Bash – Seeing all regex matches in a file in Bash

bashfilesregular expression

I'm in bash and I have a file, I want to find all matches of a regex in it and have them printed. How do I do that?

Followup: Assume I want to print only unique matches, i.e. if a match appears multiple times, it should be printed only once. How do I do that? (I want this answer in addition to the first, so I want an answer to the first question too.)

Best Answer

Try grep -o 'myregex.*stuff' file and for the second question grep -o 'myregex.*stuff' file | sort | uniq.

The -o grep switch will print only the matches instead of printing the whole line that matches the regex.

Related Solutions

Extracting subset of lines of a file based on regex for first & last line

With sed:

sed -n '/R/,$!d; /S/q; p'

Example:

$ seq 20 | sed -n '/6/,$!d; /1/q; p'
6
7
8
9

Text Processing – How to Print File Content Only if the First Line Matches a Pattern

You could do that with ed:

ed -s infile <<\IN 2>/dev/null
1s/PATTERN/&/
,p
q
IN

the trick here is to try to replace PATTERN on 1st line with itself. ed will error out if it can't find the specified pattern so ,p (print whole file) will only be executed if 1s/PATTERN/&/ is successful.

Or with sed:

sed -n '1{
/PATTERN/!q
}
p' infile

this quits if the first line does not (!) match PATTERN, otherwise it prints all lines.
Or, as pointed out by Toby Speight, with GNU sed:

sed '1{/PATTERN/!Q}' infile

Q is the same as q but it does not print the pattern space.

Best Answer

Related Solutions

Extracting subset of lines of a file based on regex for first & last line

Text Processing – How to Print File Content Only if the First Line Matches a Pattern

Related Question