Laravel：合併來自兩個表的數據

Question

我有一個與專業，技能，名稱，位置等有關係的用戶表。我可以通過此命令獲得簡單用戶信息app\users::find(1)及其技能信息，例如app\user::find(1)->skill。我如何通過一個命令獲取關於特定用戶的一般信息，職業信息和技能？

例如：

本地主機：8000 /用戶/ 1

這個環節應該給我JSON結果如下圖所示：

{ 
    "id": 0, 
    "name": "string", 
    "email": "string", 
    "gender": "string", 
    "age": 0, 
    "mobile_number": "string", 
    "company_name": "string", 
    "verification_status": 0, 
    "image_url": "string", 
    "joining_date": "2017-04-26T12:34:34.501Z", 
    "message": "string", 
    "profession": { 
    "id": 0, 
    "name": "string" 
    }, 
    "designation": { 
    "id": 0, 
    "name": "string" 
    }, 
    "location": { 
    "id": 0, 
    "longitude": 0, 
    "latitude": 0, 
    "address": "string", 
    "city": "string", 
    "country": "string" 
    }, 
    "interests": [ 
    { 
     "id": 0, 
     "name": "string", 
     "custom": "string" 
    } 
    ], 
    "skills": [ 
    { 
     "id": 0, 
     "name": "string", 
     "custom": "string" 
    } 
    ] 
} 

Answer 1

您可以加載所有關係該型號爲using eager loading，然後serialize that object to a json string。

像這樣：

app\user::where('id', 1)->with(['profession', 'designation', '...'])->first()->toJson();