XSLT：映射转换（分组）问题-解网

问：

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我需要转换上面的输入xml：

<?xml version='1.0' encoding='UTF-8'?>
<root>
    <row>
        <userId>40668825871</userId>
        <JobDescription>ANL CONTR QUALIDADE SR</JobDescription>
        <StartDate>01/01/2023</StartDate>
        <EndDate>01/08/2023</EndDate>
        <EventReason>1000</EventReason>
    </row>
    <row>
        <userId>40668825871</userId>
        <JobDescription>ANL CONTR QUALIDADE SR</JobDescription>
        <Atuacao>Interno</Atuacao>
        <StartDate>01/09/2023</StartDate>
        <EndDate>01/31/2023</EndDate>
        <EventReason>1001</EventReason>     
    </row>
    <row>
        <userId>40668825871</userId>
        <JobDescription>ANL CONTR QUALIDADE SR</JobDescription>
        <StartDate>02/01/2023</StartDate>
        <EndDate>02/28/2023</EndDate>
        <EventReason>1003</EventReason>     
    </row>
    <row>
        <userId>40668825871</userId>
        <JobDescription>ANL VALIDACAO SR</JobDescription>
        <StartDate>03/01/2023</StartDate>
        <EndDate>05/05/2023</EndDate>
        <EventReason>9000</EventReason>     
    </row>
    <row>
        <userId>40668825871</userId>
        <JobDescription>ANL VALIDACAO SR</JobDescription>
        <StartDate>05/06/2023</StartDate>
        <EndDate>12/31/2023</EndDate>
        <EventReason>1010</EventReason>     
    </row>
    <row>
        <userId>34916921801</userId>
        <JobDescription>COORD DESENV FARMACOTECNICO</JobDescription>
        <StartDate>08/01/2022</StartDate>
        <EndDate>01/31/2023</EndDate>
        <EventReason>9000</EventReason>     
    </row>
    <row>
        <userId>34916921801</userId>
        <JobDescription>COORD DESENV EMBALAGENS</JobDescription>
        <StartDate>02/01/2023</StartDate>
        <EndDate>02/28/2023</EndDate>
        <EventReason>9002</EventReason>     
    </row>
    <row>
        <userId>34916921801</userId>
        <JobDescription>COORD DESENV EMBALAGENS</JobDescription>
        <StartDate>03/01/2023</StartDate>
        <EndDate>05/05/2023</EndDate>
        <EventReason>1000</EventReason>     
    </row>
    <row>
        <userId>34916921801</userId>
        <JobDescription>COORD DESENV EMBALAGENS</JobDescription>
        <StartDate>05/06/2023</StartDate>
        <EndDate>12/31/2023</EndDate>
        <EventReason>1010</EventReason>     
    </row>
</root>

进入这个：

<?xml version='1.0' encoding='UTF-8'?>
<root>
    <employee>
    <userId>40668825871</userId>
        <job>
            <JobDescription>ANL CONTR QUALIDADE SR</JobDescription>
            <StartDate>01/01/2023</StartDate>
            <EndDate>02/28/2023</EndDate>
        </job>
        <job>
            <JobDescription>ANL VALIDACAO SR</JobDescription>
            <StartDate>03/01/2023</StartDate>
            <EndDate>12/31/2023</EndDate>
        </job>
    </employee> 
    <employee>
    <userId>34916921801</userId>
        <job>
            <JobDescription>COORD DESENV FARMACOTECNICO</JobDescription>
            <StartDate>08/01/2022</StartDate>
            <EndDate>01/31/2023</EndDate>
        </job>
        <job>
            <JobDescription>COORD DESENV EMBALAGENS</JobDescription>
            <StartDate>02/01/2023</StartDate>
            <EndDate>02/28/2023</EndDate>
        </job>
        <job>
            <JobDescription>COORD DESENV EMBALAGENS</JobDescription>
            <StartDate>03/01/2023</StartDate>
            <EndDate>12/31/2023</EndDate>
        </job>
    </employee>     
</root>

输入 XML 它基本上是特定年份的员工工作记录（因此员工可以有一条或多条记录）。

输出 xml 将是记录的分组，规则如下：

1 - 每当员工具有值为 9000 或 9002 的 eventReason 标记时，都必须使用上一条记录的 JobDescription 和 endDate 创建一个新的作业标记。startDate 将是当前员工作业最早的 startDate。

2 - 因此，在生成新的作业标记后，新作业的 startDate 将是事件为 9000 或 9002 的当前记录的 startDate，并且 JobDescription 和 endDate 将填充两种可能性：

2.1 如果存在另一个事件 9000 或 9002，则在记录事件 9000 或 9002 之前，将填充记录的 JobDescription 和 endDate 值;

2.2 如果不存在其他事件 9000 或 9002，则将用员工最后记录的值填充它们。

目前我不知道如何使用 XSLT 执行此操作。

提前致谢。

该方案如上所述。

XSLT XSLT-3.0

<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml"indent="yes"/>

<xsl:template match="/root">
    <output>
        <!-- group by employee -->
        <xsl:for-each-group select="row" group-by="userId">
            <employee>
                <xsl:copy-of select="userId"/>
                    <!-- break the group at each row with EventReason 9000 or 9002"> -->
                    <xsl:for-each-group select="current-group()" group-starting-with="row[EventReason=('9000', '9002')]">
                        <xsl:variable name="last-row" select="current-group()[last()]" />
                        <job>
                            <xsl:copy-of select="$last-row/JobDescription"/>
                            <xsl:copy-of select="StartDate"/>
                            <xsl:copy-of select="$last-row/EndDate"/>
                        </job>
                </xsl:for-each-group>
            </employee>
        </xsl:for-each-group>
    </output>
</xsl:template>

</xsl:stylesheet>

但是，我得到的结果与您显示的预期结果不同。因此，您的解释或我对它的理解中似乎缺少某些内容。但至少你应该有一个起点。

XSLT：映射转换（分组）问题

XSLT: Issue with mapping transformation (grouping)

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