在 Bash 中运行 PHP function （并将返回值保存在 ZD574D4BB40C899861791AC6A6A69 变量中）

Question

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?

Answer 1

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

Answer 2

PHP_OUT=`php -r 'echo phpinfo();'`
echo $PHP_OUT;

Answer 3

Alternatively:

php_cwd = `php -r 'echo getcwd();'`

replace the getcwd(); call with your php code as necessary.

EDIT: ninja'd by David Chan.

Answer 4

This is how you can inline PHP commands within the shell ie *sh:

#!/bin/bash

export VAR="variable_value"
php_out=$(php << 'EOF'
<?
    echo getenv("VAR"); //input
?>
EOF)
>&2 echo "php_out: $php_out"; #output

Answer 5

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

Answer 6

This is what worked for me:

VAR='/$#'
php_cwd=`/usr/bin/php << EOF
<?php echo preg_quote("$VAR"); ?>
EOF`
echo "$php_cwd"

Answer 7

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.