合并与“_LIST"共享相同名称的 XML 节点；在节点名称和根级别 [英] Merge XML nodes sharing the same name with "_LIST" in the node name and also at root level
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问题描述
下面是输入 XML，我正在寻找所需的输出 -
Below is the Input XML and I am looking for the desired output -
<xml>
<a>
<element0>987</element0>
</a>
<a>
<a_list_one>
<a_lag_one>
<element1>123</element1>
<element2>456</element2>
</a_lag_one>
</a_list_one>
<a_list_one>
<a_lag_one>
<element1>789</element1>
<element2>678</element2>
</a_lag_one>
</a_list_one>
<a_list_two>
<a_lag_two>
<a_list_three>
<a_lag_three>
<element3>570</element3>
<element4>678</element4>
</a_lag_three>
</a_list_three>
<a_list_three>
<a_lag_three>
<element3>989</element3>
<element4>231</element4>
</a_lag_three>
</a_list_three>
</a_lag_two>
<a_lag_two>
<a_list_three>
<a_lag_three>
<element3>570</element3>
<element4>678</element4>
</a_lag_three>
</a_list_three>
<a_list_three>
<a_lag_three>
<element3>9873</element3>
<element4>278</element4>
</a_lag_three>
</a_list_three>
<a_list_four>
<a_lag_four>
<element5>9121</element5>
<element6>9879</element6>
</a_lag_four>
</a_list_four>
<a_list_three>
<a_lag_four>
<element5>098</element5>
<element6>231</element6>
</a_lag_four>
</a_list_three>
</a_lag_two>
</a_list_two>
<a_list_four>
<a_lag_four>
<element5>654</element5>
<element6>7665</element6>
</a_lag_four>
</a_list_four>
</a>
<b>
<b_list_one>
<b_lag_one>
<element8>123</element8>
<element9>456</element9>
</b_lag_one>
</b_list_one>
</b>
<b>
<b_list_one>
<b_lag_one>
<element8>789</element8>
<element9>678</element9>
</b_lag_one>
</b_list_one>
</b>
</xml>
所需的 XML 是:
<xml>
<a>
<element0>987</element0>
<a_list_one>
<a_lag_one>
<element1>123</element1>
<element2>456</element2>
</a_lag_one>
<a_lag_one>
<element1>789</element1>
<element2>678</element2>
</a_lag_one>
</a_list_one>
<a_list_two>
<a_lag_two>
<a_list_three>
<a_lag_three>
<element3>570</element3>
<element4>678</element4>
</a_lag_three>
<a_lag_three>
<element3>989</element3>
<element4>231</element4>
</a_lag_three>
</a_list_three>
</a_lag_two>
<a_lag_two>
<a_list_three>
<a_lag_three>
<element3>570</element3>
<element4>678</element4>
</a_lag_three>
<a_lag_three>
<element3>9873</element3>
<element4>278</element4>
</a_lag_three>
<a_lag_four>
<element5>098</element5>
<element6>231</element6>
</a_lag_four>
</a_list_three>
<a_list_four>
<a_lag_four>
<element5>9121</element5>
<element6>9879</element6>
</a_lag_four>
</a_list_four>
</a_lag_two>
</a_list_two>
<a_list_four>
<a_lag_four>
<element5>654</element5>
<element6>7665</element6>
</a_lag_four>
</a_list_four>
</a>
<b>
<b_list_one>
<b_lag_one>
<element8>123</element8>
<element9>456</element9>
</b_lag_one>
<b_lag_one>
<element8>789</element8>
<element9>678</element9>
</b_lag_one>
</b_list_one>
</b>
</xml>
我正在寻找可以转换为所需输出的 XSL.在这里，共享相同名称并且还包含_LIST"的节点应该合并在一起.但是，此逻辑应仅在第一个_LIST"节点内发生，不应应用于内部节点.其次，在根级别也是要合并的节点.例如在这里，应该只有一个a"标签和b"标签.请帮忙.
I am looking for XSL which does the conversion to the desired output. Here, the nodes which share the same name and also contains "_LIST" should be merged together. However, this logic should happen only within the first "_LIST" node and should not apply to inner nodes. Secondly, at the root level also, the nodes to be merged. For example here, there should be only one "a" tag and "b" tag. Kindly help.
推荐答案
这里是 XSLT 1.0 的解决方案
Here is a solution for XSLT 1.0
<xsl:stylesheet version="1.0"
xmlns:msxml="urn:schemas-microsoft-com:xslt"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes" omit-xml-declaration="yes"/>
<xsl:key name="xmlChildren" match="xml/*" use="local-name()"/>
<xsl:key name="list" match="*[contains(local-name(),'_list')]" use="generate-id(..)"/>
<!-- Select the child nodes of the xml node. -->
<xsl:template match="xml/*">
<!-- Get the name of the current node. -->
<xsl:variable name="localName" select="local-name()"/>
<!-- Is this the first child of the xml node with this name? -->
<xsl:if test="generate-id(.) = generate-id(key('xmlChildren', $localName)[1])">
<xsl:copy>
<!-- Output all of the xml grandchild nodes of any xml child node with same name as the current node. -->
<xsl:apply-templates select="key('xmlChildren', $localName)/*">
<xsl:with-param name="parentName" select="$localName"/>
</xsl:apply-templates>
</xsl:copy>
</xsl:if>
</xsl:template>
<!-- Select the nodes with a local name that contains '_list'. -->
<xsl:template match="*[contains(local-name(),'_list')]">
<xsl:param name="parentName"/>
<xsl:variable name="parentID" select="generate-id(..)"/>
<!-- Get the name of the current node. -->
<xsl:variable name="localName" select="local-name()"/>
<xsl:choose>
<!-- Is this list a first generation grandchild of xml? -->
<xsl:when test="parent::*/parent::xml">
<!-- Is this the first instance of this list? -->
<xsl:if test="generate-id(.) = generate-id(key('xmlChildren', $parentName)/*[local-name()=$localName][1])">
<xsl:copy>
<xsl:apply-templates select="key('xmlChildren', $parentName)/*[local-name()=$localName]/*"/>
</xsl:copy>
</xsl:if>
</xsl:when>
<xsl:otherwise>
<!-- Is this the first instance of this list? -->
<xsl:if test="generate-id(.) = generate-id(key('list', $parentID)[local-name()=$localName][1])">
<xsl:copy>
<xsl:apply-templates select="key('list', $parentID)[local-name() = $localName]/*"/>
</xsl:copy>
</xsl:if>
</xsl:otherwise>
</xsl:choose>
</xsl:template>
<xsl:template match="node()|@*">
<xsl:copy>
<xsl:apply-templates select="node()|@*"/>
</xsl:copy>
</xsl:template>
</xsl:stylesheet>
这篇关于合并与“_LIST"共享相同名称的 XML 节点；在节点名称和根级别的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
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