英文:
Parsing RSS feed in Go
问题
我正在尝试用Go编写一个播客下载器。以下代码解析了一个RSS源，但在将解析的数据打印到标准输出时，频道的链接为空。我不知道为什么会这样。有什么建议吗？我对Go还不熟悉。
package main
import (
"encoding/xml"
"fmt"
"net/http"
)
type Enclosure struct {
Url string `xml:"url,attr"`
Length int64 `xml:"length,attr"`
Type string `xml:"type,attr"`
}
type Item struct {
Title string `xml:"title"`
Link string `xml:"link"`
Desc string `xml:"description"`
Guid string `xml:"guid"`
Enclosure Enclosure `xml:"enclosure"`
PubDate string `xml:"pubDate"`
}
type Channel struct {
Title string `xml:"title"`
Link string `xml:"link"`
Desc string `xml:"description"`
Items []Item `xml:"item"`
}
type Rss struct {
Channel Channel `xml:"channel"`
}
func main() {
resp, err := http.Get("http://www.bbc.co.uk/programmes/p02nrvz8/episodes/downloads.rss")
if err != nil {
fmt.Printf("Error GET: %v\n", err)
return
}
defer resp.Body.Close()
rss := Rss{}
decoder := xml.NewDecoder(resp.Body)
err = decoder.Decode(&rss)
if err != nil {
fmt.Printf("Error Decode: %v\n", err)
return
}
fmt.Printf("Channel title: %v\n", rss.Channel.Title)
fmt.Printf("Channel link: %v\n", rss.Channel.Link)
for i, item := range rss.Channel.Items {
fmt.Printf("%v. item title: %v\n", i, item.Title)
}
}
英文:
I am trying to write a podcast downloader in Go. The following code parses an RSS feed but the link of the channel is empty when printing the parsed data to the standard output. I don't know why. Any suggestions? I am new to Go.
<!-- language: lang-go -->
package main
import (
"encoding/xml"
"fmt"
"net/http"
)
type Enclosure struct {
Url string `xml:"url,attr"`
Length int64 `xml:"length,attr"`
Type string `xml:"type,attr"`
}
type Item struct {
Title string `xml:"title"`
Link string `xml:"link"`
Desc string `xml:"description"`
Guid string `xml:"guid"`
Enclosure Enclosure `xml:"enclosure"`
PubDate string `xml:"pubDate"`
}
type Channel struct {
Title string `xml:"title"`
Link string `xml:"link"`
Desc string `xml:"description"`
Items []Item `xml:"item"`
}
type Rss struct {
Channel Channel `xml:"channel"`
}
func main() {
resp, err := http.Get("http://www.bbc.co.uk/programmes/p02nrvz8/episodes/downloads.rss")
if err != nil {
fmt.Printf("Error GET: %v\n", err)
return
}
defer resp.Body.Close()
rss := Rss{}
decoder := xml.NewDecoder(resp.Body)
err = decoder.Decode(&rss)
if err != nil {
fmt.Printf("Error Decode: %v\n", err)
return
}
fmt.Printf("Channel title: %v\n", rss.Channel.Title)
fmt.Printf("Channel link: %v\n", rss.Channel.Link)
for i, item := range rss.Channel.Items {
fmt.Printf("%v. item title: %v\n", i, item.Title)
}
}
答案1
得分: 4
rss源中的xml具有一个包含两个子元素'link'的channel元素：'link'和'atom:link'。即使其中一个具有命名空间前缀，Go xml解组器仍然会发现冲突。请参见local name collisions fail和github上的问题。
<?xml version="1.0" encoding="UTF-8"?>
...
<channel>
<title>Forum - Sixty Second Idea to Improve the World</title>
<link>http://www.bbc.co.uk/programmes/p02nrvz8</link>
...
<atom:link href="http://www.bbc.co.uk/..." />
英文:
The xml from the rss feed has a channel element with two child 'link' elements: 'link' and 'atom:link'. Even though one has a namespace prefix, Go xml unmarshaller sees a conflict. See also local name collisions fail and issue on github.
<?xml version="1.0" encoding="UTF-8"?>
...
<channel>
<title>Forum - Sixty Second Idea to Improve the World</title>
<link>http://www.bbc.co.uk/programmes/p02nrvz8</link>
...
<atom:link href="http://www.bbc.co.uk/..." />
答案2
得分: 0
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