编组 JSON 编组兼容映射到 XML
Marshal a JSON Marshal compatible map to XML
我有一张地图:[]map[string]string
。
在 json.marshal()
兼容对象中填充结果。输出:
[
{
"key1": "val1",
"key2": "val2"
},
{
"randval3": "val1",
"randval2": "xyz1"
"randval1": "xyz3"
},
...
]
但是,当我运行xml.marshal()
。我收到 xml: unsupported type: map[string]string
。鉴于 XML 需要节点名称等事实,这似乎是合理的。所以我基本上正在寻找的是一种获取方式:
<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 'object'
您可以声明一个自定义地图并让它实现 xml.Marshaler
接口。
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>
对于 Marshalling/Unmarshalling 地图,您需要编写自己的 Marshal()
和 Unmarshal()
函数。
下面是为 type Maps []map[string]string
实现这些功能以及如何使用它们的示例。
type Maps []map[string]string
type xmlMapEntry struct {
XMLName xml.Name
Value string `xml:",chardata"`
}
func (m Maps) MarshalXML(e *xml.Encoder, start xml.StartElement) error {
if len(m) == 0 {
return nil
}
err := e.EncodeToken(start)
if err != nil {
return err
}
for _, ma := range m {
for k, v := range ma {
e.Encode(xmlMapEntry{XMLName: xml.Name{Local: k}, Value: v})
}
}
return e.EncodeToken(start.End())
}
func (m *Maps) UnmarshalXML(d *xml.Decoder, start xml.StartElement) error {
*m = Maps{}
for {
var e xmlMapEntry
err := d.Decode(&e)
if err == io.EOF {
break
} else if err != nil {
return err
}
*m = append(*m, map[string]string{e.XMLName.Local: e.Value})
}
return nil
}
func main() {
myarraymap := []map[string]string{
0: {"Key1": "val1"},
1: {"Key2": "val2"},
2: {"Randval3": "val1"},
3: {"Randval2": "xyz1"},
4: {"Randval1": "xyz2"},
}
// Encode to XML
x, _ := xml.MarshalIndent(Maps(myarraymap), "", " ")
fmt.Printf("my marshaled xml map: %v\n", string(x))
// Decode back from XML
var rm []map[string]string
xml.Unmarshal(x, (*Maps)(&rm))
fmt.Printf("my unmarshalled xml map: %v \n", rm)
}
我有一张地图:[]map[string]string
。
在 json.marshal()
兼容对象中填充结果。输出:
[
{
"key1": "val1",
"key2": "val2"
},
{
"randval3": "val1",
"randval2": "xyz1"
"randval1": "xyz3"
},
...
]
但是,当我运行xml.marshal()
。我收到 xml: unsupported type: map[string]string
。鉴于 XML 需要节点名称等事实,这似乎是合理的。所以我基本上正在寻找的是一种获取方式:
<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()
您可以声明一个自定义地图并让它实现 xml.Marshaler
接口。
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>
对于 Marshalling/Unmarshalling 地图,您需要编写自己的 Marshal()
和 Unmarshal()
函数。
下面是为 type Maps []map[string]string
实现这些功能以及如何使用它们的示例。
type Maps []map[string]string
type xmlMapEntry struct {
XMLName xml.Name
Value string `xml:",chardata"`
}
func (m Maps) MarshalXML(e *xml.Encoder, start xml.StartElement) error {
if len(m) == 0 {
return nil
}
err := e.EncodeToken(start)
if err != nil {
return err
}
for _, ma := range m {
for k, v := range ma {
e.Encode(xmlMapEntry{XMLName: xml.Name{Local: k}, Value: v})
}
}
return e.EncodeToken(start.End())
}
func (m *Maps) UnmarshalXML(d *xml.Decoder, start xml.StartElement) error {
*m = Maps{}
for {
var e xmlMapEntry
err := d.Decode(&e)
if err == io.EOF {
break
} else if err != nil {
return err
}
*m = append(*m, map[string]string{e.XMLName.Local: e.Value})
}
return nil
}
func main() {
myarraymap := []map[string]string{
0: {"Key1": "val1"},
1: {"Key2": "val2"},
2: {"Randval3": "val1"},
3: {"Randval2": "xyz1"},
4: {"Randval1": "xyz2"},
}
// Encode to XML
x, _ := xml.MarshalIndent(Maps(myarraymap), "", " ")
fmt.Printf("my marshaled xml map: %v\n", string(x))
// Decode back from XML
var rm []map[string]string
xml.Unmarshal(x, (*Maps)(&rm))
fmt.Printf("my unmarshalled xml map: %v \n", rm)
}