Question

type Struct struct {
   Value  string `json:"value"`
   Value1 string `json:"value_one"`
   Nest   Nested `json:"nest"`
}

type Nested struct {
   Something string `json:"something"`
}

我想添加不在结构定义中的元素而不创建另一个结构类型。例如

Struct.Extra1 = Nested{"yy"}
Struct.Nested.Extra2 = "zz"

将导致

{
    "Value": "xx",
    "Value1": "xx",
    "Extra1": {
      "Something", "yy"
    },
    "Nest": {
      "Something": "xx",
      "Extra2": "zz"
    }
}

解决方法1： 我想添加omitempty来实现这一目标，但它会使结构变得复杂。

type Struct struct {
   Value  string
   Value1 string
   Nest   Nested
   Extra1 Nested `json:"omitempty"`
}

type Nested struct {
   Something string
   Extra2 string `json:"omitempty"`
}

溶液2：

myextras := make(map[string]interface{})
// get Struct.Nested in map[string]interface{} format
myextras = Struct.Nest
myextras["Extra2"] = "zz"

// get Struct in map[string]interface{} format
struct["Nest"] = myextras
struct["Extra1"] = Nested{"yy"}

// solves the problem with lots of type casting but doesn't support json tag naming

是否有更好的解决方案来添加嵌套元素，这些元素在结构数据类型中没有用json-tag支持表示，可用于输出给用户。

Answer 1

基于这个答案：Can I use MarshalJSON to add arbitrary fields to a json encoding in golang?

您可以执行类似的操作（演示：http://play.golang.org/p/dDiTwxhoNn）：

package main

import (
    "encoding/json"
    "fmt"
    "log"
)

type Book struct {
    Title  string
    Author string

    // extra is used for additional dynamic element marshalling
    extra func() interface{}
}

type FakeBook Book

func (b *Book) SetExtra(fn func() interface{}) {
    b.extra = fn
}

func (b *Book) MarshalJSON() ([]byte, error) {
    if b.extra == nil {
        b.extra = func() interface{} { return *b }
    }

    return json.Marshal(b.extra())
}

func main() {
    ms := &Book{
        Title:  "Catch-22",
        Author: "Joseph Heller",
    }

    ms.SetExtra(func() interface{} {
        return struct {
            FakeBook
            Extra1 struct {
                Something string `json:"something"`
            } `json:"extra1"`
        }{
            FakeBook: FakeBook(*ms),
            Extra1: struct {
                Something string `json:"something"`
            }{
                Something: "yy",
            },
        }
    })

    out, err := json.MarshalIndent(ms, "", "  ")
    if err != nil {
        log.Fatalln(err)
    }
    fmt.Println(string(out))

    mb := &Book{
        Title:  "Vim-go",
        Author: "Fatih Arslan",
    }

    mb.SetExtra(func() interface{} {
        return struct {
            FakeBook
            Something string `json:"something"`
        }{
            FakeBook:  FakeBook(*mb),
            Something: "xx",
        }
    })

    out, err = json.MarshalIndent(mb, "", "  ")
    if err != nil {
        log.Fatalln(err)
    }
    fmt.Println(string(out))

    mc := &Book{
        Title:  "Another-Title",
        Author: "Fatih Arslan",
    }

    out, err = json.MarshalIndent(mc, "", "  ")
    if err != nil {
        log.Fatalln(err)
    }
    fmt.Println(string(out))
}

Answer 2

是肯定的。有一个类型json.Raw，它不是结构而是[]字节。你可以用任何元帅/解组的方式管理它。

Answer 3

地图方法是唯一理智的方法，其他一切（如json.RawMessage字段无论如何都需要额外的编组步骤。

Answer 4

如果某人对提供的解决方案不满意：

尝试tidwall/sjson。它提供了用于快速JSON编辑的功能，而无需定义任何结构。昨天节省了我很多时间：D

用法示例：

value, _ := sjson.Set(`{"name":{"last":"Anderson"}}`, "name.last", "Smith")
println(value)

// Output:
// {"name":{"last":"Smith"}}

Answer 5

如果您不想安装额外的包，您可以使用此代码更改或添加新值：

package main

import "fmt"
import "strings"
import "encoding/json"

func main() {
    s := []byte(`{ "level1a":"aaa", "level1b":{"level2a":"bbb", "level2b":{"level3":"aaa"} } }`)
    var j interface{}
    json.Unmarshal(s, &j)
    SetValueInJSON(j, "level1a", "new value 1a")
    SetValueInJSON(j, "level1b.level2a", "new value 2a")
    SetValueInJSON(j, "level1b.level2b.level3", "new value 3")
    SetValueInJSON(j, "level1b.level2c", "new key")
    s,_ = json.Marshal(j)
    fmt.Println(string(s))
    // result: {"level1a":"new value 1a","level1b":{"level2a":"new value 2a","level2b":{"level3":"new value 3"},"level2c":"new key"}}

}

func SetValueInJSON(iface interface{}, path string, value interface{}) interface{} {
    m := iface.(map[string]interface{})
    split := strings.Split(path, ".")
    for k, v := range m {
        if strings.EqualFold(k, split[0]) {
            if len(split) == 1 {
                m[k] = value
                return m
            }
            switch v.(type) {
            case map[string]interface{}:
                return SetValueInJSON(v, strings.Join(split[1:], "."), value)
            default:
                return m
            }
        }
    }
    // path not found -> create
    if len(split) == 1 {
        m[split[0]] = value
    } else {
        newMap := make(map[string]interface{})
        newMap[split[len(split)-1]] = value
        for i := len(split) - 2; i > 0; i-- {
            mTmp := make(map[string]interface{})
            mTmp[split[i]] = newMap
            newMap = mTmp
        }
        m[split[0]] = newMap
    }
    return m
}

Golang修改没有struct的json

5 个答案: