在 Go 中解组编号的 XML 标签

我try 了相当多的 struct 来try 对该XML进行解组，但我无法真正弄清楚:

package main

import (
    "bytes"
    "encoding/xml"
    "fmt"
)

type SettingContainer struct {
    XMLName  xml.Name  `xml:"Settings"`
    Settings []Setting `xml:",any"`
}

type Setting struct {
    XMLName          xml.Name          `xml:",any"`
    SettingRegisters []SettingRegister `xml:",any"`
}

type SettingRegister struct {
    XMLName xml.Name `xml:",any"`
    Value   string   `xml:"value,attr"`
}

func main() {
    var xmlData = bytes.NewBufferString(`
    <Settings>
        <Setting0>
            <Setting0_register value="0x20" />
            <Setting0ArbitraryName value="0x0" />
        </Setting0>
        <Setting1>
            <Setting1_register value="0x60" />
            <Setting1WhateverEnable value="0x0" />
        </Setting1>
        <Setting2>
            <Setting2_register value="0x80" />
            <Setting2blahblah value="0x1" />
        </Setting2>
        <Setting4>
            <Setting4_register value="0x2fffff8" />
            <Setting4ThisCanBeAnything value="0x0" />
        </Setting4>
    </Settings>`)

    var s []SettingContainer
    xml.Unmarshal(xmlData.Bytes(), &s)

    for _, i := range s {
        fmt.Println(i)
    }
}

我就是不能让它打印任何东西.我实际需要的字段是SettingN_REGISTER以及SettingN标记名的值attr.我如何才能将XML正确地解组到这些字段中呢？我正在考虑制作一个定制的解组函数，但我真的找不到如何正确地制作一个.我能从这个XML中得到的只有"设置"标记名.