How to pass PHP variables to Shell Script? [duplicate]

Question

Possible Duplicate:
pass parameters to php with shell

I am running a php script within shell script as

#! /bin/sh

php file.php

Shell will run the script and display the output of the php script, but how can I pass variable defined within php script into shell script to be used therein (for further processing by shell)?

For example, consider a php file of

<?php
$test = "something";
?>

How can I pass the value of $test to the shell script as

#! /bin/sh

php config.php
echo $test

UPDATE: The methods suggested are based on printing the variables. I prefer not to print out anything, as the php script has other applications too (to be included in other php scripts).

See this question: stackoverflow.com/questions/4747396/…

chiborg
– chiborg

2012-05-29 12:08:46 +00:00
Commented May 29, 2012 at 12:08 — chiborg
– chiborg, Commented May 29, 2012 at 12:08

lanzz · Accepted Answer · 2012-05-29 12:05:04Z

3

Your PHP script can print shell-style variable assignments:

print("VAR1=foo\n");
print("VAR2=bar\n");

In your shell script, you will need to evaluate these assignments, in order to import them into your environment:

. <(php file.php)

answered May 29, 2012 at 12:05

lanzz

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Dennis Williamson · Accepted Answer · 2012-05-29 12:41:43Z

1

Your PHP script can print the values and your shell script can use process or command substitution or read to retrieve the values.

PHP:

<?php
    print($test1 . "\n");
    print($test2 . "\n");
?>

Bash:

while read -r line
do
    something_with "$line"
done < <(php_script)

or

saveIFS=$IFS
IFS=$'\n'
array=($(php_script))
IFS=$saveIFS

or

saveIFS=$IFS
IFS=$'\n'
read -r -d '' var1 var2 <<< "$(php_script)"
IFS=$saveIFS

or other variations.

answered May 29, 2012 at 12:41

Dennis Williamson

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