【貌似有挑战性】shell怎么获取php中return的值

【貌似有挑战性】shell如何获取php中return的值
如题 php这个文件就是返回一个数组 

PHP code

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<?php return array (
  'a' => '1',
  'b' => '2',
  'c' => 'public',
  'd' => '',
  'e' => '',
);
?>

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PHP code

[User:root Time:07:23:08 Path:/home/liangdong/php]$ php a.php 
Array
(
    [a] => 1
    [b] => 2
    [c] => public
    [d] => 
    [e] => 
)
[User:root Time:07:23:10 Path:/home/liangdong/php]$ cat a.php 
<?php $arr = include('b.php');
print_r($arr);
?>
[User:root Time:07:23:12 Path:/home/liangdong/php]$ cat b.php 
<?php return array (
                  'a' => '1',
                    'b' => '2',
                          'c' => 'public',
                            'd' => '',
                                  'e' => '',
                );
?>
<br><font color="#e78608">------解决方案--------------------</font><br>shell不熟，估计你的需求我以后会用的，就研究了一个，<br><br>

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PHP code

#!/bin/bash
eval `cat php_data.txt |grep "=>"|tr -d " ,>\'"|xargs -I {} echo {}\;|tr "\n" " "`
echo ${a}
echo ${b}
echo ${c}
echo ${d}
<br><font color="#e78608">------解决方案--------------------</font><br>
跨语言的数据交换，需采用双方都能识别的数据结构，<br>否则就毫无意义<br><br>如果你示例 php 代码是被 zend 预编译过的，或是被威盾扰码过的（你不能说他就不是php文件了吧）<br>那你在 shell 中再如何分离出数据？ <div class="clear">
                 
              
              
        
            </div>

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