英文:
XSLT Concatenation from different nodes when a string is found
问题
我想要连接不同的节点从"Time"直到在"Sum"中提供另一个数据。大多数sum节点都是空的。我想要将下一个SUM值之前的所有节点连接到一个单一的引用字段中。
<Reference>Debit,TMU/2023/1059 A ,TMU/2023/1019 A ,TMU/2023/1020 A </Reference>
<Reference> Debit,56011846 A </Reference>
这是XML输入，其中包含SUM和TIME节点。大多数SUM节点都为空。我想要将下一个SUM值之前的所有节点连接到一个单一的引用字段中。
英文:
> I want to concatenate different nodes from "Time" until there is
> another data provided in "Sum". Most of the sum nodes are empty. I
> will like to concatenate the all nodes before the next SUM value to a
> single reference field.
<?xml version="1.0"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<!-- recursive named template -->
<xsl:template name="hm">
<xsl:param name="amount"/>
<xsl:param name="pos"/>
<xsl:for-each select="../row">
<xsl:if test="$amount = Sum and $pos = position()"></xsl:if>
<xsl:choose>
<xsl:when test="string-length(Sum) &gt; 0">
<xsl:value-of select="Time"/>
</xsl:when>
<xsl:otherwise>
<xsl:value-of select="', '"/>
<xsl:value-of select="Time"/>
</xsl:otherwise>
</xsl:choose>
</xsl:for-each>
</xsl:template>
<Reference>
<xsl:call-template name="hm">
<xsl:with-param name="amount" select="Sum"/>
<xsl:with-param name="pos" select="position()"/>
</xsl:call-template>
</Reference>
</Payment>
</xsl:if>
</xsl:for-each>
</Trailer>
</xsl:template>
</xsl:stylesheet>
> > # # <!--This is the output which gives concatenation of all the nodes instead of the nodes before the next sum-->
<Trailer>
<Payment>
<Reference>Debit,TMU/2023/1059 A ,TMU/2023/1019 A ,TMU/2023/1020 A, Debit,56011846 A
</Reference>
</Payment>
<Payment>
<Reference>Debit,TMU/2023/1059 A ,TMU/2023/1019 A ,TMU/2023/1020 A, Debit,56011846 A
</Reference>
</Payment>
</Trailer>
>
> # <!--But my expected output is with the value of all the TIME nodes before the next SUM that has a value-->
<Trailer>
<Payment>
<Reference>Debit,TMU/2023/1059 A ,TMU/2023/1019 A ,TMU/2023/1020 A </Reference>
</Payment>
<Payment>
<Reference> Debit,56011846 A </Reference>
</Payment>
</Trailer>
> > #This is the xml input which has the nodes SUM and TIME. Most of the sum nodes are empty. I will like to concatenate the all nodes
> before the next SUM value to a single reference field.
<root>
<row>
<Time>Debit</Time>
<Sum>(874,516.25)</Sum>
</row>
<row>
<Time>TMU/2023/1059 A</Time>
<Sum></Sum>
</row>
<row>
<Time>TMU/2023/1019 A</Time>
<Sum></Sum>
</row>
<row>
<Time>TMU/2023/1020 A</Time>
<Sum></Sum>
</row>
<row>
<Time>Debit</Time>
<Sum>(664,297.52)</Sum>
</row>
<row>
<Time>56011846 A</Time>
<Sum></Sum>
</row>
</root>
答案1
得分: 0
如我在评论中提到的，这似乎是一个分组问题。如果（如您所报告）您的处理器支持XSLT 2.0，您应该能够使用以下代码生成预期的输出：
XSLT 2.0
<xsl:stylesheet version="2.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes"/>
<xsl:template match="/root">
<Trailer>
<xsl:for-each-group select="row" group-starting-with="row[Sum/text()]">
<Payment>
<Reference>
<xsl:value-of select="current-group()/Time" separator=","/>
</Reference>
</Payment>
</xsl:for-each-group>
</Trailer>
</xsl:template>
</xsl:stylesheet>
英文:
As I mentioned in the comments, this seems to be a grouping question. If (as you report) your processor supports XSLT 2.0, you should be able to produce the expected output using:
XSLT 2.0
<xsl:stylesheet version="2.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes"/>
<xsl:template match="/root">
<Trailer>
<xsl:for-each-group select="row" group-starting-with="row[Sum/text()]">
<Payment>
<Reference>
<xsl:value-of select="current-group()/Time" separator=","/>
</Reference>
</Payment>
</xsl:for-each-group>
</Trailer>
</xsl:template>
</xsl:stylesheet>
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