$ar1 = array("color"=> array("favorite"=>"red","name"=>"test"));
$ar2 = array("color"=> array("favorite"=>"green","code"=>"code"));
print_r (array_merge ($ar1, $ar2));
print_r(array_merge_recursive ($ar1, $ar2));

可以看到array_merge只是將每個第二個數(shù)組的值復制到第一個數(shù)組上而已，顯然數(shù)組一favorite.name被數(shù)組二code給覆蓋了，而array_merge_recursive不會覆蓋掉相同的健名但一旦兩個數(shù)組健名相同則它會創(chuàng)建一個新數(shù)組并將兩個數(shù)組中值添加到新數(shù)組中

以下是我自定義的實現(xiàn)方法：

/**
 * @Author:   HTL
 * @Description: 根據(jù)指定的健名將$ar2中的所有健值復制到$ar1對應的健值中
 * @Return：$ar1
 */
 function array_more_merge($ar1,$ar2,$key_name){
  if(
    !$ar1
    || !is_array($ar1)
    || count($ar1)<=0
    || !$ar2
    || !is_array($ar2)
    || count($ar2)<=0
    || !$key_name
    || strlen($key_name)<=0)
  {
    return $ar1;
  }
  /*不能使用for進行循環(huán),因為如果數(shù)組中的索引不是連續(xù)的則使用for可能無法正確的完成循環(huán)
  如 $a =
      Array(
      1 => Array("id" => 1 )
      ,8 => Array("id" => 2 )
      ,4 => Array("id" => 4 )
     );
  該數(shù)組數(shù)為3所以只能找到[1],而[8][4]在循環(huán)中不會進行處理,因為$i <=count($a)不成立<br>　　*/
  for ($i=0; $i <=count($a) ; $i++) {
      print_r($a[$i]);
  }
  foreach ($ar1 as $i=>$items) {
    foreach ($ar2 as $key=>$item) {
      if($ar2[$key][$key_name]==$ar1[$i][$key_name]){
        $keys = array_keys($ar2[$key]);
        foreach ($keys as $name) {
          $ar1[$i][$name] = $ar2[$key][$name];
        }
      }
    }
  }
  return $ar1;
}
$ar1 = '{"1":{"cost":"1","id":"1","total_price":"5"},"2":{"cost":"2","id":"2","total_price":"10"},"4":{"cost":"4","id":"4","total_price":"20"}}';
$ar1 = json_decode($ar1,true);
$ar2 = '{"1":{"cost":"1","id":"1","total_price":"5"},"2":{"cost":"20","id":"2","total_price":"100"},"4":{"cost":"40","id":"4","total_price":"200"}}';
$ar2 = json_decode($ar2,true);
echo "array_merge_recursive:";
print_r(array_merge_recursive($ar1,$ar2));
echo "array_merge:";
print_r(array_merge($ar1,$ar2));
echo "array_more_merge:";
print_r($this->array_more_merge($ar1,$ar2,"id"));
die();

運行結(jié)果如下圖所示：