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.

Answer 1

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).

Answer 2

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