Linux – Escaping special characters in grep regex

bashgreplinuxregex

I am trying to run a grep regular expression on a file, where I have to exclude lines where "00" and "0" appear. I came up with this expression:

grep -a -E \"stored\"\:\ \"\*123\*(?!00)[0-9]{2,5}\#\" $filename

But when I try to run it in bash, I either get

-bash !00: event not found, or (once I have typed set +H),

-bash: syntax error near unexpected token('`

Please what do I need to do to properly escape this regular expression in bash?

Best Answer

You get an error message because you did not escape !, which has a special meaning in the shell. To use that as part of the pattern parameter of grep, you must quote it or escape it.

Instead of quoting every single special character one by one, it's a lot easier to enclose the entire pattern within single or double-quotes (depending on your purpose), for example:

grep -a -P '"stored": "*123*(?!00)[0-9]{2,5}#"'

Also note that I replaced -E with -P, because negative lookahead (?!...) doesn't work in typical implementations of grep extended regex. The -P flag enables Perl regular expressions, which should support this.

Related Solutions

Ls | grep regex, ignore files with .tab,

How I would do this would depend on the context. If this wasn't for a script, and I just wanted to see this, I would be okay with using ls:

ls -a | egrep -v '(tab|\.o$|\.bin$|~$)'

If in a script, I would use a loop:

for file in *; do
#file $file in .* *; do   -- To Include dot (hidden files)                     
   if ! egrep -q '(tab|\.o$|\.bin$|~$)' <<<$file; then
      echo $file
   fi
done

You could add another if test with -f if you only want regular files and not directories. If you want to do this recursively, you could use GNU find with a -regex test and posix-egrep. But take into account the fact that this regex will match the full pathname relative to the directory you searched in.

Another nice option is extended globs if you have support for it, see Dennis's answer.

Grep date with regex

Brackets can be used to give you a range of numbers. The carrot can match the start of a line. So..

grep "^12\/1[6-8]\/09" will give you everything that starts with 12/16/09, 12/17/09, and 12/18/09

Best Answer

Related Solutions

Ls | grep regex, ignore files with .tab,

Grep date with regex

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