从 xslt 向后工作以创建 xml [英] Work backwards from an xslt to create an xml

问题描述

所以，我有很多 XSLT 文件需要生成预览，但是我没有相应的 XML 文件.我想知道是否可以通过 XSLT 文件并为该 XSLT 创建必填字段列表?例如，

So, I have quite a few XSLT files that I need to generate a preview for, however I do not have the corresponding XML files. I was wondering if it was possible to go through an XSLT file and create a list of the required fields for that XSLT? For example,

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
  <xsl:template match="/">
      <xsl:for-each select="catalog/cd">
        <tr>
          <td><xsl:value-of select="title" /></td>
          <td><xsl:value-of select="artist" /></td>
        </tr>
    </xsl:for-each>
  </xsl:template>
</xsl:stylesheet>

我想从那个 XSLT 转到这个:"/catalog/cd/title,/catalog/cd/artist" 因为它们是这个 XSLT 所需的字段，或者为所有这些字段设置一个默认值以便它输出标题为/catalog/cd/title"，艺术家为/catalog/cd/artist".

I want to go from that XSLT to this: "/catalog/cd/title, /catalog/cd/artist" as they are the fields required by this XSLT, or set a default value for all of them so it outputs "/catalog/cd/title" for title and "/catalog/cd/artist" for artist.

推荐答案

只是试一试，这也不适用于多个模板:

just to give it a shot, also this won't work with multiple templates:

<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0" exclude-result-prefixes="xsl">
<xsl:output method="xml" indent="yes"/>
<xsl:variable name="xsl-namespace">http://www.w3.org/1999/XSL/Transform</xsl:variable>

<xsl:template match="*[namespace-uri() != $xsl-namespace]" >
    <xsl:copy>
        <xsl:apply-templates select="@*[namespace-uri() != $xsl-namespace]"/>
        <xsl:apply-templates select="*[namespace-uri() != $xsl-namespace]"/>
    </xsl:copy>
</xsl:template>

<xsl:template match="@*[namespace-uri() != $xsl-namespace]" >
    <xsl:copy/>
</xsl:template>

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