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如何从字符串表示形式中获取嵌套结构from变量?

go json

Mayukh Sarkar · 技术社区 · 7 年前

我有一个JSON文件看起来像

{
    "type": "Weekly",
    "clients": [
        "gozo",
        "dva"
    ],
    "sender": "no-reply@flowace.in",
    "recipients": {
        "gozo": [
            "a@gmail.com",
            "b@hotmail.com"
        ],
        "dva": [
            "c@gmail.com",
            "d@hotmail.com"
        ]
    },
    "features": [
        "Top5UsedApps",
        "TimeSpentOnEachL3",
        "NewlyAssignedL3",
        "HoursLogged"
    ],
    "dbCloning": [
        "dva"
    ] 
}

我已经映射了如下结构。

type receivers struct {
    Gozo []string `json:"gozo"`
    Dva  []string `json:"dva"`
    // Add more recievers
}

// Config -- The config object parsed from the JSON file
type Config struct {
    ReportType string    `json:"type"`
    Clients    []string  `json:"clients"`
    Sender     string    `json:"sender"`
    Recipients receivers `json:"recipients"`
    Cloning    []string  `json:"dbCloning"`
}

然后在另一个源文件的某个地方,我做了如下操作,

func main() {
    conf := LoadConfig(os.Args[1])
    for _, client := range conf.Clients {

        // Use the client variable of some other function calls

        fmt.Println(conf.Recipients[client]) // This will not work
}

conf.Recipients 直接。

PS:考虑一下 LoadConfig 函数将JSON解组并返回 conf

编辑1: 看起来是设计决策失误。现在来解决 map[string][]string

2 回复 | 直到 7 年前

Ullaakut 7 年前

问题是你的类型 receivers 不应该给字段命名。它应该是一个 map[string][]string

下面是一个工作示例:

package main

import (
    "encoding/json"
    "fmt"
)

type Config struct {
    ReportType string              `json:"type"`
    Clients    []string            `json:"clients"`
    Sender     string              `json:"sender"`
    Recipients map[string][]string `json:"recipients"`
    Cloning    []string            `json:"dbCloning"`
}

var data = []byte(`{
    "type": "Weekly",
    "clients": [
        "gozo",
        "dva"
    ],
    "sender": "no-reply@flowace.in",
    "recipients": {
        "gozo": [
            "a@gmail.com",
            "b@hotmail.com"
        ],
        "dva": [
            "c@gmail.com",
            "d@hotmail.com"
        ]
    },
    "features": [
        "Top5UsedApps",
        "TimeSpentOnEachL3",
        "NewlyAssignedL3",
        "HoursLogged"
    ],
    "dbCloning": [
        "dva"
    ] 
}`)

func main() {
    var conf Config
    json.Unmarshal(data, &conf)

    for _, client := range conf.Clients {
        fmt.Println(conf.Recipients[client])
    }
}

给出输出

[a@gmail.com b@hotmail.com]
[c@gmail.com d@hotmail.com]

Himanshu 7 年前

range子句提供了一种迭代数组、切片、字符串、映射或通道的方法。因此,在将struct转换为上述类型之一之前,不能在struct上设置范围。

要在结构字段上循环,可以通过为结构创建一个reflect值片段来进行反射。

package main

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

type receivers struct {
    Gozo []string `json:"gozo"`
    Dva  []string `json:"dva"`
    // Add more recievers
}

// Config -- The config object parsed from the JSON file
type Config struct {
    ReportType string    `json:"type"`
    Clients    []string  `json:"clients"`
    Sender     string    `json:"sender"`
    Recipients receivers `json:"recipients"`
    Cloning    []string  `json:"dbCloning"`
}

var data = []byte(`{
    "type": "Weekly",
    "clients": [
        "gozo",
        "dva"
    ],
    "sender": "no-reply@flowace.in",
    "recipients": {
        "gozo": [
            "a@gmail.com",
            "b@hotmail.com"
        ],
        "dva": [
            "c@gmail.com",
            "d@hotmail.com"
        ]
    },
    "features": [
        "Top5UsedApps",
        "TimeSpentOnEachL3",
        "NewlyAssignedL3",
        "HoursLogged"
    ],
    "dbCloning": [
        "dva"
    ] 
}`)

func main() {
    var conf Config
    if err := json.Unmarshal(data, &conf); err != nil{
        fmt.Println(err)
    }

    v := reflect.ValueOf(conf.Recipients)

    values := make([]interface{}, v.NumField())

    for i := 0; i < v.NumField(); i++ {
        values[i] = v.Field(i).Interface()
    }

    fmt.Println(values)
}

工作代码 Go Playground

NumField 返回结构v中的字段数。它会恐慌如果

func (v Value) NumField() int

Field 返回结构的第i个字段

func (v Value) Field(i int) Value

接口以接口{}的形式返回v的当前值。

func (v Value) Interface() (i interface{})