两个二维数组的合并

Question

合并前的数组a：

array (size=3)
  0 => 
    array (size=2)
      'id' => string '113' (length=3)
      'email' => string '[e-mail]' (length=24)
    
  1 => 
    array (size=2)
      'id' => string '111' (length=3)
      'email' => string '[e-mail]' (length=24)
  2 => 
    array (size=2)
      'id' => string '109' (length=3)
      'email' => string '[e-mail]' (length=19)

合并前的数组b：

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)

合并后的数组（合并条件，a的id==b的user_id时合并）：

array (size=3)
  0 => 
    array (size=3)
      'id' => string '113' (length=3)
      'email' => string '[e-mail]' (length=24)
      'count' => string '1' (length=1)
    
  1 => 
    array (size=3)
      'id' => string '111' (length=3)
      'email' => string '[e-mail]' (length=24)
      'count' => string '0' (length=1)
  2 => 
    array (size=3)
      'id' => string '109' (length=3)
      'email' => string '[e-mail]' (length=19)
      'count' => string '8' (length=1)

Answer 1

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）

<?php
$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
)

Answer 2

$a = array(
    array('id'=>'113','email'=>'[e-mail]'), 
    array('id'=>'111','email'=>'[e-mail]'), 
    array('id'=>'109','email'=>'[e-mail]')
    );

$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 '[e-mail]' (length=24)
      'count' => string '1' (length=1)
  1 => 
    array (size=2)
      'id' => string '111' (length=3)
      'email' => string '[e-mail]' (length=24)
  2 => 
    array (size=3)
      'id' => string '109' (length=3)
      'email' => string '[e-mail]' (length=19)
      'count' => string '8' (length=1)

Answer 3

数组A($arrA)

array (size=3)
  0 => 
    array (size=2)
      'id' => string '113' (length=3)
      'email' => string '[e-mail]' (length=24)
  1 => 
    array (size=2)
      'id' => string '111' (length=3)
      'email' => string '[e-mail]' (length=24)
  2 => 
    array (size=2)
      'id' => string '109' (length=3)
      'email' => string '[e-mail]' (length=19)

数组B($arrB)

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)

数组C($arrC)

array (size=2)
  0 => 
    array (size=2)
      'id' => string '113' (length=2)
      'count' => string '1' (length=1)
  1 => 
    array (size=2)
      'id' => string '109' (length=2)
      'count' => string '8' (length=1)

你的需求没有现成的函数可以满足，倒是可以使用array_walk()来遍历重组：
例如：

array_walk($arrA,function(&$item, $key, $prefix)use($arrB){
    foreach($arrB as $k=>$v){
        if($item['id']==$v['user_id']){
            $item['count'] = $v['count'];
        }
    }
});

实现！
另外：执行：array_replace_recursive($arrA,$arrB)得到：

array (size=3)
  0 => 
    array (size=3)
      'id' => string '113' (length=3)
      'email' => string '[e-mail]' (length=24)
      'user_id' => string '113' (length=2)
      'count' => string '1' (length=1)
    
  1 => 
    array (size=3)
      'id' => string '111' (length=3)
      'email' => string '[e-mail]' (length=24)
      'user_id' => string '109' (length=2)
      'count' => string '8' (length=1)
  2 => 
    array (size=3)
      'id' => string '109' (length=3)
      'email' => string '[e-mail]' (length=19)

执行：array_replace_recursive($arrA,$arrC)得到：

array (size=3)
  0 => 
    array (size=3)
      'id' => string '113' (length=3)
      'email' => string '[e-mail]' (length=24)
      'count' => string '1' (length=1)
  1 => 
    array (size=3)
      'id' => string '109' (length=3)
      'email' => string '[e-mail]' (length=24)
      'count' => string '8' (length=1)
  2 => 
    array (size=3)
      'id' => string '109' (length=3)
      'email' => string '[e-mail]' (length=19)

Answer 4

提供一个不使用循环的办法

$a = [
    ['id' => 1, 'age' => 18],
    ['id' => 2, 'age' => 19],
    ['id' => 3, 'age' => 20],
    ['id' => 4, 'age' => 22]
];

$b = [
    ['id' => 2, 'name' => 'aa'],
    ['id' => 4, 'name' => 'bb']
];

$a1 = array_column($a, null, 'id');
$b1 = array_column($b, null, 'id');

$keys = array_keys($a1);
$temp = array_fill_keys($keys, []);
$b1 = $b1 + $temp;
ksort($b1);
function merge($v1, $v2) {
    return array_merge($v1, $v2);
}
$new = array_map('merge', $a1, $b1);
print_r($new);

Answer 5

array_merge_recursive($a, $b);

Answer 6

array_merge_recursive， 需要注意的是此函数合并的是相同的键名，而需求是合并相同的id， 所以首先要把id提取出来作为键名， 其次此函数只合并相同的字符串键名，所以还需要把键名转换为字符串，然后就可以使用array_merge_recursive 合并了。

$a1 = $b1 = [];

foreach($a as $v){$a1['id_'.$v['id']]=$v;}

foreach($b as $v){$b1['id_'.$v['id']]=$v;}

print_r(array_values(array_merge_recursive($a1,$b1)));