Regex: Match only if string A is found and string B is not

Question

I'm trying to write a regular expression that will essentially return true if string A is found and string B is not found.

To be more specific, I'm looking for any file on my server which has the text 'base64_decode' in it, but not 'copyright'.

Thanks!

Answer 1

Use negative lookahead and lookbehind:

^(?<!.*copyright.*)(base64_decode)(?!.*copyright.*)$

Perl doesn't support this yet :-P.

Answer 2

It's not recommended to do this in one regex, but if you must, you can use lookaheads:

^(?=.*must-have)(?!.*must-not-have)

You may want to do this in single-line/dot-all mode, and the beginning anchor may be \A instead of ^.

(?=…) is positive lookahead; it asserts that a given pattern can be matched. (?!…) is negative lookahead; it asserts that a given pattern can NOT be matched.

References

• regular-expressions.info/Lookarounds, Anchors, Dot

Answer 3

Piped greps should be able to achieve that easily:

find -type f -print | xargs grep -l "base64_decode" | xargs grep -L "copyright"

Answer 4

I'm not sure your real task can be solved purely within the regex passed into grep, since grep processes files line-by-line. I would use the -l (--files-with-matches) and -L (--files-without-match) options along with command substitution backticks, like so:

grep -L copyright `grep -l base64_decode *`

grep -l base64_decode * lists the names of all the files with "base64_decode" in them, and the backticks put that list on the command line after grep -L copyright, which searches those files and lists the subset of them that doesn't contain "copyright".