根据一些匹配的值在同一个 xml 中复制和移动 xml 节点

Question

这是我的第一个。我不擅长 xml/xslt/c#/VBS 编程。我在网站上搜索，但找不到我要找的东西。

我需要在<pm-identifier>MGRTN0000004999999</pm-identifier> 内部结构<deleted-assignment> 和identifier>MGRTN0000004999999</pm-identifier> 内部结构<product> 之间找到一个匹配项，如果值相等，那么节点<article-number>000000000004999995</article-number> 应该被复制到结构<deleted-assignment> 只是在<pm-identifier>MGRTN0000004999999</pm-identifier> 行下方。

如果没有匹配，它仍应显示输出中的所有节点。如果有匹配，它应该只是复制 from to 如上所述。

存在多个具有不同值的<deleted-assignment> 和多个<Product> 节点。

<products>
    <deleted-assignment>
    <pm-identifier>MGRTN0000004999999</pm-identifier>
    <structure-system-identifier>Web Structure</structure-system-identifier>
    <structure-group-identifier>M010010080002</structure-group-identifier>
    <delete>true</delete>
    </deleted-assignment>
    <product>
    <pm-identifier>MGRTN0000004999999</pm-identifier>
    <prod-status>CREATED</prod-status>
    <val-status>Valid</val-status>
    <dup-status>Unique</dup-status>
    <con-status>New</con-status>
    <im-status>New</im-status>
    <wcs-status>New</wcs-status>
    <article-number>000000000004999995</article-number>
    <art-category>GA</art-category>
    <auto-order>No</auto-order>
    <archived>No</archived>
    </Product>
</Products>

我想要这样的输出：

<products>
    <deleted-assignment>
        <pm-identifier>MGRTN0000004999999</pm-identifier>
        <article-number>000000000004999995</article-number>
        <structure-system-identifier>Web Structure</structure-system-identifier>
        <structure-group-identifier>M010010080002</structure-group-identifier>
    <delete>true</delete>
    </deleted-assignment>
    <product>
            <pm-identifier>MGRTN0000004999999</pm-identifier>
        <prod-status>CREATED</prod-status>
        <val-status>Valid</val-status>
            <dup-status>Unique</dup-status>
        <con-status>New</con-status>
        <im-status>New</im-status>
        <wcs-status>New</wcs-status>
        <article-number>000000000004999995</article-number>
        <art-category>GA</art-category>
        <auto-order>No</auto-order>
        <archived>No</archived>
   </product>
</product>

提前致谢

Answer 1

如果（！）我理解正确，你想这样做：

XSLT 1.0

<xsl:stylesheet version="1.0" 
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes"/>
<xsl:strip-space elements="*"/>

<xsl:key name="product" match="product" use="pm-identifier" />

<!-- identity transform -->
<xsl:template match="@*|node()">
    <xsl:copy>
        <xsl:apply-templates select="@*|node()"/>
    </xsl:copy>
</xsl:template>

<xsl:template match="deleted-assignment/pm-identifier">
    <xsl:copy-of select="."/>
    <xsl:copy-of select="key('product', .)/article-number"/>
</xsl:template>

</xsl:stylesheet>