将单个字段分别编组为两个标签在golang中

huangapple
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英文:

Marshal single field into two tags in golang

问题

尝试理解一种为结构化为以下形式的XML创建自定义编组器的方法：

  1. <!-- language: lang-xml -->
  2.     <Appointment>
  3.     <Date>2004-12-22</Date>
  4.     <Time>14:00</Time>
  5.     </Appointment>

我在考虑这样的解决方案：

  1. <!-- language: lang-go -->
  2.     type Appointment struct {
  3.         DateTime time.Time `xml:"???"`
  4.     }

问题是，我应该在"???"的位置放什么，才能将一个字段保存到两个不同的XML标签中？

英文:

Trying to understand a way to create custom marshaller for xml structured as:

<!-- language: lang-xml -->

  1. &lt;Appointment&gt;
  2. &lt;Date&gt;2004-12-22&lt;/Date&gt;
  3. &lt;Time&gt;14:00&lt;/Time&gt;
  4. &lt;/Appointment&gt;

I'm thinking of something like:

<!-- language: lang-go -->

  1. type Appointment struct {
  2.     DateTime time.Time `xml:&quot;???&quot;`
  3. }

Question is, what would i put instead of ??? to have a single field be saved into two different xml tags?

答案1

得分: 3

复杂的编组/解组行为通常需要满足Marshal/Unmarshal接口（这对于XML、JSON和类似设置的Go类型都是如此）。

您需要使用MarshalXML()函数满足xml.Marshaler接口，类似以下示例：

  1. package main
  3. import (
  4.     "encoding/xml"
  5.     "fmt"
  6.     "time"
  7. )
  9. type Appointment struct {
  10.     DateTime time.Time
  11. }
  13. type appointmentExport struct {
  14.     XMLName struct{} `xml:"appointment"`
  15.     Date    string   `xml:"date"`
  16.     Time    string   `xml:"time"`
  17. }
  19. func (a *Appointment) MarshalXML(e *xml.Encoder, start xml.StartElement) error {
  20.     n := &appointmentExport{
  21.         Date: a.DateTime.Format("2006-01-02"),
  22.         Time: a.DateTime.Format("15:04"),
  23.     }
  24.     return e.Encode(n)
  25. }
  27. func main() {
  28.     a := &Appointment{time.Now()}
  29.     output, _ := xml.MarshalIndent(a, "", "    ")
  30.     fmt.Println(string(output))
  31. }
  33. // 输出：
  34. // <appointment>
  35. //     <date>2016-04-15</date>
  36. //     <time>17:43</time>
  37. // </appointment>

以上代码将输出XML格式的字符串。

英文:

Complex marshaling/unmarshaling behavior generally requires satisfying a Marshal/Unmarshal interface (this is of true of XML, JSON, and similarly setup types in go).

You'd need to satisfy the xml.Marshaler interface with a MarshalXML() function, something like the following:

  1. package main
  3. import (
  4.     &quot;encoding/xml&quot;
  5.     &quot;fmt&quot;
  6.     &quot;time&quot;
  7. )
  9. type Appointment struct {
  10.     DateTime time.Time
  11. }
  13. type appointmentExport struct {
  14.     XMLName struct{} `xml:&quot;appointment&quot;`
  15.     Date    string   `xml:&quot;date&quot;`
  16.     Time    string   `xml:&quot;time&quot;`
  17. }
  19. func (a *Appointment) MarshalXML(e *xml.Encoder, start xml.StartElement) error {
  20.     n := &amp;appointmentExport{
  21.         Date: a.DateTime.Format(&quot;2006-01-02&quot;),
  22.         Time: a.DateTime.Format(&quot;15:04&quot;),
  23.     }
  24.     return e.Encode(n)
  25. }
  27. func main() {
  28.     a := &amp;Appointment{time.Now()}
  29.     output, _ := xml.MarshalIndent(a, &quot;&quot;, &quot;    &quot;)
  30.     fmt.Println(string(output))
  31. }
  33. // prints:
  34. // &lt;appointment&gt;
  35. //     &lt;date&gt;2016-04-15&lt;/date&gt;
  36. //     &lt;time&gt;17:43&lt;/time&gt;
  37. // &lt;/appointment&gt;

答案2

得分: 0

快速猜测，你不能。你应该使用你的Appointment类型实现xml.Marshaler接口...

英文:

Quick guess, you can't. You should implement the xml.Marshaler interface with your Appointment type…

huangapple
  • 本文由 发表于 2016年4月13日 20:59:22
  • 转载请务必保留本文链接：https://go.coder-hub.com/36599180.html
  • go
  • marshalling
  • xml
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