英文:
How to convert []byte XML to JSON output in Golang
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")// POSTfunc (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JSONvar 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[]bytejsonData, 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。
英文:
Is there a way to convert XML ([]byte) to JSON output in Golang?
I've got below function where body is []byte but I want to transform this XML response to JSON after some manipulation. I've tried Unmarshal in xml package with no success:
// POSTfunc (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)}
I'm also using Go-restful package
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 mainimport ("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。
英文:
The generic answer to your question of how to convert XML input to JSON output might be something like this:
http://play.golang.org/p/7HNLEUnX-m
package mainimport ("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。
英文:
If you need to convert an XML document to JSON with an unknown struct, you can use goxml2json.
Example :
import (// Other imports ...xj "github.com/basgys/goxml2json")func (u *UserResource) authenticateUser(request *restful.Request, response *restful.Response) {// Extract data from restful.Requestxml := strings.NewReader(`<?xml version="1.0" encoding="UTF-8"?><app version="1.0"><request>1111</request></app>`)// Convertjson, err := xj.Convert(xml)if err != nil {// Oops...}// ... Use JSON ...}
> Note : I'm the author of this library.
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