如何在golang中将数据结构作为参数传递

Question

XML到Json

package main

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

type Persons struct {
    Person []struct {
        Name string
        Age  int
    }
}
type Places struct {
    Place []struct {
        Name    string
        Country string
    }
}

type Parks struct {
    Park struct {
        Name     []string
        Capacity []int
    }
}

const personXml = `
    <Persons>
        <Person><Name>Koti</Name><Age>30</Age></Person>
        <Person><Name>Kanna</Name><Age>29</Age></Person>
    </Persons>
`

const placeXml = `
    <Places>
        <Place><Name>Chennai</Name><Country>India</Country></Place>
        <Place><Name>London</Name><Country>UK</Country></Place>
    </Places>
`

const parkXml = `
    <Parks>
        <Park><Name>National Park</Name><Capacity>10000</Capacity></Park>
        <Park>Asian Park</Name><Capacity>20000</Capacity></Park>
    </Parks>
`

func WhatIamUsing() {
    var persons Persons
    xml.Unmarshal([]byte(personXml), &persons)
    per, _ := json.Marshal(persons)
    fmt.Printf("%s\n", per)

    var places Places
    xml.Unmarshal([]byte(placeXml), &places)
    pla, _ := json.Marshal(places)
    fmt.Printf("%s\n", pla)

    var parks Parks
    xml.Unmarshal([]byte(parkXml), &parks)
    par, _ := json.Marshal(parks)
    fmt.Printf("%s\n", par)
}

我想要的是一个泛型函数,它接受xml字符串和dataStruct并返回一个Json输出.但是下面的功能是抛出错误如何捏造这个？

func Xml2Json(xmlString string, DataStruct interface{}) (jsobj string, err error) {
    var dataStruct DataStruct
    xml.Unmarshal([]byte(personXml), &dataStruct)
    js, _ := json.Marshal(dataStruct)
    return fmt.Sprintf("%s\n", js), nil
}

func main() {
    jsonstring, _ := Xml2Json(personXml, Persons)
}

错误信息:

prog.go:73:DataStruct不是一种类型

prog.go:80:type Persons不是表达式

goplay链接:http://play.golang.org/p/vayb0bawKx

Answer 1

您不能Persons在界面中存储类型(如).你可以传递reflect.Type给你的功能.然后,Xml2Json(personXml, reflect.TypeOf(Persons))在我看来,你的电话看起来很难看.

更好的方法可能是:

func Xml2Json(xmlString string, value interface{}) (string, error) {
    if err := xml.Unmarshal([]byte(xmlString), value); err != nil {
        return "", err
    }
    js, err := json.Marshal(value)
    if err != nil {
        return "", err
    }
    return string(js), nil
}

Xml2Json(personXml, new(Persons))如果您对值本身不感兴趣,可以使用此函数

var persons Persons
Xml2Json(personXML, &persons)

当您还想检索struct值以供以后处理时.