Asp.net XSLT1.0子字符串,然后选择distinct
我对xslt非常陌生,因此非常感谢您的帮助。下面是我的示例xml文件Asp.net XSLT1.0子字符串,然后选择distinct,asp.net,xml,xslt,visual-studio-2012,xslt-1.0,Asp.net,Xml,Xslt,Visual Studio 2012,Xslt 1.0,我对xslt非常陌生,因此非常感谢您的帮助。下面是我的示例xml文件 <DocumentElement> <Records> <date>2014-07-01 00:00</date> </Records> <Records> <date>2014-08-03 00:00</date> </Records> <
<DocumentElement>
<Records>
<date>2014-07-01 00:00</date>
</Records>
<Records>
<date>2014-08-03 00:00</date>
</Records>
<Records>
<date>2013-08-03 00:00</date>
</Records>
<DocumentElement>
2014-07-01 00:00
2014-08-03 00:00
2013-08-03 00:00
我需要的只是从日期中选择不同的年份
目前我有下面的xslt,它带来了重复的年份
<?xml version="1.0" encoding="utf-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:msxsl="urn:schemas-microsoft-com:xslt" exclude-result-prefixes="msxsl" xmlns:ms="urn:schemas-microsoft-com:xslt">
<xsl:output method="xml" indent="yes"/>
<xsl:template match="/">
<xsl:variable name="years" select="DocumentElement/Records/date"/>
<ul>
<xsl:for-each select="$years">
<li>
<xsl:element name="a">
<xsl:attribute name="href">
<xsl:value-of select="concat('?archive=',substring( ., 1, 4))"/>
</xsl:attribute>
<xsl:value-of select="substring( ., 1, 4)"/>
</xsl:element>
</li>
</xsl:for-each>
</ul>
</xsl:template>
</xsl:stylesheet>
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结果如下:
<ul>
<li>
<a href="?archive=2014">2014</a>
</li>
<li>
<a href="?archive=2014">2014</a>
</li>
<li>
<a href="?archive=2014">2013</a>
</li>
</ul>
-
-
-
但我的预期结果应该是
<ul>
<li>
<a href="?archive=2014">2014</a>
</li>
<li>
<a href="?archive=2014">2013</a>
</li>
</ul>
-
-
我尝试了以下操作,但得到了空输出
<?xml version="1.0" encoding="utf-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:msxsl="urn:schemas-microsoft-com:xslt" exclude-result-prefixes="msxsl" xmlns:ms="urn:schemas-microsoft-com:xslt">
<xsl:output method="xml" indent="yes"/>
<xsl:template match="/">
<xsl:variable name="years" select="substring(DocumentElement/Records/date, 1, 4)"/>
<ul>
<xsl:for-each select="$years[not(.=preceding::*)]">
<li>
<xsl:element name="a">
<xsl:attribute name="href">
<xsl:value-of select="concat('?archive=',substring( ., 1, 4))"/>
</xsl:attribute>
<xsl:value-of select="substring( ., 1, 4)"/>
</xsl:element>
</li>
</xsl:for-each>
</ul>
</xsl:template>
</xsl:stylesheet>
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任何帮助都将不胜感激。谢谢。在XSLT1.0中,Muenchian分组比使用前/后/前同级/后同级轴更有效。试试这个:
<?xml version="1.0" encoding="utf-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" >
<xsl:output method="xml" indent="yes"/>
<xsl:key name="Date" match="DocumentElement/Records/date" use="substring(.,1,4)"/>
<xsl:template match="/">
<ul>
<xsl:for-each select="DocumentElement/Records/date[generate-id() = generate-id(key('Date', substring(.,1,4))[1])]">
<li>
<a href="{concat('?archive=',substring(.,1,4))}">
<xsl:value-of select="substring( ., 1, 4)"/>
</a>
</li>
</xsl:for-each>
</ul>
</xsl:template>
</xsl:stylesheet>
-
在XSLT1.0中,Muenchian分组比使用前/后/前同级/后同级轴更有效。试试这个:
<?xml version="1.0" encoding="utf-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" >
<xsl:output method="xml" indent="yes"/>
<xsl:key name="Date" match="DocumentElement/Records/date" use="substring(.,1,4)"/>
<xsl:template match="/">
<ul>
<xsl:for-each select="DocumentElement/Records/date[generate-id() = generate-id(key('Date', substring(.,1,4))[1])]">
<li>
<a href="{concat('?archive=',substring(.,1,4))}">
<xsl:value-of select="substring( ., 1, 4)"/>
</a>
</li>
</xsl:for-each>
</ul>
</xsl:template>
</xsl:stylesheet>
-
在XSLT1.0中,Muenchian分组比使用前/后/前同级/后同级轴更有效。试试这个:
<?xml version="1.0" encoding="utf-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" >
<xsl:output method="xml" indent="yes"/>
<xsl:key name="Date" match="DocumentElement/Records/date" use="substring(.,1,4)"/>
<xsl:template match="/">
<ul>
<xsl:for-each select="DocumentElement/Records/date[generate-id() = generate-id(key('Date', substring(.,1,4))[1])]">
<li>
<a href="{concat('?archive=',substring(.,1,4))}">
<xsl:value-of select="substring( ., 1, 4)"/>
</a>
</li>
</xsl:for-each>
</ul>
</xsl:template>
</xsl:stylesheet>
-
在XSLT1.0中,Muenchian分组比使用前/后/前同级/后同级轴更有效。试试这个:
<?xml version="1.0" encoding="utf-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" >
<xsl:output method="xml" indent="yes"/>
<xsl:key name="Date" match="DocumentElement/Records/date" use="substring(.,1,4)"/>
<xsl:template match="/">
<ul>
<xsl:for-each select="DocumentElement/Records/date[generate-id() = generate-id(key('Date', substring(.,1,4))[1])]">
<li>
<a href="{concat('?archive=',substring(.,1,4))}">
<xsl:value-of select="substring( ., 1, 4)"/>
</a>
</li>
</xsl:for-each>
</ul>
</xsl:template>
</xsl:stylesheet>
-