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

Answer 1

使用fgrep ，它与grep -F相同（匹配固定的字符串）。

Answer 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 .