go中map与xml互转

淡淡的烟草味﹌ 2022-05-24 07:40 338阅读 0赞

go中map与xml不能直接互转，需要自定义map类型并重写MarshalXML()和UnmarshalXML()方法。

stringmap.go

package stringmap
    
    import (
        "encoding/xml"
        "io"
    )
    type StringMap map[string]string
    
    type xmlMapEntry struct {
        XMLName xml.Name
        Value   string `xml:",chardata"`
    }
    
    func (m StringMap) MarshalXML(e *xml.Encoder, start xml.StartElement) error {
        if len(m) == 0 {
            return nil
        }
    
        err := e.EncodeToken(start)
        if err != nil {
            return err
        }
    
        for k, v := range m {
            e.Encode(xmlMapEntry{XMLName: xml.Name{Local: k}, Value: v})
        }
    
        return e.EncodeToken(start.End())
    }
    
    func (m *StringMap) UnmarshalXML(d *xml.Decoder, start xml.StartElement) error {
        *m = StringMap{}
        for {
            var e xmlMapEntry
    
            err := d.Decode(&e)
            if err == io.EOF {
                break
            } else if err != nil {
                return err
            }
    
            (*m)[e.XMLName.Local] = e.Value
        }
        return nil
    }

stringmap\_test.go

package stringmap
    
    import (
        "testing"
        "encoding/xml"
        "fmt"
    )
    
    func TestName(t *testing.T) {
        userMap := make(map[string]string)
        userMap["name"] = "Name"
        userMap["id"] = "1"
    
        buf, _ := xml.Marshal(StringMap(userMap))
        fmt.Println(string(buf))
    
        stringMap := make(map[string]string)
        err := xml.Unmarshal(buf, (*StringMap)(&stringMap))
        if err != nil {
            fmt.Println(err)
        }
        fmt.Println(stringMap)
    }

输出：

<StringMap><name>Name</name><id>1</id></StringMap>
    map[name:Name id:1]

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