使用一些已知字段名称和一些未知字段名称解组JSON [英] Unmarshal JSON with some known, and some unknown field names

问题描述

我有以下JSON

{"a":1, "b":2, "?":1, "??":1}

我知道它具有"a"和"b"字段，但是我不知道其他字段的名称.因此，我想以以下类型将其解组:

I know that it has the "a" and "b" fields, but I don't know the names of other fields. So I want to unmarshal it in the following type:

type Foo struct {
  // Known fields
  A int `json:"a"`
  B int `json:"b"`
  // Unknown fields
  X map[string]interface{} `json:???` // Rest of the fields should go here.
}

我该怎么做?

推荐答案

解组两次

一个选项是解组两次:一次进入一个Foo类型的值，一次进入一个map[string]interface{}类型的值，并删除键"a"和"b":

Unmarshal twice

One option is to unmarshal twice: once into a value of type Foo and once into a value of type map[string]interface{} and removing the keys "a" and "b":

type Foo struct {
    A int                    `json:"a"`
    B int                    `json:"b"`
    X map[string]interface{} `json:"-"` // Rest of the fields should go here.
}

func main() {
    s := `{"a":1, "b":2, "x":1, "y":1}`
    f := Foo{}
    if err := json.Unmarshal([]byte(s), &f); err != nil {
        panic(err)
    }

    if err := json.Unmarshal([]byte(s), &f.X); err != nil {
        panic(err)
    }
    delete(f.X, "a")
    delete(f.X, "b")

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

输出(在转到游乐场上尝试):

{A:1 B:2 X:map[x:1 y:1]}

解组一次并手动处理

另一种选择是将一次解组到map[string]interface{}并手动处理Foo.A和Foo.B字段:

Unmarshal once and manual handling

Another option is to unmarshal once into an map[string]interface{} and handle the Foo.A and Foo.B fields manually:

type Foo struct {
    A int                    `json:"a"`
    B int                    `json:"b"`
    X map[string]interface{} `json:"-"` // Rest of the fields should go here.
}

func main() {
    s := `{"a":1, "b":2, "x":1, "y":1}`
    f := Foo{}
    if err := json.Unmarshal([]byte(s), &f.X); err != nil {
        panic(err)
    }
    if n, ok := f.X["a"].(float64); ok {
        f.A = int(n)
    }
    if n, ok := f.X["b"].(float64); ok {
        f.B = int(n)
    }
    delete(f.X, "a")
    delete(f.X, "b")

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

输出是相同的(进入游乐场):

{A:1 B:2 X:map[x:1 y:1]}

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