Execute php script from bash , assign output to a bash variable

Question

I have a bash script which need to execute some php scripts and to get back the results e.g

#!/bin/bash
/usr/bin/php -f $HOME/lib/get_fifobuild.php

The script get_fifobuild.php returns an integer which I need to assign into a bash variable. I ll appreciate if someone help me out.

thanks :)

Edit: php show.php

<?php 
  echo phpinfo();
  exit;
?>

bash script:

#!/bin/bash
HOME=`dirname $0`;
log(){
    NEW_LOG=$HOME/logs/cloud-`date +%d_%m_%Y`.log
    echo $1 >> $NEW_LOG
}
log "Date: `date`";
data=$(/usr/bin/php -f  $HOME/lib/show.php);
log $data;

output:

Date: Fri Jun 15 19:16:00 PKT 2012
phpinfo()

no luck yet

Possible duplicate of Assigning values printed by PHP CLI to shell variables. There are many questions on SO about assigning output of a program to shell variables. Did you do a search? — Dennis Williamson
– Dennis Williamson, Commented Jun 15, 2012 at 13:24

John Lawrence · Accepted Answer · 2012-06-15 14:49:50Z

13

myvariable=$(/usr/bin/php -f $HOME/lib/get_fifobuild.php)

Will assign the output from your php script to a variable called "myvariable".

Update:

This will assign the output of the command to the variable, but as you are still having problems I can perhaps suggest a few things:

you have 'get_builds.php' and 'get_fifobuild.php' elsewhere.
check that $HOME is being set correctly. You may be better with a different variable name here as that environment variable generally is set to your home directory. This however is unlikely to be the problem as you are getting output from the script.
Is the text you gave the exact contents of your PHP file? If you have quotes around phpinfo() for example it will cause the output to just be the string "phpinfo()". In fact, you do not need the echo at all and could make the contents of your PHP file as follows.

get_fifobuild.php:

<?php
    phpinfo();
?>

Update 2:

Try changing your script to:

#!/bin/bash
HOME=`dirname $0`;
log(){
    NEW_LOG=$HOME/logs/cloud-`date +%d_%m_%Y`.log
    echo "$1" >> $NEW_LOG
}
log "Date: `date`";
data=$(/usr/bin/php -f $HOME/lib/show.php);
log "$data";

Basically adding double quotes around the variables in the 'log' and 'echo' lines. The problem you were having was that only the first line of your php output was being logged.

edited Jun 15, 2012 at 14:49

answered Jun 15, 2012 at 12:59

John Lawrence

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2 Comments

sakhunzai Over a year ago

please check question edits , I have tried this but with no luck

Victor Perov Over a year ago

VAR=$(php -r "echo 213;")

kojiro · Accepted Answer · 2012-06-15 14:27:52Z

0

foobar=`/usr/bin/php -f $HOME/lib/get_fifobuild.php`

note: these are backticks.

edited Jun 15, 2012 at 14:27

kojiro

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answered Jun 15, 2012 at 13:00

Rufinus

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2 Comments

cmh Over a year ago

Please use the $() convention over the `` convention. It's more readable and backticks cannot be nested (whereas the dollar convention can).

kojiro Over a year ago

Uh, apparently I cannot reverse my downvote because you didn't edit it enough so that it shows as having been edited. Sometimes SO is dumb. So I made a tiny edit. :-/

Vasiliy Letuyev · Accepted Answer · 2024-06-08 19:01:28Z

0

But phpinfo() doesn't need echo before them.
https://www.php.net/manual/en/function.phpinfo.php

<?php
phpinfo();

answered Jun 8, 2024 at 19:01

Vasiliy Letuyev

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