有关问题有点绕，头晕者勿进^

问题有点绕，头晕者勿进^_^
array ( 0 => array ( 'id' => '10', 'title' => 'a','type' => '1', 'bian'=>'1'),
1 => array ( 'id' => '10', 'title' => 'a','type' => '1', 'bian'=>2),
2 => array ( 'id' => '11', 'title' => 'b','type' => '2', 'bian'=>'1'),
3 => array ( 'id' => '11', 'title' => 'c','type' => '2', 'bian'=>'2'),
)
当type=1去除相同id重复并且key变成ID，当type=2时KEY变成id_bian的形式,最终转成以下数组
array ( 0 => array ( '10' => '', '11_1' => '','11_2' => ''),)
然后在与下面数组合并
array ( 0 => array ( '10' => '1','11_2' => '1'),
1 => array ( '10' => '1','11_2' => '1'),
2 => array ( '11_1' => '1','11_2' => '1'),
3 => array ( '10' => '2','11_2' => '1'),
)
最终得到
array ( 0 => array ( '10' => '1','11_1'=>'','11_2' => '1'),
1 => array ( '10' => '1', '11_1 => '', 11_2' => '1'),
2 => array ( '10' => '', '11_1' => '1', '11_2' => '1'),
3 => array ( '10' => '2', '11_1' => '', '11_2' => '1'),
)
------解决方案--------------------

$a = array (
 0 => array ( 'id' => '10', 'title' => 'a','type' => '1', 'bian'=>'1'), 
 1 => array ( 'id' => '10', 'title' => 'a','type' => '1', 'bian'=>'2'), 
 2 => array ( 'id' => '11', 'title' => 'b','type' => '2', 'bian'=>'1'),
 3 => array ( 'id' => '11', 'title' => 'c','type' => '2', 'bian'=>'2'),  
);
$b = array (
 0 => array ( '10' => '1','11_2' => '1'), 
 1 => array ( '10' => '1','11_2' => '1'),
 2 => array ( '11_1' => '1','11_2' => '1'),
 3 => array ( '10' => '2','11_2' => '1'),  
);
$t = array();
foreach($a as $item) {
  if($item['type'] == '1') {
    $t[$item['id']] = '';
  }else $t[$item['id'].'_'.$item['bian']] = '';
}
$c = array();
foreach($b as $v) {
  $r = array();
  foreach($t as $k=>$n) $r[$k] = @$v[$k];
  $c[] = $r;
}
print_r($c);

Array
(
    [0] => Array
        (
            [10] => 1
            [11_1] => 
            [11_2] => 1
        )

    [1] => Array
        (
            [10] => 1
            [11_1] => 
            [11_2] => 1
        )

    [2] => Array
        (
            [10] => 
            [11_1] => 1
            [11_2] => 1
        )

    [3] => Array
        (
            [10] => 2
            [11_1] => 
            [11_2] => 1
        )

)