Grep not as a regular expression

Question

I need to search for a PHP variable $someVar. However, Grep thinks that I am trying to run a regex and is complaining:

$ grep -ir "Something Here" * | grep $someVar Usage: grep [OPTION]... PATTERN [FILE]... Try `grep --help' for more information. $ grep -ir "Something Here" * | grep "$someVar" <<Here it returns all rows with "someVar", not only those with "$someVar">> 

I don't see an option for telling grep not to interpret the string as a regex, but to include the $ as just another string character.

Answer 1

Use fgrep (deprecated), grep -F or grep --fixed-strings, to make it treat the pattern as a list of fixed strings, instead of a regex.

For reference, the documentation mentions (excerpts):

-F --fixed-strings Interpret the pattern as a list of fixed strings (instead of regular expressions), separated by newlines, any of which is to be matched. (-F is specified by POSIX.)

fgrep is the same as grep -F. Direct invocation as fgrep is deprecated, but is provided to allow historical applications that rely on them to run unmodified.

For the complete reference, check: https://www.gnu.org/savannah-checkouts/gnu/grep/manual/grep.html

Answer 2

grep -F is a standard way to tell grep to interpret argument as a fixed string, not a pattern.