I am trying to run a PHP function inside Bash... but it is not working.
#! /bin/bash
/usr/bin/php << 'EOF'
<?php echo getcwd(); ?>
EOF
In the reality, I needed to keep the return value in a bash variable... By the way, I am using the php's getcwd() function only to illustrate the bash operation.
UPDATE: Is there a way to pass a variable?
VAR='/$#'
php_cwd=`/usr/bin/php << 'EOF'
<?php echo preg_quote($VAR); ?>
EOF`
echo "$php_cwd"
Any ideas?
php_cwd=`/usr/bin/php << 'EOF'
<?php echo getcwd(); ?>
EOF`
echo "$php_cwd" # Or do something else with it
PHP_OUT=`php -r 'echo phpinfo();'`
echo $PHP_OUT;
Alternatively:
php_cwd = `php -r 'echo getcwd();'`
replace the getcwd(); call with your php code as necessary.
EDIT: ninja'd by David Chan.
This is how you can inline PHP commands within the shell i.e. *sh:
#!/bin/bash
export VAR="variable_value"
php_out=$(php << 'EOF'
<?
echo getenv("VAR"); //input
?>
EOF)
>&2 echo "php_out: $php_out"; #output
Use '-R' of php command line. It has a build-in variable that reads inputs.
VAR='/$#'
php_cwd=$(echo $VAR | php -R 'echo preg_quote($argn);')
echo $php_cwd
This is what worked for me:
VAR='/$#'
php_cwd=`/usr/bin/php << EOF
<?php echo preg_quote("$VAR"); ?>
EOF`
echo "$php_cwd"
I have a question - why don't you use functions to print current working directory in bash? Like:
#!/bin/bash
pwd # prints current working directory.
Or
#!/bin/bash
variable=`pwd`
echo $variable
Edited: Code above changed to be working without problems.
