How to get only the version number of PHP using shell commands?

Question

I want to get only the version of php installed on CentOS.

Output of php -v

PHP 7.1.16 (cli) (built: Mar 28 2018 13:19:29) ( NTS )
Copyright (c) 1997-2018 The PHP Group
Zend Engine v3.1.0, Copyright (c) 1998-2018 Zend Technologies

I tried this following:

php -v | grep PHP | awk '{print $2}'

But the output I got was:

7.1.16
(c)

How can I get only 7.1.16?

Answer 1

Extending Jeff Schaller's answer, skip the pipeline altogether and just ask for the internal constant representation:

$ php -r 'echo PHP_VERSION;'
7.1.15

You can extend this pattern to get more, or less, information:

$ php -r 'echo PHP_MAJOR_VERSION;'
7

See the PHP list of pre-defined constants for all available.

The major benefit: it doesn't rely on a defined output format of php -v. Given it's about the same performance as a pipeline solution, then it seems a more robust choice.

---

If your objective is to test for the version, then you can also use this pattern. For example, this code will exit 0 if PHP >= 7, and 1 otherwise:

php -r 'exit((int)version_compare(PHP_VERSION, "7.0.0", "<"));'

---

For reference, here are timings for various test cases, ordered fastest first:

$ time for (( i=0; i<1000; i++ )); do php -v | awk '/^PHP [0-9]/ { print $2; }' >/dev/null; done

real    0m13.368s
user    0m8.064s
sys     0m4.036s

$ time for (( i=0; i<1000; i++ )); do php -r 'echo PHP_VERSION;' >/dev/null; done

real    0m13.624s
user    0m8.408s
sys     0m3.836s

$ time for (( i=0; i<1000; i++ )); do php -v | head -1 | cut -f2 -d' ' >/dev/null; done

real    0m13.942s
user    0m8.180s
sys     0m4.160s

Answer 2

If you've installed php via the package manager (e.g. RPM or yum), then you can query the version from there:

rpm -q --queryformat="%{VERSION}" php

Alternatively, you can ask php to tell you its version directly:

php -r 'echo phpversion();'

Answer 3

On my system:

$> php -v | grep ^PHP | cut -d' ' -f2
7.0.32-0ubuntu0.16.04.1

as grep PHP matches every PHP string it encounters.

The ^PHP means "match only the string 'PHP' when it is at the start of a line".

Obviously, this works if the output format of php -v is consistent across versions/builds.

For reference, the whole output was:

PHP 7.0.32-0ubuntu0.16.04.1 (cli) ( NTS )
Copyright (c) 1997-2017 The PHP Group
Zend Engine v3.0.0, Copyright (c) 1998-2017 Zend Technologies
    with Zend OPcache v7.0.32-0ubuntu0.16.04.1, Copyright (c) 1999-2017, by Zend Technologies

Answer 4

There are different ways, I like to use look behind:

php -v | grep -Po '(?<=^PHP )[^ ]+'

or

php -v | grep -Po '(?<=PHP )([0-9.]+)'

Answer 5

Since you started with awk, here's an awk solution:

$ php -v | awk '/^PHP/{print $2}'
7.2.10

Answer 6

php -v | awk NR==1'{print $2}'

Or

php -v | awk '{print $2; exit}'

Answer 7

If you want to do it with just a single function being piped, you can try using sed like this:

php -v | sed -e '/^PHP/!d' -e 's/.* \([0-9]\+\.[0-9]\+\.[0-9]\+\).*$/\1/'

First it deletes any line that doesn't begin with PHP, then it clips the version from that line assuming it is the first sequence in the form of x.y.z.

Or, if you want something closer to your original script, simply put ^ in the front of your grep pattern to only look for lines that start with PHP:

php -v | grep ^PHP | awk '{print $2}'

Answer 8

I think this is more clean and elegant solution (should also work in Windows):

php -r "echo PHP_MAJOR_VERSION.'.'.PHP_MINOR_VERSION.'.'.PHP_RELEASE_VERSION;"

The output would be like 7.2.24.

Answer 9

php -v | grep ^PHP | awk '{print $1, $2}'

Output will be PHP 7.1.7

Tested on Ubuntu 18.04LTS