I want to use "awk" or sed to print all the lines that start with "comm=" in a file

Question

I want to use "awk" or "sed" to print all the lines that start with comm= from the file filex, Note that each line contains "comm=somthing"

for example : comm=rm , comm=ll, comm=ls  ....

How can i achieve that ?

Solution 1

For lines that start with comm=

sed -n '/^comm=/p' filex

awk '/^comm=/' filex

If comm= is anywhere in the line then

sed -n '/comm=/p' filex

awk '/comm=/' filex

Solution 2

You could use grep also :

grep comm= filex

this will display all the lines containing comm=.

Solution 3

Here's an approach using grep:

grep -o '\<comm=[[:alnum:]]*\>'

This treats a word as consisting of alphanumeric characters; extend the character class as needed.

Comments

[wael]

that only works it comm is at the beginning of the line, how to make it work if comm is at any place in each line, thanks for ur help Jaypa

[Michael J. Barber]

@bob Given that you've stated in a comment that it's just one comm= per line, this solution is overkill; use the answer from Cédric Julien. Additionally, to say "thanks" on Stackoverflow, you can just upvote the useful answers rather than commenting.

[wael]

this is returning the whole line i just want to return the match, and this is how it can be done : grep -o '\<comm=[[:alnum:]]*\>' , thanks for your help

[potong]

For brevity you might like sed '/\<comm=/!d'

[vefthym]

or grep "^comm=" filex to match lines starting with comm= ?