如何在laravel中返回响应对象json

Es *_*a M 0 php json laravel laravel-5

我已经得到回应,但我不需要这样

{
  "status": "200",
  "id": 1,
  "email": "a@a.a",
  "mobile": "",
  "source": "",
  "source_id": "",
  "message": "Bad Request : Already Logged In"
}
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我需要在另一个对象中返回这些数据,例如:

{
 {
  "status": "200",
 }
 { 
  "id": 1,
  "email": "a@a.a",
  "mobile": "",
  "source": "",
  "source_id": "",
  "message": "Bad Request : Already Logged In"
 }
}
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Alp*_*sus 8

更新(2021 年 2 月 9 日):如果适用,将模型序列化为 JSON 的最佳方法是使用Eloquent 的 API 资源。这在您构建 RESTful API 时特别有用。


使用response()json()方法。

$data = [
    "status" => "200",
    "details" => [
        "id": 1,
        "email": "a@a.a",
        "mobile": "",
        "source": "",
        "source_id": "",
        "message": "Bad Request : Already Logged In"            
    ]
];

return response()->json($data);
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这将返回以下 JSON:

{
    "status": "200",
    "details":  { 
        "id": 1,
        "email": "a@a.a",
        "mobile": "",
        "source": "",
        "source_id": "",
        "message": "Bad Request : Already Logged In"
    }
}
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