如何从在线 XML 文件中解析 XML 数据

huangapple
go评论120阅读模式
英文:

How to unmarshal xml data from online xml file

问题

我有一个假设的 XML 文件，位于 https://www.notre-shop.com/sitemap_products_1.xml，我想在我的 Go 代码中对这个 XML 进行解析，所以我做了以下操作：

  1. package main
  3. import (
  4.     "encoding/xml"
  5.     "fmt"
  6.     "io/ioutil"
  7.     "log"
  8.     "net/http"
  9. )
  11. var Product struct {
  12.     Locs []string `xml:"url>loc"`
  13.     Name []string `xml:"url>image>title"`
  14. }
  16. func main() {
  17.     res, err := http.Get("https://www.notre-shop.com/sitemap_products_1.xml")
  18.     if err != nil {
  19.         log.Fatal(err)
  20.     }
  22.     data, err := ioutil.ReadAll(res.Body)
  23.     if err != nil {
  24.         log.Fatal(err)
  25.     }
  26.     defer res.Body.Close()
  28.     err = xml.Unmarshal(data, &Product)
  29.     if err != nil {
  30.         log.Fatal(err)
  31.     }
  32.     for x := range Product.Name {
  33.         fmt.Println(Product.Name[x], Product.Locs[x])
  34.     }
  35. }

但是这段代码没有输出任何内容。我做错了什么？

这是完整的代码：https://play.golang.org/p/pZ6j4-lSEz。

英文:

I have an xml file on suppose https://www.notre-shop.com/sitemap_products_1.xml and I want to unmarshal this xml in my go code so I did this

  1. package main
  3. import (
  4.     "encoding/xml"
  5.     "fmt"
  6.     "io/ioutil"
  7.     "log"
  8.     "net/http"
  9. )
  11. var Product struct {
  12.     Locs []string `xml:"url>loc"`
  13.     Name []string `xml:"url>image:title"`
  14. }
  16. func main() {
  17.     res, err := http.Get("https://www.notre-shop.com/sitemap_products_1.xml")
  18.     if err!=nil{
  19.         log.Fatal(err)
  20.     }
  22.     data, err := ioutil.ReadAll(res.Body)
  23.     if err!=nil{
  24.         log.Fatal(err)
  25.     }
  26.     defer res.Body.Close()
  28.     err = xml.Unmarshal(data, &Product)
  29.     if err!=nil{
  30.         log.Fatal(err)
  31.     }
  32.     for x, _ := range Product.Name {
  33.         fmt.Println(Product.Name[x], Product.Locs[x])
  34.     }
  35. }

But this doesn't print anything. What am I doing wrong?

Here is the complete code https://play.golang.org/p/pZ6j4-lSEz on play.

答案1

得分: 3

请尝试以下代码，这段代码对我有效（注意：你也可以使用ioutil.ReadAll和xml.Unmarshal，而不是xml.Decode）：

  1. package main
  3. import (
  4. 	"encoding/xml"
  5. 	"fmt"
  6. 	"log"
  7. 	"net/http"
  8. )
  10. type URLSet struct {
  11. 	XMLName string `xml:"urlset"`
  13. 	URLs []URL `xml:"url"`
  14. }
  16. type URL struct {
  17. 	Loc   string `xml:"loc"`
  18. 	Image Image  `xml:"image"`
  19. }
  21. type Image struct {
  22. 	Title string `xml:"title"`
  23. }
  25. func main() {
  26. 	resp, err := http.Get("https://www.notre-shop.com/sitemap_products_1.xml")
  27. 	if err != nil {
  28. 		log.Fatalln(err)
  29. 	}
  30. 	defer resp.Body.Close()
  32. 	var urlSet URLSet
  33. 	if err = xml.NewDecoder(resp.Body).Decode(&urlSet); err != nil {
  34. 		log.Fatalln(err)
  35. 	}
  37. 	for _, url := range urlSet.URLs {
  38. 		fmt.Println(url.Loc, url.Image.Title)
  39. 	}
  40. }

这段代码可以从指定的URL获取XML数据，并解析为结构体。然后，它会遍历解析后的结构体，并打印每个URL和对应的标题。

英文:

Please try the following code which works for me (note: you could also use ioutil.ReadAll and xml.Unmarshal as you had before, instead of xml.Decode):

  1. package main
  3. import (
  4. 	"encoding/xml"
  5. 	"fmt"
  6. 	"log"
  7. 	"net/http"
  8. )
  10. // <urlset xmlns="http://www.sitemaps.org/schemas/sitemap/0.9" xmlns:image="http://www.google.com/schemas/sitemap-image/1.1">
  11. //    <url>
  12. //        <loc>
  13. //            https://www.notre-shop.com/products/test-product-releasing-soon-2
  14. //        </loc>
  15. //        <lastmod>2017-01-17T08:04:44Z</lastmod>
  16. //        <changefreq>daily</changefreq>
  17. //        <image:image>
  18. //            <image:loc>
  19. //                https://cdn.shopify.com/s/files/1/0624/0605/products/NOTRE-CHICAGO-QK9C9548_fde37b05-495e-47b0-8dd1-b053c9ed3545.jpg?v=1481853712
  20. //            </image:loc>
  21. //            <image:title>Test Product Releasing Soon 2</image:title>
  22. //        </image:image>
  23. //    </url>
  24. // </urlset>
  25. type URLSet struct {
  26. 	XMLName string `xml:"urlset"`
  28. 	URLs []URL `xml:"url"`
  29. }
  31. type URL struct {
  32. 	Loc   string `xml:"loc"`
  33. 	Image Image  `xml:"image"`
  34. }
  36. type Image struct {
  37. 	Title string `xml:"title"`
  38. }
  40. func main() {
  41. 	resp, err := http.Get("https://www.notre-shop.com/sitemap_products_1.xml")
  42. 	if err != nil {
  43. 		log.Fatalln(err) // log.Fatal always exits the program, need to check err != nil first
  44. 	}
  45. 	defer resp.Body.Close()
  47. 	var urlSet URLSet
  48. 	if err = xml.NewDecoder(resp.Body).Decode(&urlSet); err != nil {
  49. 		log.Fatalln(err)
  50. 	}
  52. 	for _, url := range urlSet.URLs {
  53. 		fmt.Println(url.Loc, url.Image.Title)
  54. 	}
  55. }

