[英]Please help in grouping by Title within for each catalog using Muenchian grouping XSLT 1.0
Please help with a code to group based on TITLE within its each separate CATALOG tag.请帮助在其每个单独的 CATALOG 标签中根据 TITLE 分组的代码。 I am using XSLT 1.0 version.我正在使用 XSLT 1.0 版本。
<?xml version="1.0" encoding="UTF-8"?>
<hello>
<catalog>
<cd>
<title>Empire Burlesque</title>
<artist>Bob Dylan</artist>
<country>USA</country>
<company>Columbia</company>
<price>10.90</price>
<year>1985</year>
</cd>
<cd>
<title>Empire Burlesque</title>
<artist>Bob Dylan</artist>
<country>USA</country>
<company>Columbia</company>
<price>10.90</price>
<year>2000</year>
</cd>
<cd>
<title>Unchain my heart</title>
<artist>Joe Cocker</artist>
<country>USA</country>
<company>EMI</company>
<price>8.20</price>
<year>1987</year>
</cd>
</catalog>
<catalog>
<cd>
<title>Ring My Bells</title>
<artist>Enrique</artist>
<country>USA</country>
<company>EMI</company>
<price>8.20</price>
<year>1987</year>
</cd>
<cd>
<title>Empire Burlesque</title>
<artist>Bob Dylan</artist>
<country>USA</country>
<company>Columbia</company>
<price>10.90</price>
<year>1985</year>
</cd>
</catalog>
</hello>
Excepted Output异常输出
<Song>
<Desc>From First Catalog</Desc>
<nameofAlbum>Empire Burlesque</nameofAlbum>
<nameofAlbum>Unchain my heart</nameofAlbum>
</Song>
<Song>
<Desc>From Second Catalog</Desc>
<nameofAlbum>Ring My Bells</nameofAlbum>
<nameofAlbum>Empire Burlesque</nameofAlbum>
</Song>
Please help as the requirement is to group based on title of each CatLog only.请帮助,因为要求仅根据每个 CatLog 的标题进行分组。
I have tried using the Muenchian grouping but it was grouping with all the catalogs tags where as I require as induvial grouping between catalogs我曾尝试使用 Muenchian 分组,但它与所有目录标签分组,因为我需要在目录之间进行独立分组
Since you want to dedup within the catalog
, I would use generate-id()
and create a composite key with the generated ID for the catalog element and the cd/title
.由于您想在catalog
进行重复数据删除,我将使用generate-id()
并使用为 catalog 元素和cd/title
生成的 ID 创建一个复合键。
Use that composite key for the xsl:key
matching on cd
, then you can iterate over each /hello/catalog
and then for each cd
under that catalog
you can use the key for Muenchian grouping:使用该复合键进行cd
上的xsl:key
匹配,然后您可以遍历每个/hello/catalog
,然后对于该catalog
下的每个cd
,您可以使用 Muenchian 分组的键:
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">
<xsl:output indent="yes"/>
<xsl:variable name="labels">
<label>First</label>
<label>Second</label>
</xsl:variable>
<xsl:key name="cd-by-catalog-and-title" match="cd" use="concat(generate-id(..), title)"/>
<xsl:template match="/">
<xsl:for-each select="/hello/catalog">
<xsl:variable name="pos" select="position()"/>
<Song>
<Desc>From <xsl:value-of select="$labels/*[position() = $pos]"/> Catalog</Desc>
<xsl:for-each select="cd[count(. | key('cd-by-catalog-and-title', concat(generate-id(..), title))[1]) = 1]">
<nameofAlbum><xsl:value-of select="title"/></nameofAlbum>
</xsl:for-each>
</Song>
</xsl:for-each>
</xsl:template>
</xsl:stylesheet>
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