如何在Golang中开发基础的API接口返回统一格式_Golang API返回格式项目实战汇总

type Response struct {
    Code    int         `json:"code"`
    Message string      `json:"message"`
    Data    interface{} `json:"data,omitempty"` // 没有数据时可以省略
}

func Success(data interface{}, message string, w http.ResponseWriter) {
    w.Header().Set("Content-Type", "application/json")
    json.NewEncoder(w).Encode(Response{
        Code:    200,
        Message: message,
        Data:    data,
    })
}

func Fail(code int, message string, w http.ResponseWriter) {
    w.Header().Set("Content-Type", "application/json")
    json.NewEncoder(w).Encode(Response{
        Code:    code,
        Message: message,
        Data:    nil,
    })
}

func getUser(w http.ResponseWriter, r *http.Request) {
    user := map[string]interface{}{
        "id":   1,
        "name": "Alice",
    }
    Success(user, "获取用户成功", w)
}

const (
    SUCCESS        = 200
    ERROR          = 500
    INVALID_PARAMS = 400
    UNAUTHORIZED   = 401
)

Fail(INVALID_PARAMS, "参数校验失败", w)

func Success(c *gin.Context, data interface{}, msg string) {
    c.JSON(http.StatusOK, Response{
        Code:    200,
        Message: msg,
        Data:    data,
    })
}

func Fail(c *gin.Context, code int, msg string) {
    c.JSON(http.StatusOK, Response{
        Code:    code,
        Message: msg,
        Data:    nil,
    })
}

router.GET("/user", func(c *gin.Context) {
    Success(c, userModel, "查询成功")
})