golang json unmarshal map of map [string] interface {}

Question

我有以下的代码,试图解组此JSON文件,但是行json.Unmarshal([]字节(MSG ["餐厅"]),餐厅)总是给人一个错误.如何让Unmarshal忽略"餐馆"或仅将"餐馆"数据传递给Unmarshal功能？

谢谢!

{
  "restaurant": {
    "name": "Tickets",
    "owner": {
      "name": "Ferran"
    }
  }
}


    file, e := ioutil.ReadFile("./rest_read.json")
    if e != nil {
            fmt.Println("file error")
            os.Exit(1)
    }

    var data interface{}
    json.Unmarshal(file, &data)

    msg := data.(map[string]interface{})
    log.Println(msg)
    log.Println(msg["restaurant"])
    log.Println(reflect.TypeOf(msg["restaurant"]))

    var restaurant Restaurant
    json.Unmarshal([]byte(msg["restaurant"]), &restaurant)

    log.Println("RName: ", restaurant.Name)
    log.Println("Name: ", restaurant.Owner.Name)

Answer 1

可以通过解码到接口然后从结果中提取顶级映射来执行通用解组ala gson,例如:

var msgMapTemplate interface{}
err := json.Unmarshal([]byte(t.ResponseBody), &msgMapTemplate)
t.AssertEqual(err, nil)
msgMap := msgMapTemplate.(map[string]interface{})

有关详细信息,请参阅http://blog.golang.org/json-and-go中的 "解码任意数据" .

Answer 2

我建议为您的数据构建一个合适的模型.这将使您能够将数据干净地解组为Go结构.

package main

import (
    "encoding/json"
    "fmt"
)

type Restaurant struct {
    Restaurant RestaurantData `json:"restaurant"`
}

type RestaurantData struct {
    Name  string `json:"name"`
    Owner Owner  `json:"owner"`
}

type Owner struct {
    Name string `json:"name"`
}

func main() {
    data := `{"restaurant":{"name":"Tickets","owner":{"name":"Ferran"}}}`
    r := Restaurant{}
    json.Unmarshal([]byte(data), &r)

    fmt.Printf("%+v", r)
}