Unmarshalling XML using Go: How to find attributes with the same value?

huangapple
go评论186阅读模式
英文:

Unmarshalling XML using Go: How to find attributes with the same value?

问题

我正在尝试解析下面的XML，如何找到所有<info>节点中type="Genres"的节点，并将它们的值存储在[]Genre中？

  1. <manga id="4199" type="manga" name="Jinki: Extend" precision="manga">
  2.     <info type="Main title" lang="EN">Jinki: Extend</info>
  3.     <info type="Genres">action</info>
  4.     <info type="Genres">science fiction</info>
  5.     <info type="Themes">mecha</info>
  6.     <info type="Number of tankoubon">9</info>
  7.     <info type="Number of pages">186</info>
  8. </manga>

我希望将这些值存储在类似以下结构的结构体中：

  1. // Manga 结构体
  2. type Manga struct {
  3.     WorkID    int     `xml:"id,attr"`
  4.     Name      string  `xml:"name,attr"`
  5.     Precision string  `xml:"precision,attr"`
  6.     Genres    []Genre // 这部分我需要帮助
  7. }
  9. // Genre 结构体
  10. type Genre struct {
  11.     Value string
  12. }

我知道这个XML不是理想的，但这是我必须处理的，希望你们能帮助我。

提前感谢。

英文:

I'm having trouble unmarshalling the XML below, how do I find all &lt;info&gt; nodes with type=&quot;Genres&quot; and store their values inside a []Genre?

  1. &lt;manga id=&quot;4199&quot; type=&quot;manga&quot; name=&quot;Jinki: Extend&quot; precision=&quot;manga&quot;&gt;
  2.     &lt;info type=&quot;Main title&quot; lang=&quot;EN&quot;&gt;Jinki: Extend&lt;/info&gt;
  3.     &lt;info type=&quot;Genres&quot;&gt;action&lt;/info&gt;
  4.     &lt;info type=&quot;Genres&quot;&gt;science fiction&lt;/info&gt;
  5.     &lt;info type=&quot;Themes&quot;&gt;mecha&lt;/info&gt;
  6.     &lt;info type=&quot;Number of tankoubon&quot;&gt;9&lt;/info&gt;
  7.     &lt;info type=&quot;Number of pages&quot;&gt;186&lt;/info&gt;
  8. &lt;/manga&gt;

I'm looking to store the values in structs similar to these:

  1. // Manga struct
  2. type Manga struct {
  3.     WorkID    int     `xml:&quot;id,attr&quot;`
  4.     Name      string  `xml:&quot;name,attr&quot;`
  5.     Precision string  `xml:&quot;precision,attr&quot;`
  6.     Genres    []Genre `[this is the part I need help on]`
  7. }
  9. // Genre struct
  10. type Genre struct {
  11.     Value string
  12. }

I know the XML is not ideal, but it is what I have to work with, I hope you guys can help me with this.

Thanks in advance.

答案1

得分: 3

由于<manga>包含了一系列的<info>元素，因此更合理的做法是使用Info结构体的列表，而不是尝试将<info>元素翻译成各种类型。我会定义如下的数据结构：

  1. type Manga struct {
  2.     WorkID    int    `xml:"id,attr"`
  3.     Name      string `xml:"name,attr"`
  4.     Precision string `xml:"precision,attr"`
  5.     Info      []Info `xml:"info"`
  6. }
  8. type Info struct {
  9.     Type  string `xml:"type,attr"`
  10.     Value string `xml:",chardata"`
  11. }

输出结果（以JSON编码方便查看）如下：

  1. {
  2.   "WorkID": 4199,
  3.   "Name": "Jinki: Extend",
  4.   "Precision": "manga",
  5.   "Info": [
  6.     {
  7.       "Type": "Main title",
  8.       "Value": "Jinki: Extend"
  9.     },
  10.     {
  11.       "Type": "Genres",
  12.       "Value": "action"
  13.     },
  14.     {
  15.       "Type": "Genres",
  16.       "Value": "science fiction"
  17.     },
  18.     {
  19.       "Type": "Themes",
  20.       "Value": "mecha"
  21.     },
  22.     {
  23.       "Type": "Number of tankoubon",
  24.       "Value": "9"
  25.     },
  26.     {
  27.       "Type": "Number of pages",
  28.       "Value": "186"
  29.     }
  30.   ]
  31. }
英文:

