Unix grep command

Question

I have an utility script, that displays an information about deployed java app. Here is an example output of this script:

Name: TestAPP

Version : SNAPSHOT

Type : ear, ejb, webservices, web

Source path : /G/bin/app/TESTAPP_LIVE_1.1.9.1.1.ear

Status : enabled

Is it possible to grep Version and source path values using grep command? Right now im able to do this using following command:

| grep Version

But it outputs the whole string (e.g. Version: Snapshot) when i am need only a values (e.g Snapshot to use in further script commands)

You can also use awk: |grep Version|awk '{$print $3}' (I'm using $3 and not $2 because there is a space between ':' and the other words) — MMM
– MMM, Commented Apr 26, 2013 at 13:02

Satya · Accepted Answer · 2013-04-26 13:00:31Z

2

grep Version | cut -d ':' -f 2

answered Apr 26, 2013 at 13:00

Satya

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Comments

Benjamin Toueg · Accepted Answer · 2013-04-26 13:10:58Z

2

Here is a pure grep solution.

Use the -P option for regex mode, and -o option for retrieving only what is matching.

grep -Po "(?<=^Version : ).*"

Here is what you would do for Source:

grep -Po "(?<=^Source : ).*"

It uses a postive lookbehind.

answered Apr 26, 2013 at 12:50

Benjamin Toueg

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MMM · Accepted Answer · 2013-04-26 13:06:50Z

0

Here's a solution using awk if you're interested:

grep Version | awk '{print $3}'

$3 means to print the third word from that line.

Note that:

This displays one word only
This assumes you have spaces between the colon (and therefore the version is actually the third "word"). If you don't, use $2 instead.

answered Apr 26, 2013 at 13:06

MMM

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