如何在PHP命令（CLI）中传递bash脚本变量值

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

Answer 1

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