Golang在XML结构中指定顶级标签

huangapple
go评论105阅读模式
英文:

Golang specify top-level tag in xml structure

问题

所以我正在尝试将一些XML解组或解码为类型（我还不完全清楚它们之间的区别），但似乎无法指定最外层的类型（在这种情况下是<people>）。当我尝试指定此标签时，不会出现错误，但编组值不包含我期望的任何内容。如何指定最外层的标签，为什么第二个赋值没有预期的行为？

  1. package main
  3. import "fmt"
  4. import "encoding/xml"
  5. import "log"
  7. var data string = `
  8. <people>
  9.   <person>
  10.     <id>46</id>
  11.     <name>John Smith</name>
  12.   </person>
  13.   <person>
  14.     <id>3007</id>
  15.     <name>Joe Smith</name>
  16.   </person>
  17. </people>
  18. `
  20. type Person struct {
  21.     Id   int    `xml:"id"`
  22.     Name string `xml:"name"`
  23. }
  25. type People struct {
  26.     PersonList []Person `xml:"person"`
  27. }
  29. type Response struct {
  30.     PeopleItem People `xml:"people"`
  31. }
  33. func main() {
  34.     // 解析People
  35.     // 无法指定最外层标签<people></people>
  36.     var people People
  37.     err := xml.Unmarshal([]byte(data), &people)
  39.     if err != nil {
  40.         log.Fatal(err)
  41.     }
  43.     fmt.Println(people)
  44.     // 输出"{[{46 John Smith} {3007 Joe Smith}]}"
  45.     // 这是合理的
  47.     // 尝试解析整个响应，得到的是一个空切片的结构体的结构体
  48.     var response Response
  49.     err = xml.Unmarshal([]byte(data), &response)
  50.     if err != nil {
  51.         log.Fatal(err)
  52.     }
  54.     // response是一个空数组的结构体的结构体
  55.     // 为什么会这样？
  56.     fmt.Println(response)
  57.     // 为什么会输出"{{[]}}"？
  58. }

以上是你提供的代码的翻译。

英文:

So I am trying to Unmarshal or Decode some xml into a type (I am still not entirely clear on what the difference is), and I don't seem to be able to specify the outermost type (in this case &lt;people&gt;). When I try to specify this tag, instead of getting an error, the Marshalled value does not contain any of the content I am expecting. How do you specify the outermost tag and why doesn't the second assignment have the expected behavior?

  1. package main
  3. import &quot;fmt&quot;
  4. import &quot;encoding/xml&quot;
  5. import &quot;log&quot;
  7. var data string = `
  8. &lt;people&gt;
  9.   &lt;person&gt;
  10.     &lt;id&gt;46&lt;/id&gt;
  11.     &lt;name&gt;John Smith&lt;/name&gt;
  12.   &lt;/person&gt;
  13.   &lt;person&gt;
  14.     &lt;id&gt;3007&lt;/id&gt;
  15.     &lt;name&gt;Joe Smith&lt;/name&gt;
  16.   &lt;/person&gt;
  17. &lt;/people&gt;
  18. `
  20. type Person struct {
  21. 	Id int `xml:&quot;id&quot;`
  22. 	Name string `xml:&quot;name&quot;`
  23. }
  25. type People struct {
  26. 	PersonList []Person `xml:&quot;person&quot;`
  27. }
  29. type Response struct {
  30.     PeopleItem People `xml:&quot;people&quot;`
  31. }
  33. func main() {
  34.     // parsing People
  35.     // cannot specify outermost tag &lt;people&gt;&lt;/people&gt;
  36. 	var people People
  37. 	err := xml.Unmarshal([]byte(data), &amp;people)
  39. 	if err != nil {
  40. 		log.Fatal(err)
  41. 	}
  43. 	fmt.Println(people)
  44.     // prints &quot;{[{46 John Smith} {3007 Joe Smith}]}&quot;
  45.     // which is reasonable
  47.     // attempting to parse entire response, yields struct of struct of empty slice
  48.     var response Response
  49.     err = xml.Unmarshal([]byte(data), &amp;response)
  50.     if err != nil {
  51.         log.Fatal(err)
  52.     }
  54.     // response is struct of struct of empty array
  55.     // why does this happen?
  56.     fmt.Println(response)
  57.     // why does this print &quot;{{[]}}&quot; ?
  58. }

