[英]Grouping XML nodes by attribute value in XSLT
Im quite new to xslt transforms and I need help with one kind of transformation. 我对xslt转换非常陌生,我需要一种转换的帮助。 I need to group all nodes of certain type by one of it atributes and list parents of every kind of attribute.
我需要将某种类型的所有节点按属性之一分组,并列出每种属性的父级。 It is a kind of making summary of usage of certain things in the document.
这是对文档中某些内容的用法进行汇总的一种。 The I will present simplified example.
我将提供简化的示例。
Input: 输入:
<root>
<node name="node1">
<somechild child-id="1">
</node>
<node name="node2">
<somechild child-id="2">
</node>
<node name="node3">
<somechild child-id="1">
</node>
<node name="node4">
<somechild child-id="2">
</node>
<node name="node5">
<somechild child-id="3">
</node>
</root>
Desired output: 所需的输出:
<root>
<somechild child-id="1">
<is-child-of>
<node name="node1" />
<node name="node3" />
</is-child-of>
</somechild>
<somechild child-id="2">
<is-child-of>
<node name="node2" />
<node name="node4" />
</is-child-of>
</somechild>
<somechild child-id="3">
<is-child-of>
<node name="node5" />
</is-child-of>
</somechild>
</root>
Idea is that if is the same element in many nodes they have same child-id. 想法是,如果许多节点中的元素相同,则它们具有相同的子ID。 I need to find all used by every .
我需要找到每个人都用过的东西。 I found this question XSLT transformation to xml, grouping by key which is kind of similar but there is a declaration of all authors on the begining and I don't have such, is allways only a child of .
我发现这个问题从XSLT到xml的转换,按键进行分组,这有点相似,但是一开始就声明了所有作者,但我没有这样的声明,始终只是的子级。
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" indent="yes"/>
<xsl:key name="k" match="somechild" use="@child-id"/>
<xsl:key name="n" match="node" use="somechild/@child-id"/>
<xsl:template match="root">
<xsl:copy>
<xsl:apply-templates
select="//somechild[generate-id(.) = generate-id(key('k', @child-id))]"/>
</xsl:copy>
</xsl:template>
<xsl:template match="somechild">
<xsl:copy>
<xsl:apply-templates select="@*"/>
<is-child-of>
<xsl:apply-templates select="key('n', @child-id)"/>
</is-child-of>
</xsl:copy>
</xsl:template>
<xsl:template match="node">
<xsl:copy>
<xsl:apply-templates select="@*"/>
</xsl:copy>
</xsl:template>
<xsl:template match="@*">
<xsl:copy>
<xsl:apply-templates select="@*"/>
</xsl:copy>
</xsl:template>
</xsl:stylesheet>
Output: 输出:
<root>
<somechild child-id="1">
<is-child-of>
<node name="node1" />
<node name="node3" />
</is-child-of>
</somechild>
<somechild child-id="2">
<is-child-of>
<node name="node2" />
<node name="node4" />
</is-child-of>
</somechild>
<somechild child-id="3">
<is-child-of>
<node name="node5" />
</is-child-of>
</somechild>
</root>
You could try the following approach? 您可以尝试以下方法吗?
<!-- select the current child id to filter by -->
<xsl:variable name="id" select="somechild/@child-id"/>
<!-- select the nodes which have a somechild element with the child-id to look for -->
<xsl:for-each select="/root//some-child[@child-id = $id]/..">
<!-- for each such node, do something -->
</xsl:for-each>
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