私はXMLからJSONへの変換について多くの記事を見てきましたが、最近これを行うプログラムを書いていますが、JSONからXMLに変換する方法についても興味がありました?JSONからXMLへの変換
サンプルJSON:
"version":"0.1",
"termsofService":"http://www.wunderground.com/weather/api/d/terms.html",
"features": {
"conditions": 1
}
}
, "current_observation": {
"image": {
"url":"http://icons.wxug.com/graphics/wu2/logo_130x80.png",
"title":"Weather Underground",
"link":"http://www.wunderground.com"
},
"display_location": {
"full":"Kearney, MO",
"city":"Kearney",
"state":"MO",
"state_name":"Missouri",
私はそれはあなたにどんな使用するだろうかどうかわからないんだけど、私はXMLプログラムに私のJSONを投稿します。
package main
import (
"fmt"
"net/url"
"encoding/xml"
"net/http"
"log"
"io/ioutil"
"encoding/json"
)
type reportType struct{
Version xml.CharData `xml:"version"`
TermsOfService xml.CharData `xml:"termsofService"
`
Features xml.CharData `xml:"features>feature"`
Full string `xml:"current_observation>display_location>full"`
StateName string `xml:"current_observation>display_location>state_name"`
WindGust string `xml:"current_observation>observation_location>full"`
Problem myErrorType `xml:"error"`
}
type myErrorType struct{
TypeOfError xml.CharData `xml:"type"`
Desciption xml.CharData `xml:"description"`
}
type reportTypeJson struct{
Version string `json:"version"`;
TermsOfService string `json:"termsofService"`;
Features map[string]string `json:"features"`;
CurrentObservation map[string]map[string]string `json:"current_observation"`
}
func main() {
fmt.Println("data is from WeatherUnderground.")
fmt.Println("https://www.wunderground.com/")
var state, city string
str1 := "What is your state?"
str2 := "What is your city?"
fmt.Println(str1)
fmt.Scanf("%s", &state)
fmt.Println(str2)
fmt.Scanf("%s", &city)
baseURL := "http://api.wunderground.com/api/";
apiKey := "Nunna"
var query string
//set up the query
query = baseURL+apiKey +
"/conditions/q/"+
url.QueryEscape(state)+ "/"+
url.QueryEscape(city)+ ".xml"
fmt.Println("The escaped query: "+query)
response, err := http.Get(query)
doErr(err, "After the GET")
var body []byte
body, err = ioutil.ReadAll(response.Body)
doErr(err, "After Readall")
fmt.Println(body);
fmt.Printf("The body: %s\n",body)
//Unmarshalling
var report reportType
xml.Unmarshal(body, &report)
fmt.Printf("The Report: %s\n", report)
fmt.Printf("The description is [%s]\n",report.Problem.Desciption)
//Now marshal the data out in JSON
var data []byte
var output reportTypeJson
output.Version = string(report.Version);
output.TermsOfService = string(report.TermsOfService)
output.Features= map[string]string{"feature":string(report.Features)} // allocate a map, add the 'features' value to it and assign it to output.Features
output.CurrentObservation = map[string]map[string]string {
"display_location": map[string]string {
"full": report.Full,
"state_name": report.StateName,
},"observation_location":map[string]string {"full": report.WindGust},
}
data,err = json.MarshalIndent(output,""," ")
doErr(err, "From marshalIndent")
fmt.Printf("JSON output nicely formatted: \n%s\n",data)
}
func doErr(err error, message string){
if err != nil{
log.Panicf("ERROR: %s %s \n", message, err.Error())
}
}
//OUTPUT:
//JSON output nicely formatted:
//{
// "version": "0.1",
// "termsofService": "http://www.wunderground.com/weather/api/d/terms.html",
// "features": {
// "feature": "conditions"
// },
// "current_observation": {
// "display_location": {
// "full": "Kearney, MO",
// "state_name": "Missouri"
// },
// "observation_location": {
// "full": "HOMESTEAD HILLS, Kearney, Missouri"
// }
// }
//}
仕様をコードよりもはるかに興味深いものにするでしょう。 JSONをある種のXMLに変換するのは簡単です。 「通常の」ケースで使用できない結果を生成することなく、エッジケース(XMLで有効でない文字を含むJSONなど)を処理できる変換を設計することが課題です。 –