I want to replace the end of a line on line with ';' but only on lines that do not already end in ';'. I came up with the following to put in the replace dialog:
Find What = '[^\;]\r\n'
Replace with = '\;\r\n'
My problem is that this also selects the last character of a line. I don't want to replace the last character... just the carriage return.
Any ideas on how to correct this?
3 Answers
Your regex as it stands will replace the whole end of the line, so the best bet is to take the whole line, check it for a semi, and only replace the section that needs it (eg add a semi).
(.*)([^\;])(\r\n)
is an expression with 3 capture groups:
(.*)-- the first part of the line([^\;])-- the check for the missing semi(\r\n)-- the line ending
So, we can take the first group, add a semi, and then take the last group to get the full line, with the only difference being the semi itself.
the replace expression \1\;\3 will concatenate together the first group, a semi, and the last group.
Note that the \# syntax is common to Notepad++ and some other tools, but is not universal (many use $1, $2,...$n). be sure to check your editors documentation.
Use
Find What: ([^\;])\r\n
Replace with: $1;\r\n
It will replace the last character with the same character + ';' unless it is ';'
Also, in 'replace' '\' is not needed before ';'
- Ctrl+H
- Find what:
(?<!;)$ - Replace with:
; - check Wrap around
- check Regular expression
- Replace all
Explanation:
(?<! # negative lookbehind, zero-length assertion that make sure we haven't before current position:
; # a semicolon
) # end lookbehind
$ # end of line
