问题

在Golang中，有一种方法可以将XML（[]byte）转换为JSON输出。您可以尝试使用encoding/xml包中的Unmarshal函数进行转换。以下是您提供的代码示例的修改版本：

import (
	"encoding/xml"
	"encoding/json"
	"io/ioutil"
	"net/http"
	"bytes"
	"github.com/emicklei/go-restful"
)

// POST 
func (u *UserResource) authenticateUser(request *restful.Request, response *restful.Response) {
	App := new(Api)
	App.url = "http://api.com/api"
	usr := new(User)
	err := request.ReadEntity(usr)
	if err != nil {
		response.AddHeader("Content-Type", "application/json")
		response.WriteErrorString(http.StatusInternalServerError, err.Error())
		return
	}

	buf := []byte("<app version=\"1.0\"><request>1111</request></app>")
	r, err := http.Post(App.url, "text/plain", bytes.NewBuffer(buf))
	if err != nil {
		response.AddHeader("Content-Type", "application/json")
		response.WriteErrorString(http.StatusInternalServerError, err.Error())
		return
	}
	defer r.Body.Close()
	body, err := ioutil.ReadAll(r.Body)
	if err != nil {
		response.AddHeader("Content-Type", "application/json")
		response.WriteErrorString(http.StatusInternalServerError, err.Error())
		return
	}

	// 将XML转换为JSON
	var result map[string]interface{}
	err = xml.Unmarshal(body, &result)
	if err != nil {
		response.AddHeader("Content-Type", "application/json")
		response.WriteErrorString(http.StatusInternalServerError, err.Error())
		return
	}

	// 将JSON转换为[]byte
	jsonData, err := json.Marshal(result)
	if err != nil {
		response.AddHeader("Content-Type", "application/json")
		response.WriteErrorString(http.StatusInternalServerError, err.Error())
		return
	}

	response.AddHeader("Content-Type", "application/json")
	response.WriteHeader(http.StatusCreated)
	response.Write(jsonData)
}

以上代码将XML响应转换为JSON，并在进行一些操作后输出。您需要导入encoding/xml和encoding/json包，并对XML和JSON进行相应的解析和编码。

// POST 
func (u *UserResource) authenticateUser(request *restful.Request, response *restful.Response) {
App := new(Api)
App.url = "http://api.com/api"
usr := new(User)
err := request.ReadEntity(usr)
if err != nil {
response.AddHeader("Content-Type", "application/json")
response.WriteErrorString(http.StatusInternalServerError, err.Error())
return
}
buf := []byte("<app version=\"1.0\"><request>1111</request></app>")
r, err := http.Post(App.url, "text/plain", bytes.NewBuffer(buf))
if err != nil {
response.AddHeader("Content-Type", "application/json")
response.WriteErrorString(http.StatusInternalServerError, err.Error())
return
}
defer r.Body.Close()
body, err := ioutil.ReadAll(r.Body)
response.AddHeader("Content-Type", "application/json")
response.WriteHeader(http.StatusCreated)
//	err = xml.Unmarshal(body, &usr)
//	if err != nil {
//		fmt.Printf("error: %v", err)
//		return
//	}
response.Write(body)
//	fmt.Print(&usr.userName)
}

答案1

得分: 13

将XML输入转换为JSON输出的通用答案可能如下所示：

package main

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

type DataFormat struct {
	ProductList []struct {
		Sku      string `xml:"sku" json:"sku"`
		Quantity int    `xml:"quantity" json:"quantity"`
	} `xml:"Product" json:"products"`
}

func main() {
	xmlData := []byte(`<?xml version="1.0" encoding="UTF-8"?>
<ProductList>
    <Product>
        <sku>ABC123</sku>
        <quantity>2</quantity>
    </Product>
    <Product>
        <sku>ABC123</sku>
        <quantity>2</quantity>
    </Product>
</ProductList>`)

	data := &DataFormat{}
	err := xml.Unmarshal(xmlData, data)
	if nil != err {
		fmt.Println("Error unmarshalling from XML", err)
		return
	}

	result, err := json.Marshal(data)
	if nil != err {
		fmt.Println("Error marshalling to JSON", err)
		return
	}

	fmt.Printf("%s\n", result)
}

这段代码使用Go语言将XML数据解析为结构体，并将其转换为JSON格式。你可以将XML数据存储在xmlData变量中，然后通过xml.Unmarshal函数将其解析为DataFormat结构体。接下来，使用json.Marshal函数将结构体转换为JSON格式，并将结果打印出来。

package main
import (
"encoding/json"
"encoding/xml"
"fmt"
)
type DataFormat struct {
ProductList []struct {
Sku      string `xml:"sku" json:"sku"`
Quantity int    `xml:"quantity" json:"quantity"`
} `xml:"Product" json:"products"`
}
func main() {
xmlData := []byte(`<?xml version="1.0" encoding="UTF-8" ?>
<ProductList>
<Product>
<sku>ABC123</sku>
<quantity>2</quantity>
</Product>
<Product>
<sku>ABC123</sku>
<quantity>2</quantity>
</Product>
</ProductList>`)
data := &DataFormat{}
err := xml.Unmarshal(xmlData, data)
if nil != err {
fmt.Println("Error unmarshalling from XML", err)
return
}
result, err := json.Marshal(data)
if nil != err {
fmt.Println("Error marshalling to JSON", err)
return
}
fmt.Printf("%s\n", result)
}

答案2

得分: 11

如果您需要将一个具有未知结构的XML文档转换为JSON，您可以使用goxml2json。

示例：

import (
  // 其他导入...
  xj "github.com/basgys/goxml2json"
)

func (u *UserResource) authenticateUser(request *restful.Request, response *restful.Response) {
  // 从restful.Request中提取数据
  xml := strings.NewReader(`<?xml version="1.0" encoding="UTF-8"?><app version="1.0"><request>1111</request></app>`)

  // 转换
  json, err := xj.Convert(xml)
  if err != nil {
    // 出错了...
  }

  // ... 使用JSON ...
}

注意：我是这个库的作者。

import (
// Other imports ...
xj "github.com/basgys/goxml2json"
)
func (u *UserResource) authenticateUser(request *restful.Request, response *restful.Response) {
// Extract data from restful.Request
xml := strings.NewReader(`<?xml version="1.0" encoding="UTF-8"?><app version="1.0"><request>1111</request></app>`)
// Convert
json, err := xj.Convert(xml)
if err != nil {
// Oops...
}
// ... Use JSON ...
}