Home >Backend Development >PHP Tutorial >两个二维数组的合并

两个二维数组的合并

WBOY
WBOYOriginal
2016-06-06 20:28:292379browse

合并前的数组a:

<code>array (size=3)
  0 => 
    array (size=2)
      'id' => string '113' (length=3)
      'email' => string 'yintx_1292342352@163.com' (length=24)
    
  1 => 
    array (size=2)
      'id' => string '111' (length=3)
      'email' => string 'yintx_1293456456@163.com' (length=24)
  2 => 
    array (size=2)
      'id' => string '109' (length=3)
      'email' => string 'yintx_129@99999.com' (length=19)
</code>

合并前的数组b:

<code>array (size=2)
  0 => 
    array (size=2)
      'user_id' => string '113' (length=2)
      'count' => string '1' (length=1)
  1 => 
    array (size=2)
      'user_id' => string '109' (length=2)
      'count' => string '8' (length=1)
</code>

合并后的数组(合并条件,a的id==b的user_id时合并):

<code>array (size=3)
  0 => 
    array (size=3)
      'id' => string '113' (length=3)
      'email' => string 'yintx_1292342352@163.com' (length=24)
      'count' => string '1' (length=1)
    
  1 => 
    array (size=3)
      'id' => string '111' (length=3)
      'email' => string 'yintx_1293456456@163.com' (length=24)
      'count' => string '0' (length=1)
  2 => 
    array (size=3)
      'id' => string '109' (length=3)
      'email' => string 'yintx_129@99999.com' (length=19)
      'count' => string '8' (length=1)</code>

回复内容:

合并前的数组a:

<code>array (size=3)
  0 => 
    array (size=2)
      'id' => string '113' (length=3)
      'email' => string 'yintx_1292342352@163.com' (length=24)
    
  1 => 
    array (size=2)
      'id' => string '111' (length=3)
      'email' => string 'yintx_1293456456@163.com' (length=24)
  2 => 
    array (size=2)
      'id' => string '109' (length=3)
      'email' => string 'yintx_129@99999.com' (length=19)
</code>

合并前的数组b:

<code>array (size=2)
  0 => 
    array (size=2)
      'user_id' => string '113' (length=2)
      'count' => string '1' (length=1)
  1 => 
    array (size=2)
      'user_id' => string '109' (length=2)
      'count' => string '8' (length=1)
</code>

合并后的数组(合并条件,a的id==b的user_id时合并):

<code>array (size=3)
  0 => 
    array (size=3)
      'id' => string '113' (length=3)
      'email' => string 'yintx_1292342352@163.com' (length=24)
      'count' => string '1' (length=1)
    
  1 => 
    array (size=3)
      'id' => string '111' (length=3)
      'email' => string 'yintx_1293456456@163.com' (length=24)
      'count' => string '0' (length=1)
  2 => 
    array (size=3)
      'id' => string '109' (length=3)
      'email' => string 'yintx_129@99999.com' (length=19)
      'count' => string '8' (length=1)</code>

array_merge_recursive() 函数与 array_merge() 函数 一样,将一个或多个数组的元素的合并起来,一个数组中的值附加在前一个数组的后面。并返回作为结果的数组。
但是,与 array_merge() 不同的是,当有重复的键名时,值不会被覆盖,而是将多个相同键名的值递归组成一个数组。(参见例子 1)

$a1=array("a"=>"Horse","b"=>"Dog");
$a2=array("c"=>"Cow","b"=>"Cat");
print_r(array_merge_recursive($a1,$a2));
?>
输出:
Array (
[a] => Horse
[b] => Array ( [0] => Dog [1] => Cat )
[c] => Cow
)

<code>$a = array(
    array('id'=>'113','email'=>'yintx_1292342352@163.com'), 
    array('id'=>'111','email'=>'yintx_1293456456@163.com'), 
    array('id'=>'109','email'=>'yintx_129@99999.com')
    );

$b = array(
    array('user_id'=>'113','count'=>'1'), 
    array('user_id'=>'109','count'=>'8')
    );

foreach($a as $ka=>$va){
    foreach($b as $kb=>$vb){
        if($va['id'] == $vb['user_id']){
            $a[$ka]['count'] = $vb['count'];
        }
    }
}

var_dump($a);

----------

array (size=3)
  0 => 
    array (size=3)
      'id' => string '113' (length=3)
      'email' => string 'yintx_1292342352@163.com' (length=24)
      'count' => string '1' (length=1)
  1 => 
    array (size=2)
      'id' => string '111' (length=3)
      'email' => string 'yintx_1293456456@163.com' (length=24)
  2 => 
    array (size=3)
      'id' => string '109' (length=3)
      'email' => string 'yintx_129@99999.com' (length=19)
      'count' => string '8' (length=1)</code>

array_merge_recursive($a, $b);

Statement:
The content of this article is voluntarily contributed by netizens, and the copyright belongs to the original author. This site does not assume corresponding legal responsibility. If you find any content suspected of plagiarism or infringement, please contact admin@php.cn