XSLT - 嵌套 For-each with core-if

Question

如果使用 resourceID 属性在 TaskLabor 节点中不存在 Resource 节点，我想从 XML 中删除它们。 我不知道如何访问 for-each 内部的变量。 任何人都可以解决这个问题吗？

我在下面提供了 XSL 转换。 而不是for-each，可以使用apply-templates来完成吗？

输入 XML：

<Project projectID="project1" name="My Project">
    <Resources>
        <Resource resourceID="abc@mycompany.com"/>
        <Resource resourceID="def@mycompany.com"/> <!-- Need to Remove this -->
        <Resource resourceID="xyz@mycompany.com"/>
        <Resource resourceID="test@mycompany.com"/> <!-- Need to Remove this -->
    </Resources>
    <Tasks>
        <Task name="Task1"
        taskID="100-150221">
            <Assignments>
                <TaskLabor finish="2020-12-31"
                   resourceID="abc@mycompany.com"
                   start="2020-01-01">
                </TaskLabor>
                <TaskLabor finish="2020-12-31"
                   resourceID="xyz@mycompany.com"
                   start="2020-01-01">
                </TaskLabor>
            </Assignments>
        </Task>
        <Task name="Task2"
        taskID="100-12313">
            <Assignments>
                <TaskLabor finish="2020-12-31"
                   resourceID="abc@mycompany.com"
                   start="2020-01-01">
                </TaskLabor>
            </Assignments>
        </Task>
    </Tasks>
</Project>

XSL 转换：

<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:template match="/">
        <xsl:variable name="resources" select="Project/Resources/Resource"/>
        <xsl:variable name="tasks" select="Project/Tasks/Task"/>
        <xsl:for-each select="$resources">
            <xsl:variable name="resID" select="./@resourceID"/>
            <xsl:for-each select="$tasks">
                <xsl:for-each select="./Assignments/TaskLabor">
                     <xsl:if test="$resID = ./@resourceID">
                        <xsl:value-of select="Project/Resources/Resource"/>

                     </xsl:if>
                </xsl:for-each>
            </xsl:for-each>
        </xsl:for-each>
     </xsl:template>
</xsl:stylesheet>

预期产出

<Project projectID="project1" name="My Project">
    <Resources>
        <Resource resourceID="abc@mycompany.com"/>
        <Resource resourceID="xyz@mycompany.com"/>
    </Resources>
    <Tasks>
        <Task name="Task1"
        taskID="100-150221">
            <Assignments>
                <TaskLabor finish="2020-12-31"
                   resourceID="abc@mycompany.com"
                   start="2020-01-01">
                </TaskLabor>
                <TaskLabor finish="2020-12-31"
                   resourceID="xyz@mycompany.com"
                   start="2020-01-01">
                </TaskLabor>
            </Assignments>
        </Task>
        <Task name="Task2"
        taskID="100-12313">
            <Assignments>
                <TaskLabor finish="2020-12-31"
                   resourceID="abc@mycompany.com"
                   start="2020-01-01">
                </TaskLabor>
            </Assignments>
        </Task>
    </Tasks>
</Project>

Answer 1

声明一个键<xsl:key name="task-ref" match="TaskLabor" use="@resourceID"/>然后你只需要阻止不引用 TaskLabor 的元素被身份转换复制为他们使用空模板：

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet
    xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
    version="1.0">

  <xsl:key name="task-ref" match="TaskLabor" use="@resourceID"/>

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

  <xsl:template match="Resource[not(key('task-ref', @resourceID))]"/>

</xsl:stylesheet>

https://xsltfiddle.liberty-development.net/3NSTbfg