将JSON与Marshal兼容的映射封送至XML [英] Marshal a JSON Marshal compatible map to XML
问题描述
我有一张地图: [] map [string] string
.
将结果填充到 json.marshal()
兼容对象中.输出:
Populating the results in a json.marshal()
compatible object. Outputing:
[
{
"key1": "val1",
"key2": "val2"
},
{
"randval3": "val1",
"randval2": "xyz1"
"randval1": "xyz3"
},
...
]
但是,当我运行 xml.marshal()
时.我收到 xml:不支持的类型:map [string] string
.考虑到XML需要节点名等事实,这似乎是合理的.因此,我基本上要寻找的是一种获取方法:
However, when I run xml.marshal()
. I receive a xml: unsupported type: map[string]string
. Which seems reasonable given the fact that XML needs node names etc. So what I'm basically looking for is a way to get:
<rootElement>
<child>
<key1>val1</key1>
<key2>val1</key2>
</child>
<child>
<randval3>val1</randval3>
<randval2>xyz1</randval2>
<randval1>xyz1</randval1>
</child>
</rootElement>
但是我一直无法获得与 xml.unmarshal()
But I'm getting stuck with getting an 'object' compatible with xml.unmarshal()
推荐答案
您可以声明一个自定义映射,并使其实现 xml.Marshaler
接口.
You could declare a custom map and have it implement the xml.Marshaler
interface.
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>
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