如何省略外部XML元素的名称

Question

我需要生成这样的XML文件：

<package>
<dc:contributor id="creator1">John Doe</dc:contributor>
<meta refines="creator1" property="role" scheme="marc:relators">aut</meta>
<meta refines="creator1" property="role" scheme="onix:codelist17">A01</meta>
<meta refines="creator1" property="file-as">Doe, John</meta>
<meta refines="creator1" property="diplay-seq">1</meta>
<dc:contributor id="reviewer1">Mary Poppins</dc:contributor>
<meta refines="reviewer1" property="role" scheme="marc:relators">rev</meta>
<meta refines="reviewer1" property="file-as">Poppins, Mary</meta>
<meta refines="reviewer1" property="diplay-seq">1</meta>
<dc:contributor id="translator1">Alfred Weasley</dc:contributor>
<meta refines="translator1" property="role" scheme="marc:relators">trl</meta>
<meta refines="translator1" property="role" scheme="onix:codelist17">B06</meta>
<meta refines="translator1" property="file-as">Weasley, Alfred</meta>
<meta refines="translator1" property="diplay-seq">1</meta>
</package>

这里的订单很重要。

我使用这种结构：

type Pack struct {
XMLName struct{}        `xml:"package"`
People  []PersonWrapper `xml:""` // <-- is this the problem?
}
type Person struct {
Id       string `xml:"id,attr"`
Name     string `xml:"-"`
Surname  string `xml:"-"`
FullName string `xml:",innerxml"`
}
type Meta struct {
Refines  string `xml:"refines,attr"`
Property string `xml:"property,attr"`
Scheme   string `xml:"scheme,attr,omitempty"`
Value    string `xml:",innerxml"`
}
type PersonWrapper struct {
Person Person `xml:"dc:contributor"`
Metas  []Meta `xml:"meta"`
}

有了这个，我得到了：

<package>
<People>
<dc:contributor id="creator1">John Doe</dc:contributor>
<meta refines="creator1" property="role" scheme="marc:relators">aut</meta>
<meta refines="creator1" property="role" scheme="onix:codelist17">A01</meta>
<meta refines="creator1" property="file-as">Doe, John</meta>
<meta refines="creator1" property="diplay-seq">1</meta>
</People>
<People>
<dc:contributor id="reviewer1">Mary Poppins</dc:contributor>
<meta refines="reviewer1" property="role" scheme="marc:relators">rev</meta>
<meta refines="reviewer1" property="file-as">Poppins, Mary</meta>
<meta refines="reviewer1" property="diplay-seq">1</meta>
</People>
<People>
<dc:contributor id="translator1">Ben Weasley</dc:contributor>
<meta refines="translator1" property="role" scheme="marc:relators">trl</meta>
<meta refines="translator1" property="role" scheme="onix:codelist17">B06</meta>
<meta refines="translator1" property="file-as">Weasley, Ben</meta>
<meta refines="translator1" property="diplay-seq">1</meta>
</People>
</package>

如何去除<People/>外部标签？我试图将内部结构封装到更高级别的其他结构中，但无法达到所需的结构。

Answer 1

您可以为PersonWrapper实现xml.Marshaler接口(并将xmlname添加到嵌套结构中(。

type Pack struct {
XMLName struct{} `xml:"package"`
People  []PersonWrapper
}
type Person struct {
XMLName  xml.Name `xml:"dc:contributor"`
Id       string   `xml:"id,attr"`
Name     string   `xml:"-"`
Surname  string   `xml:"-"`
FullName string   `xml:",innerxml"`
}
type Meta struct {
XMLName  xml.Name `xml:"meta"`
Refines  string   `xml:"refines,attr"`
Property string   `xml:"property,attr"`
Scheme   string   `xml:"scheme,attr,omitempty"`
Value    string   `xml:",innerxml"`
}
type PersonWrapper struct {
Person Person `xml:"dc:contributor"`
Metas  []Meta `xml:"meta"`
}
func (pw PersonWrapper) MarshalXML(e *xml.Encoder, start xml.StartElement) error {
if err := e.Encode(pw.Person); err != nil {
return err
}
if err := e.Encode(pw.Metas); err != nil {
return err
}
return nil
}

https://play.golang.org/p/hzrBdHXHwOB