Golang : Decode XML data from RSS feed

Was looking for a way to decode RSS feed XML data today and here is the code for the tutorial on how to decode XML data from RSS feed with Golang.

Enjoy!

 package main

 import (
 "bytes"
 "encoding/xml"
 "fmt"
 "io/ioutil"
 "net/http"
 "os"
 )

 type Rss struct {
 Channel Channel `xml:"channel"`
 }

 type Item struct {
 Title string `xml:"title"`
 Link string `xml:"link"`
 Description string `xml:"description"`
 }

 type Channel struct {
 Title string `xml:"title"`
 Link string `xml:"link"`
 Description string `xml:"description"`
 Items []Item `xml:"item"`
 }

 func main() {

 response, err := http.Get("http://www.thestar.com.my/RSS/Metro/Community/")

 if err != nil {
 fmt.Println(err)
 os.Exit(1)
 }

 defer response.Body.Close()

 XMLdata, err := ioutil.ReadAll(response.Body)

 if err != nil {
 fmt.Println(err)
 os.Exit(1)
 }

 rss := new(Rss)

 buffer := bytes.NewBuffer(XMLdata)

 decoded := xml.NewDecoder(buffer)

 err = decoded.Decode(rss)

 if err != nil {
 fmt.Println(err)
 os.Exit(1)
 }

 fmt.Printf("Title : %s\n", rss.Channel.Title)
 fmt.Printf("Description : %s\n", rss.Channel.Description)
 fmt.Printf("Link : %s\n", rss.Channel.Link)

 total := len(rss.Channel.Items)

 fmt.Printf("Total items : %v\n", total)

 for i := 0; i < total; i++ {
 fmt.Printf("[%d] item title : %s\n", i, rss.Channel.Items[i].Title)
 fmt.Printf("[%d] item description : %s\n", i, rss.Channel.Items[i].Description)
 fmt.Printf("[%d] item link : %s\n\n", i, rss.Channel.Items[i].Link)
 }

 }