Question

我正在尝试根据匹配字段＆＃34; employerPayeReference＆＃34;组合以下数组。以下数组

数组1：

{
    "employments": [
    {
       "employerPayeReference": "123/AB456",
       "payFromEmployment": 100.00
    },
    {
       "employerPayeReference": "456/AB456",
       "payFromEmployment": 100.00
    }
]
}

数组2：

{
  "employments": [
    {
      "employerPayeReference": "123/AB456",
      "taxTakenOffPay": 10.00
    },
    {
      "employerPayeReference": "456/AB456",
      "taxTakenOffPay": 15.00
    }
  ]
}

我怎样才能得到一个看起来像这样的数组

{
  "employments": [
    {
      "employerPayeReference": "123/AB456",
      "payFromEmployment": 100.00
      "taxTakenOffPay": 10.00
    },
    {
      "employerPayeReference": "456/AB456",
      "payFromEmployment": 100.00
      "taxTakenOffPay": 15.00
    }
  ]
}

（即）它基于雇主参考匹配进行组合。任何帮助表示赞赏

Answer 1

您可以使用array_replace_recursive来获得结果。但是，因为它适用于数组，您需要使用json_decode将json转换为数组，然后使用json_encode进行编码以获得所需的结果。

$arr1 = '{
    "employments": [
    {
       "employerPayeReference": "123/AB456",
       "payFromEmployment": 100.00
    },
    {
       "employerPayeReference": "456/AB456",
       "payFromEmployment": 100.00
    }
]
}';

$arr2 = '{
  "employments": [
    {
      "employerPayeReference": "123/AB456",
      "taxTakenOffPay": 10.00
    },
    {
      "employerPayeReference": "456/AB456",
      "taxTakenOffPay": 15.00
    }
  ]
}';

$arr1 = json_decode($arr1, true);
$arr2 = json_decode($arr2, true);

$finalArr = array_replace_recursive($arr1, $arr2);

echo json_encode($finalArr);

单线：

echo json_encode(array_replace_recursive(json_decode($arr1, true), json_decode($arr2, true)));

有一种类似的方法来处理来自另一个SO answer

的数组

组合多个JSON数组

1 个答案: