使用 Go 解组 SOAP 消息

我对围棋这门语言还比较陌生.

我在try 解组一条SOAP消息时遇到了问题.我的try 是抽象Body元素的内容，避免静态地定义XML struct ，因为它会根据请求的操作而变化. 不幸的是，我找不到一种方法来正确地做到这一点.在该示例中，GetContent函数应接收指向包含内容的 struct 的指针，并将其动态添加到正文中，以便进行填充.但结果并不是预期的那样.

package main

import (
    "encoding/xml"
    "fmt"
)

type Message interface{}

type EnvelopeResponse struct {
    XMLName xml.Name `xml:"http://schemas.xmlsoap.org/soap/envelope/ Envelope"`
    Body    Message  `xml:"http://schemas.xmlsoap.org/soap/envelope/ Body"`
}

type Body struct {
    XMLName xml.Name `xml:"http://schemas.xmlsoap.org/soap/envelope/ Body"`

    Fault               *Fault  `xml:",omitempty"`
    Content             Message `xml:",omitempty"`
    SOAPBodyContentType string  `xml:"-"`
}

type Fault struct {
    XMLName xml.Name `xml:"http://schemas.xmlsoap.org/soap/envelope/ Fault"`

    Code   string `xml:"faultcode,omitempty"`
    String string `xml:"faultstring,omitempty"`
    Actor  string `xml:"faultactor,omitempty"`
    Detail string `xml:"detail,omitempty"`
}

type GetHostNumberOfEntriesResponse struct {
    XMLName                xml.Name `xml:"urn:dslforum-org:service:Hosts:1 GetHostNumberOfEntriesResponse"`
    NewHostNumberOfEntries int64    `xml:"NewHostNumberOfEntries"`
}

func GetContent(rawXml []byte, content interface{}) {
    envelope := EnvelopeResponse{Body: Body{Content: content}}
    xml.Unmarshal(rawXml, &envelope)
}

func main() {
    b := []byte(`
<?xml version="1.0"?>
<s:Envelope xmlns:s="http://schemas.xmlsoap.org/soap/envelope/" s:encodingStyle="http://schemas.xmlsoap.org/soap/encoding/">
<s:Body>
<u:GetHostNumberOfEntriesResponse xmlns:u="urn:dslforum-org:service:Hosts:1">
<NewHostNumberOfEntries>47</NewHostNumberOfEntries>
</u:GetHostNumberOfEntriesResponse>
</s:Body>
</s:Envelope>
`)
    content := &GetHostNumberOfEntriesResponse{}
    GetContent(b, content)
    fmt.Println(*content)
}

下面是操场上的例子:

https://go.dev/play/p/BBR4vEXiPbc个