Searching for literal “> \” using ack-grep

escapinggrepregex

I am looking for lines that literally have a greater than character (a ">") followed by a space followed by a backslash character (a "\")
i.e., a line with this: > \

I thought escaping would allow this, and for the greater-than it does:

  $ ack-grep "\> "

returns lines that have "> " in them.

But when I try to escape the backslash as well I get:

  $ ack-grep "\> \\"

ack-grep: Invalid regex '\> \':
Trailing \ in regex m/\> /

Best Answer

Wow, I was so close ... single quotes:

$ ack-grep '\> \\'

Figured this out after confirming that my regex match was valid using: http://regexpal.com/ and just happened to have had single quotes from trying something else.

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.

Getting the last match in a file using grep

You could try

grep pattern file | tail -1

tac file | grep pattern | head -1

tac file | grep -m1 pattern

Best Answer

Related Solutions

Ls | grep regex, ignore files with .tab,

Getting the last match in a file using grep

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