How do I get a part of the output of a command in Linux Bash?

Question

How do I get a part of the output of a command in Bash?

For example, the command php -v outputs:

PHP 5.3.28 (cli) (built: Jun 23 2014 16:25:09)
Copyright (c) 1997-2013 The PHP Group
Zend Engine v2.3.0, Copyright (c) 1998-2013 Zend Technologies with the ionCube PHP Loader v4.6.1, Copyright (c) 2002-2014, by ionCube Ltd.

And I only want to output the PHP 5.3.28 (cli) part. How do I do that?

I've tried php -v | grep 'PHP 5.3.28', but that outputs: PHP 5.3.28 (cli) (built: Jun 23 2014 16:25:09) and that's not what I want.

Peter Mortensen · Accepted Answer · 2020-11-10 02:00:53Z

10

You could try the below AWK command,

$ php -v | awk 'NR==1{print $1,$2,$3}'
PHP 5.3.28 (cli)

It prints the first three columns from the first line of input.

NR==1 (condition)ie, execute the statements within {} only if the value of NR variable is 1.
{print $1,$2,$3} Print col1,col2,col3. , in the print statement means OFS (output field separator).

edited Nov 10, 2020 at 2:00

Peter Mortensen

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answered Aug 4, 2014 at 10:22

Avinash Raj

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Peter Mortensen · Accepted Answer · 2020-11-10 00:36:14Z

4

In pure Bash you can do

echo 'PHP 5.3.28 (cli) (built: Jun 23 2014 16:25:09)' | cut -d '(' -f 1,2

Output:

PHP 5.3.28 (cli)

Or using space as the delimiter:

echo 'PHP 5.3.28 (cli) (built: Jun 23 2014 16:25:09)' | cut -d ' ' -f 1,2,3

edited Nov 10, 2020 at 0:36

Peter Mortensen

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answered Aug 4, 2014 at 10:28

slavik

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Peter Mortensen · Accepted Answer · 2020-11-10 01:52:26Z

a classic "million ways to skin a cat" question...

These methods seem to filter by spaces... If the versions/notes contain spaces, this fails.

The ( brackets, however, seem consistent across all my platforms so I've used the following:

For example, on Debian:

root@host:~# php -v  | head -1
PHP 5.3.28-1~dotdeb.0 with Suhosin-Patch (cli) (built: Dec 13 2013 01:38:56)
root@host:~# php -v  | head -1 | cut -d " " -f 1-2
PHP 5.3.28-1~dotdeb.0

So here I trim everything before the second (:

root@host:~# php -v  | head -1 | cut -d "(" -f 1-2
PHP 5.3.28-1~dotdeb.0 with Suhosin-Patch (cli)

Note: there will be a trailing white-space (blank space at the end)

Alternatively, you could always use your package manager to determine this (recommended):

root@debian-or-ubuntu-host:~# dpkg -s php5 | grep 'Version'
Version: 5.3.28-1~dotdeb.0

...or on a CentOS, Red Hat Linux, or Scientific Linux distribution:

[root@rpm-based-host ~]# rpm -qa | grep php-5
php-5.4.28-1.el6.remi.x86_64

Peter Mortensen · Accepted Answer · 2020-11-10 00:36:53Z

0

If you want all the lines that contain "php", do this:

 $ php -v | grep -i "php"

Then if you want the first three words within those, you can add another pipe as @Avinash suggested:

$ php -v | grep -i "php" | awk 'NR==1{print $1,$2,$3}'

edited Nov 10, 2020 at 0:36

Peter Mortensen

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answered Aug 4, 2014 at 10:29

nafas

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