如何用冒号解组 XML 属性? [英] How to unmarshal XML attributes with colons?

问题描述

我正在使用的一些 SVG/XML 文件的属性名称中有破折号和冒号 - 例如:

Some SVG/XML files I'm working with have dashes and colons in attribute names - for example:

<g>
  <a xlink:href="http://example.com" data-bind="121">...</a>
</g>

我试图弄清楚如何使用 golang 解组这些属性encoding/xml 包.虽然虚线属性有效，但带有冒号的属性无效:

I'm trying to figure out how to unmarshal these attributes using golang's encoding/xml package. While the dashed attributes works, the ones with the colon doesn't:

package main

import (
    "encoding/xml"
    "fmt"
)

var data = `
<g>
    <a xlink:href="http://example.com" data-bind="121">lala</a>
</g>
`

type Anchor struct {
    DataBind  int    `xml:"data-bind,attr"`  // this works
    XlinkHref string `xml:"xlink:href,attr"` // this fails
}

type Group struct {
    A Anchor `xml:"a"`
}

func main() {
    group := Group{}
    _ = xml.Unmarshal([]byte(data), &group)

    fmt.Printf("%#v
", group.A)
}

这些是看似合法的属性名称；知道如何提取 xlink:href 吗?谢谢.

These are seemingly legal attribute names; any idea how to extract the xlink:href one? thanks.

推荐答案

您的示例片段并不完全正确，因为它不包含 XML 命名空间 绑定.你可能想要的是:

Your example fragment is not quite correct, since it does not include an XML namespace binding for the xlink: prefix. What you probably want is:

<g xmlns:xlink="http://www.w3.org/1999/xlink">
  <a xlink:href="http://example.com" data-bind="121">lala</a>
</g>

您可以使用命名空间 URL 解组此属性:

You can unmarshal this attribute using the namespace URL:

XlinkHref string `xml:"http://www.w3.org/1999/xlink href,attr"`

这里是您的示例程序的更新副本，其中包含命名空间修复.

Here is an updated copy of your example program with the namespace fix.

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