我的websocket服务器将接收和解组JSON数据。此数据将始终包装在具有键/值对的对象中。key-string将作为值标识符,告诉Go服务器它是哪种值。通过了解值的类型,我可以继续进行JSON将值解组为正确类型的结构。
每个json-object可能包含多个键/值对。
JSON示例:
{
"sendMsg":{"user":"ANisus","msg":"Trying to send a message"},
"say":"Hello"
}
有什么简单的方法可以使用"encoding/json"
包来做到这一点吗?
package main
import (
"encoding/json"
"fmt"
)
// the struct for the value of a "sendMsg"-command
type sendMsg struct {
user string
msg string
}
// The type for the value of a "say"-command
type say string
func main(){
data := []byte(`{"sendMsg":{"user":"ANisus","msg":"Trying to send a message"},"say":"Hello"}`)
// This won't work because json.MapObject([]byte) doesn't exist
objmap, err := json.MapObject(data)
// This is what I wish the objmap to contain
//var objmap = map[string][]byte {
// "sendMsg": []byte(`{"user":"ANisus","msg":"Trying to send a message"}`),
// "say": []byte(`"hello"`),
//}
fmt.Printf("%v", objmap)
}
感谢您的任何建议/帮助!
发布于 2012-06-17 05:15:19
这可以通过解组到一个map[string]json.RawMessage
来完成。
var objmap map[string]json.RawMessage
err := json.Unmarshal(data, &objmap)
要进一步解析sendMsg
,您可以执行如下操作:
var s sendMsg
err = json.Unmarshal(objmap["sendMsg"], &s)
对于say
,您可以执行相同的操作,并将其解组为字符串:
var str string
err = json.Unmarshal(objmap["say"], &str)
编辑:请记住,您还需要导出sendMsg结构中的变量才能正确解组。所以你的结构定义应该是:
type sendMsg struct {
User string
Msg string
}
发布于 2020-01-13 17:21:35
这里有一个优雅的方法来做类似的事情。但是为什么要对JSON进行部分解组?那也太没道理了。
请看下面的工作代码。复制并粘贴它。
import (
"bytes"
"encoding/json" // Encoding and Decoding Package
"fmt"
)
var messeging = `{
"say":"Hello",
"sendMsg":{
"user":"ANisus",
"msg":"Trying to send a message"
}
}`
type SendMsg struct {
User string `json:"user"`
Msg string `json:"msg"`
}
type Chat struct {
Say string `json:"say"`
SendMsg *SendMsg `json:"sendMsg"`
}
func main() {
/** Clean way to solve Json Decoding in Go */
/** Excellent solution */
var chat Chat
r := bytes.NewReader([]byte(messeging))
chatErr := json.NewDecoder(r).Decode(&chat)
errHandler(chatErr)
fmt.Println(chat.Say)
fmt.Println(chat.SendMsg.User)
fmt.Println(chat.SendMsg.Msg)
}
func errHandler(err error) {
if err != nil {
fmt.Println(err)
return
}
}
发布于 2016-11-04 02:36:09
根据Stephen Weinberg的回答,我后来实现了一个名为iojson的方便工具,它有助于轻松地将数据填充到现有对象中,并将现有对象编码为JSON字符串。还提供了一个iojson中间件来与其他中间件协同工作。有关更多示例,请访问https://github.com/junhsieh/iojson
示例:
func main() {
jsonStr := `{"Status":true,"ErrArr":[],"ObjArr":[{"Name":"My luxury car","ItemArr":[{"Name":"Bag"},{"Name":"Pen"}]}],"ObjMap":{}}`
car := NewCar()
i := iojson.NewIOJSON()
if err := i.Decode(strings.NewReader(jsonStr)); err != nil {
fmt.Printf("err: %s\n", err.Error())
}
// populating data to a live car object.
if v, err := i.GetObjFromArr(0, car); err != nil {
fmt.Printf("err: %s\n", err.Error())
} else {
fmt.Printf("car (original): %s\n", car.GetName())
fmt.Printf("car (returned): %s\n", v.(*Car).GetName())
for k, item := range car.ItemArr {
fmt.Printf("ItemArr[%d] of car (original): %s\n", k, item.GetName())
}
for k, item := range v.(*Car).ItemArr {
fmt.Printf("ItemArr[%d] of car (returned): %s\n", k, item.GetName())
}
}
}
示例输出:
car (original): My luxury car
car (returned): My luxury car
ItemArr[0] of car (original): Bag
ItemArr[1] of car (original): Pen
ItemArr[0] of car (returned): Bag
ItemArr[1] of car (returned): Pen
https://stackoverflow.com/questions/11066946
复制相似问题