How to pass bash script variable value in PHP command (CLI)

Question

I need help in passing the Linux bash-script variable to execute the PHP CLI command.

I am working on the bash-script that executes PHP CLI command that gets input variable from the bash-script like folder-path to include a file for accessing a class of the PHP file.

But what happens while executing the bash-script is that the variable passed to the PHP CLI-command act as a variable for PHP.

Below is the sample-code my script

file="$folderpath"somefile.php
mage_version=$(php -r 'include "$file";print_r(Mage::getVersionInfo());')

Below is the error I am getting

PHP Notice:  Undefined variable: file in Command line code on line 1
PHP Warning:  include(): Filename cannot be empty in Command line code on line 1

Try using curly brackets ${folderpath} and ${file}

user8034901
– user8034901

2019-09-17 09:37:46 +00:00
Commented Sep 17, 2019 at 9:37 — user8034901
– user8034901, Commented Sep 17, 2019 at 9:37

0stone0 · Accepted Answer · 2019-09-17 09:46:36Z

3

Bash needs double (") quotes to parse variable, otherwise they will stay $var and so php will read them;

file="$folderpath"somefile.php
mage_version=$(php -r "include \"${file}\";print_r(Mage::getVersionInfo());")

More inline bash info

If you need a multi line solution, you can do something like: Run PHP function inside Bash (and keep the return in a bash variable)

php_cwd=`/usr/bin/php << 'EOF'
<?php echo getcwd(); ?>
EOF`
echo "$php_cwd" # Or do something else with it

edited Sep 17, 2019 at 9:46

answered Sep 17, 2019 at 9:35

0stone0

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1 Comment

DKebler Over a year ago

php -r 'require '\"${loc}\"'; echo implode(".",$OC_Version);' I had to use single quotes so that $OC_Version was not pre evaluated but taken literally as that is a variable for php not bash

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