Php 如何将不同的索引组合成一个索引

我正在尝试将不同的索引组合成一个索引。给定的代码是示例

Array(
[0] => stdClass Object
    (
        [player_id] => 92
        [player_name] => XYZ
    )

[1] => stdClass Object
    (
        [player_type_id] => 4
        [type] => All-Rounder
    ))

预期的答案是

Array([0] => stdClass Object
     ( 
      [player_id] => 92
      [player_name] => XYZ
      [player_type_id] => 4
      [type] => All-Rounder
     )

---

试试这个：

$obj_merge=（对象）数组_merge（（数组）$obj1，（数组）$obj2）

请尝试以下操作：

$objArr1 =  (array)$yourArr[0];
$objArr2 =  (array)$yourArr[1];

$mergedArr = (object)array_merge($objArr1,$objArr2);

---

你可以通过两种方式获得成功

1） 使用

array\u merge

函数

2） 使用

+

运算符

请参阅以下示例：

$obj1 = new StdClass();
$obj1->player_id = 92;
$obj1->player_name = 'Test Name';


$obj2 = new StdClass();
$obj2->player_type_id = 92;
$obj2->type = 'Test Name';

$array = array($obj1, $obj2);

$merged_array = (object) ((array) $obj1 + (array) $obj2);

print_r($merged_array);

echo '--------------------------------------- <br />';
$obj_merged = (object) array_merge((array) $obj1, (array) $obj2);

print_r($obj_merged);

使用

foreach

循环的另一种方法：

foreach($obj2 as $k => $v){
  $obj1->$k = $v;
}

print_r($obj1);

输出：

stdClass Object
(
    [player_id] => 92
    [player_name] => Test Name
    [player_type_id] => 92
    [type] => Test Name
)
--------------------------------------- 
stdClass Object
(
    [player_id] => 92
    [player_name] => Test Name
    [player_type_id] => 92
    [type] => Test Name
)

stdClass Object
(
    [player_id] => 92
    [player_name] => Test Name
    [player_type_id] => 92
    [type] => Test Name
)