我被告知要合并出现在同一地点的公交车站的所有常用名称，例如Aberdeen有三个常用站点，但它们位于不同的单元格中。我如何将这些结合起来，以便所有常见的句点出现在我的XSL样式表中的一个地点Aberdeen下？
下面是我的样式表：
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:template match="/">
<html>
<body>
<h2>BusStops</h2>
<table border="1">
<tr bgcolor="#9acd32">
<th>Locality</th>
<th>BusStop</th>
<th> Co-ordinates </th>
</tr>
<xsl:for-each select="BusStops/BusStopDetails">
<xsl:sort select="LocalityName"/>
<tr>
<td> <xsl:value-of select="LocalityName"/> (<xsl:value-of select="ParentLocalityName"/>) </td>
<td><xsl:value-of select="CommonName"/></td>
<td>
<a href="https://maps.google.com/maps?ll={Latitude},{Longitude}&q={Latitude},{Longitude}&hl=en&t=m&z=14">
(<xsl:value-of select="Latitude"/>) , (<xsl:value-of select="Longitude"/>)
</a>
</td>
</tr>
</xsl:for-each>
</table>
</body>
</html>
</xsl:template>
</xsl:stylesheet>
这是我的XML示例：
<BusStopDetails>
<AtcoCode>639070021
</AtcoCode>
<CommonName>Union Square Bus Station
</CommonName>
<LocalityName>Aberdeen
</LocalityName>
<ParentLocalityName/>
<Latitude>57.1445763077
</Latitude>
<Longitude>-2.0963111958
</Longitude>
</BusStopDetails>
<BusStopDetails>
<AtcoCode>639080008
</AtcoCode>
<CommonName>Aberdeen Railway Station
</CommonName>
<LocalityName>Aberdeen
</LocalityName>
<ParentLocalityName/>
<Latitude>57.1437934462
</Latitude>
<Longitude>-2.0980112638
</Longitude>
</BusStopDetails>
-<BusStopDetails>
<AtcoCode>9300ABA
</AtcoCode>
<CommonName>Aberdeen Ferry Terminal
</CommonName>
<LocalityName>Aberdeen
</LocalityName>
<ParentLocalityName/>
<Latitude>57.1447953603
</Latitude>
<Longitude>-2.0917010936
</Longitude>
</BusStopDetails>
答案 0 :(得分：1)
这是另一个也使用xsl:key ...
XML输入（添加了另一个BusStopDetails以获得更好的示例）
<BusStops>
<BusStopDetails>
<AtcoCode>639070021</AtcoCode>
<CommonName>Some common name</CommonName>
<LocalityName>Stackoverflow</LocalityName>
<ParentLocalityName>Some Parent Loc</ParentLocalityName>
<Latitude>57.1445763077</Latitude>
<Longitude>-2.0963111958</Longitude>
</BusStopDetails>
<BusStopDetails>
<AtcoCode>639070021</AtcoCode>
<CommonName>Union Square Bus Station</CommonName>
<LocalityName>Aberdeen</LocalityName>
<ParentLocalityName/>
<Latitude>57.1445763077</Latitude>
<Longitude>-2.0963111958</Longitude>
</BusStopDetails>
<BusStopDetails>
<AtcoCode>639080008</AtcoCode>
<CommonName>Aberdeen Railway Station</CommonName>
<LocalityName>Aberdeen
</LocalityName>
<ParentLocalityName/>
<Latitude>57.1437934462</Latitude>
<Longitude>-2.0980112638</Longitude>
</BusStopDetails>
<BusStopDetails>
<AtcoCode>9300ABA</AtcoCode>
<CommonName>Aberdeen Ferry Terminal</CommonName>
<LocalityName>Aberdeen </LocalityName>
<ParentLocalityName/>
<Latitude>57.1447953603</Latitude>
<Longitude>-2.0917010936</Longitude>
</BusStopDetails>
</BusStops>
XSLT 1.0
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output indent="yes"/>
<xsl:strip-space elements="*"/>
<xsl:key name="detailsByLocality" match="BusStopDetails" use="normalize-space(LocalityName)"/>
<xsl:template match="/BusStops">
<html>
<body>
<h2>BusStops</h2>
<table border="1">
<tr bgcolor="#9acd32">
<th>Locality</th>
<th>BusStop</th>
<th>Co-ordinates</th>
</tr>
<xsl:for-each select="BusStopDetails[count(.|key('detailsByLocality',normalize-space(LocalityName))[1])=1]">
