How to grep exact literal string (no regex)

Question

Is there a way to grep (or use another command) to find exact strings, using NO regex?

For example, if I want to search for (literally):

/some/file"that/has'lots\of"invalid"chars/and.triggers$(#2)[*~.old][3].html

I don't want to go through and escape every single "escapable". Essentially, I want to pass it through, like I would with echo:

$ echo "/some/file\"that/has'lots\of\"invalid\"chars/and.triggers$(#2)[*~.old][3].html"
/some/file"that/has'lots\of"invalid"chars/and.triggers$(#2)[*~.old][3].html

Solution 1

Use fgrep, it's the same as grep -F (matches a fixed string).

Solution 2

Well, you can put the information you want to match, each in a line, and then use grep:

grep -F -f patterns.txt file.txt

Notice the usage of the flag -F, which causes grep to consider each line of the file patterns.txt as a fixed-string to be searched in file.txt.

Comments

[Nathan J.B.]

Worked perfectly with fgrep 'preg_replace("/([a-z]+)([A-Z])/"' . -R (which failed with grep). You will still have to escape $, but that's livable.

[Nathan J.B.]

It's worth noting that this is the same as fgrep -f patterns.txt file.txt. By using a file of patterns, $ does not need to be escaped. I ultimately decided to accept the other because it's the least time consuming.

[Lily Finley]

grep -F seems to be preferred now. "Direct invocation as either egrep or fgrep is deprecated" fgrep(1) - Linux man page