我需要一些帮助。我对这种转变感到绝望。我必须使用 XSLT 1.0。
我有这个输入:
<?xml version="1.0" encoding="utf-8" ?>
<export>
<ds>
<cds>
<cd id_c="1">
<cs id="123" >
<acc>
<payments>
<payment id_p="9" price="6.90"/>
</payments>
</acc>
<acc>
<payments>
<payment id_p="9" price="6.50"/>
</payments>
</acc>
</cs>
</cd>
<cd id_c="2">
<cs id="456" >
<acc>
<payments>
<payment id_p="1" price="4.30"/>
</payments>
</acc>
<acc>
<payments>
<payment id_p="9" price="11.20"/>
</payments>
</acc>
<acc>
<payments>
<payment id_p="1" price="2.10"/>
</payments>
</acc>
</cs>
</cd>
<cd id_c="3">
<cs id="789" >
<acc>
<payments>
<payment id_p="2" price="8.90"/>
</payments>
</acc>
<acc>
<payments>
<payment id_p="9" price="5.70"/>
</payments>
</acc>
</cs>
</cd>
</cds>
</ds>
</export>
我创造了这个转变
<xsl:key name="group_payment" match="acc/payments" use="payment/@id_p"/>
<xsl:template match="export">
<document>
<xsl:for-each select=".//ds/cds/cd">
<record>
<header>
<number><xsl:value-of select=".//@id" /></number>
</header>
<items>
<xsl:for-each select=".//payments[generate-id(.)=generate-id(key('group_payment', payment/@id_p))]/payment/@id_p">
<xsl:sort select="." order="ascending" case-order="lower-first" />
<item>
<text>Payment ID: <xsl:value-of select="."/></text>
<price><xsl:value-of select="sum(key('group_payment', .)//@price)"/></price>
</item>
</xsl:for-each>
</items>
</record>
</xsl:for-each>
</document>
</xsl:template>
但这对我来说效果不佳。它将每笔付款分组为“第一匹配”。我需要在每个 payment/@id_p 元素内按 <cd> 对付款进行分组。
这是我的输出:
<?xml version="1.0" encoding="utf-8"?>
<document>
<record>
<header>
<number>123</number>
</header>
<items>
<item>
<text>Payment ID: 9</text>
<price>30.3</price>
</item>
</items>
</record>
<record>
<header>
<number>456</number>
</header>
<items>
<item>
<text>Payment ID: 1</text>
<price>6.4</price>
</item>
</items>
</record>
<record>
<header>
<number>789</number>
</header>
<items>
<item>
<text>Payment ID: 2</text>
<price>8.9</price>
</item>
</items>
</record>
</document>
我想要这个输出:
<?xml version="1.0" encoding="utf-8"?>
<document>
<record>
<header>
<number>123</number>
</header>
<items>
<item>
<text>Payment ID: 9</text>
<price>13.40</price>
</item>
</items>
</record>
<record>
<header>
<number>456</number>
</header>
<items>
<item>
<text>Payment ID: 1</text>
<price>6.40</price>
</item>
<item>
<text>Payment ID: 9</text>
<price>11.20</price>
</item>
</items>
</record>
<record>
<header>
<number>789</number>
</header>
<items>
<item>
<text>Payment ID: 2</text>
<price>8.90</price>
</item>
<item>
<text>Payment ID: 9</text>
<price>5.70</price>
</item>
</items>
</record>
</document>
我尝试了不同的分组，但总是以相同的结果结束。有人可以帮我吗？
最佳答案
AFAICT，你想做的事:
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:key name="pmt" match="payment" use="concat(@id_p, '|', ancestor::cs/@id)"/>
<xsl:template match="export">
<document>
<xsl:for-each select="ds/cds/cd/cs">
<xsl:variable name="cs_id" select="@id" />
<record>
<header>
<number>
<xsl:value-of select="$cs_id" />
</number>
</header>
<items>
<xsl:for-each select="acc/payments/payment[generate-id() = generate-id(key('pmt', concat(@id_p, '|', $cs_id))[1])]">
<item>
<text>
<xsl:text>Payment ID: </xsl:text>
<xsl:value-of select="@id_p"/>
</text>
<price>
<xsl:value-of select="sum(key('pmt', concat(@id_p, '|', $cs_id))/@price)"/>
</price>
</item>
</xsl:for-each>
</items>
</record>
</xsl:for-each>
</document>
</xsl:template>
</xsl:stylesheet>
关于xml - XSL 1.0 分组，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/72630935/