<xsl:sort select="LocalityName"/>
<tr>
<td rowspan="{count(key('detailsByLocality',normalize-space(LocalityName))) + 1}">
<xsl:value-of select="normalize-space(LocalityName)"/>
<xsl:apply-templates select="ParentLocalityName"/>
</td>
</tr>
<xsl:apply-templates select="key('detailsByLocality',normalize-space(LocalityName))"/>
</xsl:for-each>
</table>
</body>
</html>
</xsl:template>
<xsl:template match="BusStopDetails">
<tr>
<td><xsl:value-of select="normalize-space(CommonName)"/></td>
<td>
<a
href="https://maps.google.com/maps?ll={Latitude},{Longitude}&q={Latitude},{Longitude}&hl=en&t=m&z=14">
<xsl:value-of select="concat('(',Latitude,' , ',Longitude,')')"/>
</a>
</td>
</tr>
</xsl:template>
<xsl:template match="ParentLocalityName[string()]">
<xsl:value-of select="concat(' (',normalize-space(),')')"/>
</xsl:template>
</xsl:stylesheet>
<强>输出
<html>
<body>
<h2>BusStops</h2>
<table border="1">
<tr bgcolor="#9acd32">
<th>Locality</th>
<th>BusStop</th>
<th>Co-ordinates</th>
</tr>
<tr>
<td rowspan="4">Aberdeen</td>
</tr>
<tr>
<td>Union Square Bus Station</td>
<td><a href="https://maps.google.com/maps?ll=57.1445763077,-2.0963111958&q=57.1445763077,-2.0963111958&hl=en&t=m&z=14">(57.1445763077 , -2.0963111958)</a></td>
</tr>
<tr>
<td>Aberdeen Railway Station</td>
<td><a href="https://maps.google.com/maps?ll=57.1437934462,-2.0980112638&q=57.1437934462,-2.0980112638&hl=en&t=m&z=14">(57.1437934462 , -2.0980112638)</a></td>
</tr>
<tr>
<td>Aberdeen Ferry Terminal</td>
<td><a href="https://maps.google.com/maps?ll=57.1447953603,-2.0917010936&q=57.1447953603,-2.0917010936&hl=en&t=m&z=14">(57.1447953603 , -2.0917010936)</a></td>
</tr>
<tr>
<td rowspan="2">Stackoverflow (Some Parent Loc)</td>
</tr>
<tr>
<td>Some common name</td>
<td><a href="https://maps.google.com/maps?ll=57.1445763077,-2.0963111958&q=57.1445763077,-2.0963111958&hl=en&t=m&z=14">(57.1445763077 , -2.0963111958)</a></td>
</tr>
</table>
</body>
</html>
答案 1 :(得分：0)
这可以通过以下样式表来实现。我还没有修改你的格式，我只是调整了模板的布局。您可能应该使用/root或类似的内容替换第一个模板中的BusStops。我无法详细说明这一点，因为您没有提供完整的XML文件。但这对你来说是微不足道的。我还将已排序的for-each替换为已排序的apply-templates。它运作良好。如果您不希望结果按CommonName排序，而是按LocalityName排序，请将其替换。或者添加第二个图层，按CommonName排序，然后按LocalityName排序。这个解决方案应该具有指导意义。
XML文件
<?xml version="1.0" encoding="utf-8"?>
<?xml-stylesheet type="text/xsl" href="a.xslt"?>
<root>
<BusStopDetails>
<AtcoCode>639070021
</AtcoCode>
<CommonName>Union Square Bus Station
</CommonName>
<LocalityName>Aberdeen
</LocalityName>
<ParentLocalityName/>
<Latitude>57.1445763077
</Latitude>
<Longitude>-2.0963111958
</Longitude>
</BusStopDetails>
<BusStopDetails>
<AtcoCode>639080008
</AtcoCode>
<CommonName>Aberdeen Railway Station
</CommonName>
<LocalityName>Aberdeen
</LocalityName>
<ParentLocalityName/>
<Latitude>57.1437934462
</Latitude>
<Longitude>-2.0980112638
</Longitude>
</BusStopDetails>
-<BusStopDetails>
<AtcoCode>9300ABA
</AtcoCode>
<CommonName>Aberdeen Ferry Terminal
</CommonName>
<LocalityName>Aberdeen
</LocalityName>
<ParentLocalityName/>
<Latitude>57.1447953603
</Latitude>
<Longitude>-2.0917010936
</Longitude>
</BusStopDetails>
</root>
XSLT文件
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:template match="/root">
<html>
<body>
<h2>BusStops</h2>
<table border="1">
