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Golang/gin parse JSON from gin.Context

I learned from the gin doc that you can bind json to a struct like

type Login struct {
    User     string `form:"user" json:"user" binding:"required"`
    Password string `form:"password" json:"password" binding:"required"`
}

func main() {
    router := gin.Default()

    // Example for binding JSON ({"user": "manu", "password": "123"})
    router.POST("/loginJSON", func(c *gin.Context) {
        var json Login
        if c.BindJSON(&json) == nil {
            if json.User == "manu" && json.Password == "123" {
                c.JSON(http.StatusOK, gin.H{"status": "you are logged in"})
            } else {
                c.JSON(http.StatusUnauthorized, gin.H{"status": "unauthorized"})
            }
        }
    })
}

You always have to build a struct to bind JSON.

But if there is a very complex JSON data, I only what to get part of it, to create a complex struct is a big burden. Can avoid id and parse it directly?

like image 641
Lucas Liu Avatar asked Aug 11 '17 06:08

Lucas Liu


1 Answers

You don't need to create a struct for all the fields present in the JSON response, only the fields you are interested in. Any other fields present in the response will be ignored when unmarshaling.

You can also unmarshal to a generic map[string]interface{} but this is only really useful for truly dynamic data. If you know the format of the response ahead of time you will nearly always be best to create a custom struct to get type safety and avoid continual nil checks when accessing the map. Additionally, a targeted struct avoids storing unnecessary values when JSON in unmarshalled.

You can use the JSON to Go tool to help quickly create a struct definition from a JSON response. You could then easily strip out all the fields you don't need.

like image 191
Martin Campbell Avatar answered Oct 11 '22 03:10

Martin Campbell