Since &lt;manga&gt; contains a list of &lt;info&gt; elements, it makes more sense to have a list of Info structs rather than trying to translate the &lt;info&gt; elements into various types. I would define the data structures like:

  1. type Manga struct {
  2. 	WorkID    int    `xml:&quot;id,attr&quot;`
  3. 	Name      string `xml:&quot;name,attr&quot;`
  4. 	Precision string `xml:&quot;precision,attr&quot;`
  5. 	Info      []Info `xml:&quot;info&quot;`
  6. }
  8. type Info struct {
  9. 	Type  string `xml:&quot;type,attr&quot;`
  10. 	Value string `xml:&quot;,chardata&quot;`
  11. }

The output (json encoded for convenience) looks like:

  1. {
  2.   &quot;WorkID&quot;: 4199,
  3.   &quot;Name&quot;: &quot;Jinki: Extend&quot;,
  4.   &quot;Precision&quot;: &quot;manga&quot;,
  5.   &quot;Info&quot;: [
  6.     {
  7.       &quot;Type&quot;: &quot;Main title&quot;,
  8.       &quot;Value&quot;: &quot;Jinki: Extend&quot;
  9.     },
  10.     {
  11.       &quot;Type&quot;: &quot;Genres&quot;,
  12.       &quot;Value&quot;: &quot;action&quot;
  13.     },
  14.     {
  15.       &quot;Type&quot;: &quot;Genres&quot;,
  16.       &quot;Value&quot;: &quot;science fiction&quot;
  17.     },
  18.     {
  19.       &quot;Type&quot;: &quot;Themes&quot;,
  20.       &quot;Value&quot;: &quot;mecha&quot;
  21.     },
  22.     {
  23.       &quot;Type&quot;: &quot;Number of tankoubon&quot;,
  24.       &quot;Value&quot;: &quot;9&quot;
  25.     },
  26.     {
  27.       &quot;Type&quot;: &quot;Number of pages&quot;,
  28.       &quot;Value&quot;: &quot;186&quot;
  29.     }
  30.   ]
  31. }

答案2

得分: 0

如果XML文件不是很大，我会写一个实用函数，类似于：

  1. type Manga struct {
  2.     WorkID    int    `xml:"id,attr"`
  3.     Name      string `xml:"name,attr"`
  4.     Precision string `xml:"precision,attr"`
  5.     Info      []Info `xml:"info"`
  6. }
  8. type Info struct {
  9.     Type string `xml:"type,attr"`
  10.     Data string `xml:",chardata"`
  11. }
  13. func (m *Manga) Genres() []Info {
  14.     var g []Info
  16.     for _, v := range m.Info {
  17.         if v.Type == "Genres" {
  18.             g = append(g, v)
  19.         }
  20.     }
  21.     return g
  22. }

在这里查看它的运行效果：https://play.golang.org/p/bebUwwbSwG

英文:

In case the XML won't not very large, I would just write an utility function like:

  1. type Manga struct {
  2. 	WorkID    int    `xml:&quot;id,attr&quot;`
  3. 	Name      string `xml:&quot;name,attr&quot;`
  4. 	Precision string `xml:&quot;precision,attr&quot;`
  5. 	Info      []Info `xml:&quot;info&quot;`
  6. }
  8. type Info struct {
  9. 	Type string `xml:&quot;type,attr&quot;`
  10. 	Data string `xml:&quot;,chardata&quot;`
  11. }
  13. func (m *Manga) Genres() []Info {
  14. 	var g []Info
  16. 	for _, v := range m.Info {
  17. 		if v.Type == &quot;Genres&quot; {
  18. 			g = append(g, v)
  19. 		}
  20. 	}
  21. 	return g
  23. }

See it in action: https://play.golang.org/p/bebUwwbSwG

huangapple
  • 本文由 发表于 2016年8月5日 22:24:02
  • 转载请务必保留本文链接：https://go.coder-hub.com/38791598.html
  • go
  • xml
匿名

发表评论

匿名网友

:?: :razz: :sad: :evil: :!: :smile: :oops: :grin: :eek: :shock: :???: :cool: :lol: :mad: :twisted: :roll: :wink: :idea: :arrow: :neutral: :cry: :mrgreen:

确定