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

问题

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 ?

回答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

回答2:

You could use grep also :

grep comm= filex

this will display all the lines containing comm=.

回答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.

回答4:

If grep is ok to use, you could give a try to:

grep -E "^comm=" file

来源：https://stackoverflow.com/questions/8326817/i-want-to-use-awk-or-sed-to-print-all-the-lines-that-start-with-comm-in-a-f