line ^, the form (?=.{10}\.), also called a positive look-ahead, verifies if the condition Does exist a DOT character, in column 11, of the current line, is true.
The important thing to remember about look-arounds is that, as soon as the condition is evaluated, the regex engine position is reset to the position before evaluating the look-around condition. So, in our case, to the position just before the first character of each line
Consequently, if this condition is true, the following part of the regex .+ matches all the standard characters of the current line
Finally, the \R syntax matches any kind of EOL characters ( \r\n ) in Windows files, ( \n ) in Unix files or ( \r ) in Mac files
Now, as the replacement string is empty, that means that any complete line ( with a dot character at column 11 ) will be, automatically, deleted !
Best Regards,
guy038
