Run `grep` excluding a file in a specific path

Question

I want to exclude the file ./test/main.cpp from my search.

Here's what I'm seeing:

$ grep -r pattern --exclude=./test/main.cpp
./test/main.cpp:pattern
./lib/main.cpp:pattern
./src/main.cpp:pattern

I know it is possible to get the output that I want by using multiple commands in a pipes-and-filters arrangement, but is there some quoting/escaping that will make grep understand what I want natively?

Answer 1

grep can't do this for file in one certain directory if you have more files with the same name in different directories, use find instead:

find . -type f \! -path './test/main.cpp' -exec grep pattern {} \+

Answer 2

I don't think it's possible with GNU grep. You don't need pipes though.

With find:

find . ! -path ./test/main.cpp -type f -exec grep pattern {} +

With zsh:

grep pattern ./**/*~./test/main.cpp(.)

(excludes hidden files, just as well to exclude the .git, .svn...).

Answer 3

I could write a book : "The lost art of xargs". The find ... -exec … '; launches a grep for each file (but the variant with -exec … + doesn't). Well, we're wasting CPU cycles these days so why not, right? But if performance and memory and power is an issue: use xargs:

find . -type f \! -path 'EXCLUDE-FILE' -print0 | xargs -r0 grep 'PATTERN'

GNU's find's -print0 will NUL-terminate its output and xargs' -0 option honors that format as input. This ensures whatever funny characters your file has, the pipeline won't get confused. The -r option makes sure there's no error in case find finds nothing.

Note, you can now do things like:

find . -type f -print0 | grep -z -v "FILENAME EXCLUDE PATTERN" | 
  xargs -r0 grep 'PATTERN'

GNU grep's -z does the same thing as xargs' -0.

Answer 4

If your find supports -path which was added to POSIX in 2008 but still missing in Solaris:

find . ! -path ./test/main.cpp -type f -exec grep pattern /dev/null {} +

Answer 5

For the record, here's the approach that I prefer:

grep pattern $(find . -type f ! -path './test/main.cpp')

By keeping the grep at the beginning of the command, I think this is a little more clear -- plus it doesn't disable grep's color highlighting. In a sense, using find in a command-substitution is just a way of extending/replacing the (limited) file-search subset of grep's functionality.

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To me, the find -exec syntax is kind of arcane. One complexity with find -exec is the (sometimes) need for escaping various characters (notably if \; is used under Bash). Just for the purposes of putting things into familiar contexts, the following two commands are basically equivalent:

find . ! -path ./test/main.cpp -type f -exec grep pattern {} +
find . ! -path ./test/main.cpp -type f -print0 |xargs -0 grep pattern

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If you want to exclude subdirectories, it may be necessary to use a wildcard. I don't fully understand the schema here -- talk about arcane:

grep pattern $(find . -type f ! -path './test/main.cpp' ! -path './lib/*' )

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One further note to generalize find-based solutions for use in scripts: The grep command-line should include the -H/--with-filename option. Otherwise it will change the output formatting under the circumstance that there happens to be only one filename in the search results from find. This is notable because it doesn't appear to be necessary if using grep's native file-search (with the -r option).

...Even better, though, is to include /dev/null as a first file to search. This solves two problems:

• It ensures that if there is one file to search, grep will think there are two and use the multiple-file output mode.
• It ensures that if there are no files to search, grep will think there is one file and not hang waiting on stdin.

So the final answer is:

grep pattern /dev/null $(find . -type f ! -path './test/main.cpp')