我有一张 map :[]map[string]string
.
将结果填充到 json.marshal()
兼容的对象。输出:
[
{
"key1": "val1",
"key2": "val2"
},
{
"randval3": "val1",
"randval2": "xyz1"
"randval1": "xyz3"
},
...
]
但是,当我运行
xml.marshal()
.我收到了 xml: unsupported type: map[string]string
.考虑到 XML 需要节点名称等这一事实,这似乎是合理的。所以我基本上要寻找的是一种获得方法:<rootElement>
<child>
<key1>val1</key1>
<key2>val1</key2>
</child>
<child>
<randval3>val1</randval3>
<randval2>xyz1</randval2>
<randval1>xyz1</randval1>
</child>
</rootElement>
但我无法获得与
xml.unmarshal()
兼容的“对象”。
最佳答案
您可以声明一个自定义映射并让它实现 xml.Marshaler
界面。
type mymap map[string]string
func (m mymap) MarshalXML(e *xml.Encoder, start xml.StartElement) error {
if err := e.EncodeToken(start); err != nil {
return err
}
for key, val := range m {
s := xml.StartElement{Name: xml.Name{Local: key}}
if err := e.EncodeElement(val, s); err != nil {
return err
}
}
return e.EncodeToken(start.End())
}
type RootElement struct {
XMLName xml.Name `xml:"rootElement"`
Children []mymap `xml:"child"`
}
https://play.golang.com/p/0_qA9UUvhKV
func main() {
root := RootElement{Children: []mymap{
{"key1": "val1", "key2": "val2"},
{"randval1": "val1", "randval2": "xyz1", "randval3": "abc3"},
}}
data, err := xml.MarshalIndent(root, "", " ")
if err != nil {
panic(err)
}
fmt.Println(string(data))
}
输出:
<rootElement>
<child>
<key2>val2</key2>
<key1>val1</key1>
</child>
<child>
<randval3>abc3</randval3>
<randval1>val1</randval1>
<randval2>xyz1</randval2>
</child>
</rootElement>
关于arrays - Marshal 一个 JSON Marshal 兼容映射到 XML,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/61266933/