Unix Command to List files containing string but *NOT* containing another string

Question

How do I recursively view a list of files that has one string and specifically doesn't have another string? Also, I mean to evaluate the text of the files, not the filenames.

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Conclusion:

As per comments, I ended up using:

find . -name "*.html" -exec grep -lR 'base\-maps' {} \; | xargs grep -L 'base\-maps\-bot' 

This returned files with "base-maps" and not "base-maps-bot". Thank you!!

Answer 1

Try this:

grep -rl <string-to-match> | xargs grep -L <string-not-to-match> 

Explanation: grep -lr makes grep recursively (r) output a list (l) of all files that contain <string-to-match>. xargs loops over these files, calling grep -L on each one of them. grep -L will only output the filename when the file does not contain <string-not-to-match>.

Answer 2

The use of xargs in the answers above is not necessary; you can achieve the same thing like this:

find . -type f -exec grep -q <string-to-match> {} \; -not -exec grep -q <string-not-to-match> {} \; -print 

grep -q means run quietly but return an exit code indicating whether a match was found; find can then use that exit code to determine whether to keep executing the rest of its options. If -exec grep -q <string-to-match> {} \; returns 0, then it will go on to execute -not -exec grep -q <string-not-to-match>{} \;. If that also returns 0, it will go on to execute -print, which prints the name of the file.

As another answer has noted, using find in this way has major advantages over grep -Rl where you only want to search files of a certain type. If, on the other hand, you really want to search all files, grep -Rl is probably quicker, as it uses one grep process to perform the first filter for all files, instead of a separate grep process for each file.