答案1

得分: 2

你可以在不修改传入数据的情况下完成这个操作，你可以使用特殊的XMLName xml.Name字段来设置外部标签，然后使用xml:",chardata"来访问其内容。

以下是一个示例：在Go Playground上试一试

  1. package main
  3. import (
  4. 	"encoding/xml"
  5. 	"fmt"
  6. )
  8. type simpleInt struct {
  9. 	XMLName xml.Name `xml:"integer"`
  10. 	Int     int      `xml:",chardata"`
  11. }
  13. type simpleString struct {
  14. 	XMLName xml.Name `xml:"stringtag"`
  15. 	String  string   `xml:",chardata"`
  16. }
  18. type simpleBoolean struct {
  19. 	XMLName xml.Name `xml:"boolean"`
  20. 	Boolean bool     `xml:",chardata"`
  21. }
  23. func main() {
  25. 	bint := []byte("<integer>1138</integer>")
  26. 	bstring := []byte("<stringtag>Caimeo</stringtag>")
  27. 	btrue := []byte("<boolean>true</boolean>")
  28. 	bfalse := []byte("<boolean>false</boolean>")
  30. 	i := simpleInt{}
  31. 	xml.Unmarshal(bint, &i)
  32. 	fmt.Println(i, i.Int)
  34. 	s := simpleString{}
  35. 	xml.Unmarshal(bstring, &s)
  36. 	fmt.Println(s, s.String)
  38. 	m := simpleBoolean{}
  39. 	xml.Unmarshal(btrue, &m)
  40. 	fmt.Println(m, m.Boolean)
  42. 	xml.Unmarshal(bfalse, &m)
  43. 	fmt.Println(m, m.Boolean)
  44. }
英文:

You can do this without hacking the incoming data, you use the special XMLName xml.Name field to set the outside tag and then use xml:&quot;,chardata&quot; to access it's contents.

Here is an example: Try it on the go playground

  1. package main
  3. import (
  4. 	&quot;encoding/xml&quot;
  5. 	&quot;fmt&quot;
  6. )
  8. type simpleInt struct {
  9. 	XMLName xml.Name `xml:&quot;integer&quot;`
  10. 	Int     int      `xml:&quot;,chardata&quot;`
  11. }
  13. type simpleString struct {
  14. 	XMLName xml.Name `xml:&quot;stringtag&quot;`
  15. 	String  string   `xml:&quot;,chardata&quot;`
  16. }
  18. type simpleBoolean struct {
  19. 	XMLName xml.Name `xml:&quot;boolean&quot;`
  20. 	Boolean bool     `xml:&quot;,chardata&quot;`
  21. }
  23. func main() {
  25. 	bint := []byte(&quot;&lt;integer&gt;1138&lt;/integer&gt;&quot;)
  26. 	bstring := []byte(&quot;&lt;stringtag&gt;Caimeo&lt;/stringtag&gt;&quot;)
  27. 	btrue := []byte(&quot;&lt;boolean&gt;true&lt;/boolean&gt;&quot;)
  28. 	bfalse := []byte(&quot;&lt;boolean&gt;false&lt;/boolean&gt;&quot;)
  30. 	i := simpleInt{}
  31. 	xml.Unmarshal(bint, &amp;i)
  32. 	fmt.Println(i, i.Int)
  34. 	s := simpleString{}
  35. 	xml.Unmarshal(bstring, &amp;s)
  36. 	fmt.Println(s, s.String)
  38. 	m := simpleBoolean{}
  39. 	xml.Unmarshal(btrue, &amp;m)
  40. 	fmt.Println(m, m.Boolean)
  42. 	xml.Unmarshal(bfalse, &amp;m)
  43. 	fmt.Println(m, m.Boolean)
  44. }

答案2

得分: 1

这里有一个简单的解决方案：添加一个新的独特的顶级元素：

  1. var response Response
  2. err = xml.Unmarshal([]byte("<foo>"+data+"</foo>"), &response)
英文:

Here's a lazy solution: add a new, unique top level element:

  1. var response Response
  2. err = xml.Unmarshal([]byte(&quot;&lt;foo&gt;&quot;+data+&quot;&lt;/foo&gt;&quot;), &amp;response)

huangapple
  • 本文由 发表于 2015年10月29日 08:05:39
  • 转载请务必保留本文链接：https://go.coder-hub.com/33403746.html
  • go
  • xml
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