我一直在使用RenderX XEP中的XSL-FO生成简单的PDF,但我对XPath中涉及的功能相对较新。
我有一些看起来像这样的XML:
<?xml version="1.0" encoding="UTF-8"?>
<report>
<ticket>
<size>A</size>
<department>Dept 1</department>
<category>Cat 1</category>
<product>Product 1</product>
</ticket>
<ticket>
<size>A</size>
<department>Dept 1</department>
<category>Cat 2</category>
<product>Product 2</product>
</ticket>
<ticket>
<size>B</size>
<department>Dept 1</department>
<category>Cat 2</category>
<product>Product 3</product>
</ticket>
<ticket>
<size>B</size>
<department>Dept 1</department>
<category>Cat 1</category>
<product>Product 1</product>
</ticket>
<ticket>
<size>C</size>
<department>Dept 1</department>
<category>Cat 2</category>
<product>Product 2</product>
</ticket>
<ticket>
<size>D</size>
<department>Dept 3</department>
<category>Cat 1</category>
<product>Product 4</product>
</ticket>
<ticket>
<size>D</size>
<department>Dept 1</department>
<category>Cat 1</category>
<product>Product 1</product>
</ticket>
</report>
我需要能够使用XSLT1.0格式化对象将这些数据排序到PDF中。所需的格式如下:
所有相同尺寸的门票都在单独的表格中(按字母顺序排列在页面上),每张门票都列出了相关类别和部门。 例如:
我已经开始实施Muenchian分组,但是当使用大型XML提取时,错误的类别属于部门之下,而且我还没有能够列出每个类别下面的相关产品。
我可能比较不整洁的XSL:
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:fo="http://www.w3.org/1999/XSL/Format">
<xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes"/>
<xsl:key name="keySize" match="ticket" use="size" />
<xsl:key name="keyDepartment" match="ticket" use="concat(size, ' ', department)" />
<xsl:key name="keyCategory" match="ticket" use="concat(size, ' ', department, ' ', category)" />
<xsl:template match="/report">
<fo:root xmlns:fo="http://www.w3.org/1999/XSL/Format">
<fo:layout-master-set>
<fo:simple-page-master master-name="simple" page-height="29.7cm" page-width="21cm" margin-top="2cm" margin-bottom="2cm" margin-left="2cm" margin-right="2cm">
<fo:region-body />
</fo:simple-page-master>
</fo:layout-master-set>
<fo:page-sequence master-reference="simple">
<fo:flow flow-name="xsl-region-body">
<xsl:for-each select="ticket[count(. | key('keySize', size)[1]) = 1]">
<xsl:sort select="size" order="ascending" />
<fo:block>
Size <xsl:value-of select="size" />
</fo:block>
<fo:block>
<xsl:variable name="sizes" select="key('keySize', size)" />
<xsl:for-each select="$sizes[generate-id() = generate-id(key('keyDepartment', concat(size, ' ', department))[1])]">
<xsl:sort select="department" order="ascending" />
<!--TABLES START-->
<fo:table-and-caption>
<fo:table border="0.2mm solid black" width="100%">
<fo:table-header>
<fo:table-row>
<fo:table-cell>
<fo:block font-weight="bold" margin="1mm 0mm 0mm 1mm">
<xsl:text>Department</xsl:text>
</fo:block>
</fo:table-cell>
<fo:table-cell>
<fo:block font-weight="bold" margin="1mm 0mm 0mm 1mm">
<xsl:value-of select="department" />
</fo:block>
</fo:table-cell>
</fo:table-row>
</fo:table-header>
<fo:table-body>
<fo:table-cell>
<fo:block margin="1mm 0mm 0mm 1mm" />
</fo:table-cell>
</fo:table-body>
</fo:table>
</fo:table-and-caption>
<!--BRINGS IN WRONG CATEGORIES-->
<xsl:for-each select="$sizes[generate-id() = generate-id(key('keyCategory', concat(size, ' ', department, ' ', category))[1])]">
<xsl:sort select="category" order="ascending" />
<fo:table-and-caption>
<fo:table border="0.2mm solid black" width="100%">
<fo:table-header>
<fo:table-row>
<fo:table-cell>
<fo:block font-weight="bold" margin="1mm 0mm 0mm 1mm">
<xsl:text>Category</xsl:text>
</fo:block>
</fo:table-cell>
<fo:table-cell>
<fo:block font-weight="bold" margin="1mm 0mm 0mm 1mm">
<xsl:value-of select="category" />
</fo:block>
</fo:table-cell>
</fo:table-row>
</fo:table-header>
<fo:table-body>
<fo:table-row>
<fo:table-cell>
<fo:block margin="1mm 0mm 0mm 1mm"><xsl:value-of select="description" /></fo:block>
</fo:table-cell>
</fo:table-row>
<!--
DESCRIPTIONS GO HERE?
-->
</fo:table-body>
</fo:table>
</fo:table-and-caption>
</xsl:for-each>
</xsl:for-each>
</fo:block>
</xsl:for-each>
</fo:flow>
</fo:page-sequence>
</fo:root>
</xsl:template>
</xsl:stylesheet>
任何有关引入产品的帮助将不胜感激。
答案 0 :(得分:0)
我认为问题在于这一行,您尝试迭代某个部门的所有类别
<xsl:for-each select="$sizes[generate-id() = generate-id(key('keyCategory', concat(size, ' ', department, ' ', category))[1])]">
问题是变量尺寸包含不同部门的详细信息,而不仅仅是您感兴趣的部门,导致重复。
请尝试此操作,将其限制为当前部门
<xsl:variable name="depts" select="key('keyDepartment', concat(size, ' ', department))" />
<xsl:for-each select="$depts[generate-id() = generate-id(key('keyCategory', concat(size, ' ', department, ' ', category))[1])]">