答案2

得分: 0

这是XML规范中的内容：

XML规范中的"Namespaces in XML Recommendation"（XML命名空间推荐）为包含冒号字符的名称赋予了特殊含义。因此，作者在XML名称中除了用于命名空间目的之外，不应该使用冒号。但是XML处理器必须接受冒号作为名称字符。

这是XML命名空间推荐的内容：

XML命名空间推荐以与XML 1.0兼容的间接方式表示通用名称。实际上，XML命名空间推荐定义了一种从XML 1.0树到可以是通用名称的树的映射。该映射基于前缀的概念。如果元素类型名称或属性名称包含冒号，则映射将名称中冒号之前的部分视为前缀，将名称中冒号之后的部分视为本地名称。前缀"foo"引用xmlns:foo属性值中指定的URI。

你不能使用带有冒号（:）的名称来获取内部元素，而是可以避免使用前缀，这是重写后的代码：

对于性能/内存方面的考虑，由于你得到的是io.Reader，你可以使用xml.Decoder而不是xml.Unmarshal。

  1. package main
  3. import (
  4. 	"encoding/xml"
  5. 	"fmt"
  6. 	"log"
  7. 	"net/http"
  8. )
  10. var Product struct {
  11. 	Locs []string `xml:"url>loc"`
  12. 	Name []Image  `xml:"url>image"`
  13. }
  15. type Image struct {
  16. 	Title string `xml:"title"`
  17. }
  19. func main() {
  20. 	res, err := http.Get("https://www.notre-shop.com/sitemap_products_1.xml")
  21. 	if err != nil {
  22. 		log.Fatal(err)
  23. 	}
  24. 	defer res.Body.Close()
  25. 	decoder := xml.NewDecoder(res.Body)
  26. 	err = decoder.Decode(&Product)
  27. 	if err != nil {
  28. 		log.Fatal(err)
  29. 	}
  30. 	for x, _ := range Product.Name {
  31. 		fmt.Println(Product.Name[x].Title, Product.Locs[x])
  32. 	}
  33. }

这是play链接：play

英文:

This is what the XML Specification says

> The Namespaces in XML Recommendation [XML Names] assigns a meaning
> to names containing colon characters. Therefore, authors should not
> use the colon in XML names except for namespace purposes, but XML
> processors must accept the colon as a name character.

This is the XML Namespace Recommendation says

> The XML Namespaces Recommendation expresses universal names in an
> indirect way that is compatible with XML 1.0. In effect the XML
> Namespaces Recommendation defines a mapping from an XML 1.0 tree where
> element type names and attribute names are local names into a tree
> where element type names and attribute names can be universal names.
> The mapping is based on the idea of a prefix. If an element type name
> or attribute name contains a colon, then the mapping treats the part
> of the name before the colon as a prefix, and the part of the name
> after the colon as the local name. A prefix foo refers to the URI
> specified in the value of the xmlns:foo attribute.

You may not use name with colon(:) to get the inner elements rather you can avoid the pefix ,here is your code re-written

And for performance/memory considerations since you are getting a io.Reader you may use xml.Decoder instead of xml.Unmarshal.

  1. package main
  3. import (
  4. 	"encoding/xml"
  5. 	"fmt"
  6. 	"log"
  7. 	"net/http"
  8. )
  10. var Product struct {
  11. 	Locs []string `xml:"url>loc"`
  12. 	Name []Image  `xml:"url>image"`
  13. }
  15. type Image struct {
  16. 	Title string `xml:"title"`
  17. }
  19. func main() {
  20. 	res, err := http.Get("https://www.notre-shop.com/sitemap_products_1.xml")
  21. 	if err != nil {
  22. 		log.Fatal(err)
  23. 	}
  24. 	defer res.Body.Close()
  25. 	decoder := xml.NewDecoder(res.Body)
  26. 	err = decoder.Decode(&Product)
  27. 	if err != nil {
  28. 		log.Fatal(err)
  29. 	}
  30. 	for x, _ := range Product.Name {
  31. 		fmt.Println(Product.Name[x].Title, Product.Locs[x])
  32. 	}
  33. }

Here is play link play

huangapple
  • 本文由 发表于 2017年1月24日 14:05:37
  • 转载请务必保留本文链接：https://go.coder-hub.com/41821088.html
  • go
  • xml
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