在Go中解组通用json [英] unmarshal generic json in Go

问题描述

我是一名新的Go程序员(来自Java)，我想重现一种通用的方法，该方法很容易在Java中使用.

I'm a new Go programmer (From Java) and I would like to reproduce a generic way which is esay to use in Java.

我想创建一些函数，使我可以对JSON字符串进行解组，以避免代码重复.

I want to create some function which allow me to do an Unmarshal on a JSON string in order to avoid code duplicity.

这是我当前的无效代码:

This is my current code which is not working :

type myStruct1 struct {
    id string
    name string
}

func (obj myStruct1) toString() string {
    var result bytes.Buffer
    result.WriteString("id : ")
    result.WriteString(obj.id)
    result.WriteString("\n")
    result.WriteString("name : ")
    result.WriteString(obj.name)

    return result.String()
}

func main() {

    content := `{id:"id1",name="myName"}`
    object := myStruct1{}
    parseJSON(content, object)

    fmt.Println(object.toString()) 
}

func parseJSON(content string, object interface{}) {
    var parsed interface{}
    json.Unmarshal([]byte(content), &parsed)
}

此代码在运行时会向我返回此:

This code, on run, returns me this :

id : 
name : 

你有什么主意吗?

谢谢

推荐答案

问题是您要写入通用类型吗?您可能想要一个字符串映射.无论如何，这都可以与BSON一起使用:

The issue is you want to write to a generic type? You probably want a string map. This works with BSON anyways:

var anyJson map[string]interface{}
json.Unmarshal(bytes, &anyJson)

您将能够像这样访问字段:

You'll be able to access the fields like so:

anyJson["id"].(string)

别忘了输入断言您的值，并且它们必须是正确的类型，否则会出现恐慌. (您可以在golang网站上了解有关类型断言的更多信息)

Don't forget to type assert your values, and they must be the correct type or they'll panic. (You can read more about type assertions on the golang site)

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