How to unmarshall an array of different types correctly?

好久不见. 提交于 2019-11-29 03:37:44

Go official blog has a nice article about encoding/json: JSON and GO. It's possible to "Decode arbitrary data" into an interface{} and use type assertion to determine the type dynamically.

Your code can be probably modified to this:

package main

import (
    "encoding/json"
    "fmt"
)

var my_json string = `{
    "an_array":[
    "with_a string",
    {
        "and":"some_more",
        "different":["nested", "types"]
    }
    ]
}`

func WTHisThisJSON(f interface{}) {
    switch vf := f.(type) {
    case map[string]interface{}:
        fmt.Println("is a map:")
        for k, v := range vf {
            switch vv := v.(type) {
            case string:
                fmt.Printf("%v: is string - %q\n", k, vv)
            case int:
                fmt.Printf("%v: is int - %q\n", k, vv)
            default:
                fmt.Printf("%v: ", k)
                WTHisThisJSON(v)
            }

        }
    case []interface{}:
        fmt.Println("is an array:")
        for k, v := range vf {
            switch vv := v.(type) {
            case string:
                fmt.Printf("%v: is string - %q\n", k, vv)
            case int:
                fmt.Printf("%v: is int - %q\n", k, vv)
            default:
                fmt.Printf("%v: ", k)
                WTHisThisJSON(v)
            }

        }
    }
}

func main() {

    fmt.Println("JSON:\n", my_json, "\n")

    var f interface{}
    err := json.Unmarshal([]byte(my_json), &f)
    if err != nil {
        fmt.Println(err)
    } else {
        fmt.Printf("JSON: ")
        WTHisThisJSON(f)
    }
}

It gives output as follows:

JSON:
 {
    "an_array":[
    "with_a string",
    {
        "and":"some_more",
        "different":["nested", "types"]
    }
    ]
} 

JSON: is a map:
an_array: is an array:
0: is string - "with_a string"
1: is a map:
and: is string - "some_more"
different: is an array:
0: is string - "nested"
1: is string - "types"

It's not complete yet, but shows how it's gonna work.

易学教程内所有资源均来自网络或用户发布的内容,如有违反法律规定的内容欢迎反馈
该文章没有解决你所遇到的问题?点击提问,说说你的问题,让更多的人一起探讨吧!