Using grep to search for a string that has a dot in it

Question

grep uses regexes; . means "any character" in a regex. If you want a literal string, use grep -F, fgrep, or escape the . to \..

Don't forget to wrap your string in double quotes. Or else you should use \\.

So, your command would need to be:

grep -r "0\.49" *

or

grep -r 0\\.49 *

or

grep -Fr 0.49 *

grep -F -r '0.49' * treats 0.49 as a "fixed" string instead of a regular expression. This makes . lose its special meaning.

You need to escape the . as "0\.49".

A . is a regex meta-character to match any character(except newline). To match a literal period, you need to escape it.

There are so many answers here suggesting to escape the dot with \. but I have been running into this issue over and over again: \. gives me the same result as .

However, these two expressions work for me:

$ grep -r 0\\.49 *

And:

$ grep -r 0[.]49 *

I'm using a "normal" bash shell on Ubuntu and Archlinux.

Edit, or, according to comments:

$ grep -r '0\.49' *

Note, the single-quotes doing the difference here.

You can also use "[.]"

grep -r "0[.]49"