<tr bgcolor="#9acd32">
<th>Locality</th>
<th>BusStop</th>
<th>Coordinates </th>
</tr>
<xsl:apply-templates select="BusStopDetails">
<xsl:sort select="LocalityName" order="ascending" />
<xsl:sort select="CommonName" order="ascending" />
</xsl:apply-templates>
</table>
</body>
</html>
</xsl:template>
<xsl:template match="BusStopDetails"> <!-- in your setup you may use 'BusStops/BusStopDetails' -->
<tr>
<td> <xsl:value-of select="LocalityName"/> (<xsl:value-of select="ParentLocalityName"/>) </td>
<td><xsl:value-of select="CommonName"/></td>
<td>
<a href="https://maps.google.com/maps?ll={Latitude},{Longitude}&q={Latitude},{Longitude}&hl=en&t=m&z=14">
(<xsl:value-of select="Latitude"/>) , (<xsl:value-of select="Longitude"/>)
</a>
</td>
</tr>
</xsl:template>
</xsl:stylesheet>
答案 2 :(得分：0)
这是一个简单的XSL 1.0解决方案/想法，使用for-each和key。 1.生成密钥：
<xsl:key name="kLocalityName" match="BusStopDetails" use="LocalityName"/>
2遍历小组：
<xsl:for-each select="BusStopDetails[generate-id() =
generate-id(key('kLocalityName', LocalityName)[1])]" >
遍历每个组成员：
<xsl:for-each select="key('kLocalityName', $this)" >
更新
试试这个（没有html输出 - 因为我不知道它应该如何锁定）
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" indent="yes" />
<xsl:key name="kLocalityName" match="BusStopDetails" use="LocalityName"/>
<xsl:template match="/*">
<!--for-each select="BusStops/BusStopDetails" group >-->
<xsl:for-each select="BusStopDetails[generate-id() =
generate-id(key('kLocalityName', LocalityName)[1])]" >
<xsl:sort select="LocalityName"/>
<BusStopGroup>
<gname>
<xsl:value-of select="LocalityName"/>
</gname>
<xsl:variable name="this" select="LocalityName" />
<xsl:for-each select="key('kLocalityName', $this)" >
<BusStopDetails>
<xsl:value-of select="CommonName"/>
</BusStopDetails>
</xsl:for-each>
</BusStopGroup>
</xsl:for-each>
</xsl:template>
</xsl:stylesheet>
将生成：
<BusStopGroup>
<gname>Aberdeen</gname>
<BusStopDetails>Union Square Bus Station</BusStopDetails>
<BusStopDetails>Aberdeen Railway Station</BusStopDetails>
<BusStopDetails>Aberdeen Ferry Terminal</BusStopDetails>
</BusStopGroup>
现在使用html更新2：
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" indent="yes" />
<xsl:key name="kLocalityName" match="BusStopDetails" use="LocalityName"/>
<xsl:template match="/*">
<html>
<body>
<h2>BusStops</h2>
<table border="1">
<tr bgcolor="#9acd32">
<th>Locality</th>
<th>BusStop</th>
<th> Co-ordinates </th>
</tr>
<!--for-each select="BusStops/BusStopDetails" group >-->
<xsl:for-each select="BusStopDetails[generate-id() =
generate-id(key('kLocalityName', LocalityName)[1])]" >
<xsl:sort select="LocalityName"/>
<xsl:variable name="this" select="LocalityName" />
<xsl:for-each select="key('kLocalityName', $this)" >
<tr>
<xsl:if test="position()=1">
<td rowspan="{count(key('kLocalityName', $this))}">
<xsl:value-of select="LocalityName"/> (<xsl:value-of select="ParentLocalityName"/>) </td>
</xsl:if>
<td><xsl:value-of select="CommonName"/></td>
<td>
<a href="https://maps.google.com/maps?ll={Latitude},{Longitude}&q={Latitude},{Longitude}&hl=en&t=m&z=14">
(<xsl:value-of select="Latitude"/>) , (<xsl:value-of select="Longitude"/>)
</a>
</td>
</tr>
</xsl:for-each>
</xsl:for-each>
</table>
</body>
</html>
</xsl:template>
</xsl:stylesheet>