Golang 发送带有状态的 json 响应的正确方法

Question

如何发送json响应正文中带有状态代码的响应。

我的代码

func getUser(w http.ResponseWriter, r *http.Request) {
    w.Header().Set("Content-Type", "application/json")
    var user []User
    result := db.Find(&user)
    json.NewEncoder(w).Encode(result)
}

我现在的结果：

[
    {
        "name" : "test",
        "age" : "28",
        "email":"test@gmail.com"
    },
    {
        "name" : "sss",
        "age" : "60",
        "email":"ss@gmail.com"
    },
    {
        "name" : "ddd",
        "age" : "30",
        "email":"ddd@gmail.com"
    },
]

status但我需要用这样的代码发送响应

{
    status : "success",
    statusCode : 200,
    data : [
        {
            "name" : "test",
            "age" : "28",
            "email":"test@gmail.com"
        },
        {
            "name" : "sss",
            "age" : "60",
            "email":"ss@gmail.com"
        },
        {
            "name" : "ddd",
            "age" : "30",
            "email":"ddd@gmail.com"
        },
    ]
}

Answer 1

如果您想要不同的 json，请将不同的对象传递给Encode.

type Response struct {
    Status       string `json:"status"`
    StatucCode   int    `json:"statusCode"`
    Data         []User `json:"data"`
}

func getUser(w http.ResponseWriter, r *http.Request) {
    w.Header().Set("Content-Type", "application/json")
    var user []User
    result := db.Find(&user)
    json.NewEncoder(w).Encode(&Response{"success", 200, result})
}

或者使用map：

json.NewEncoder(w).Encode(map[string]interface{}{
    "status": "success", 
    "statusCode": 200, 
    "data": result,
})