 [求助]新人求助如何用xslt 去显示有schema 的 xml 请各位教教小第!!!
请各位教教小第! 如何用xslt 去显示有schema 的 xml 。 我用vs 2005 做了xml, xslt 同 schema file , 但不能显示到xslt资料(table , css )。 ///-------------------students.xml-------------------------------//////
<?xml version="1.0" encoding="utf-8"?>
<?xml-stylesheet type="text/xsl" href="students.xsl"?>
<students xmlns="http://www.w3schools.com" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:noNamespaceSchemaLocation="students.xsd">
<student>
<firstname>Chung Ting</firstname>
<lastname>Au Yeung</lastname>
<ulid>ctauyeung</ulid>
<account>xn352104</account>
<class>SENIOR</class>
</student>
<student>
<firstname>Ho Ting</firstname>
<lastname>Chan</lastname>
<ulid>htchan</ulid>
<account>xn352111</account>
<class>JENIOR</class>
</student>
</students>
///-------------------------students.xsd------------------------------------///
<?xml version="1.0" encoding="utf-8"?>
<xs:schema attributeFormDefault="unqualified" elementFormDefault="qualified" xmlns:xs="http://www.w3.org/2001/XMLSchema">
<xs:element name="students">
<xs:complexType>
<xs:sequence>
<xs:element maxOccurs="unbounded" name="student">
<xs:complexType>
<xs:sequence>
<xs:element name="firstname" type="xs:string" />
<xs:element name="lastname" type="xs:string" />
<xs:element name="ulid" type="xs:string" />
<xs:element name="account" type="xs:string" />
<xs:element name="class" type="xs:string" />
</xs:sequence>
</xs:complexType>
</xs:element>
</xs:sequence>
</xs:complexType>
</xs:element>
</xs:schema> //---------------------------students.xsl---------------------------------///
<?xml version="1.0" encoding="utf-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:xs="http://www.w3.org/2001/XMLSchema">
<xsl output method="html" encoding="utf-8" indent="no"/>
<!--<xsl:import-schema schema-location="students.xsd"/>-->
<xsl:template match="xs:element[@name='students']">
<html>
<body>
<table cellspacing="0">
<tr bgcolor="#FFFFFF">
<th>FIRSTNAME</th>
<th>LASTNAME</th>
<th>ULID</th>
<th>ACCOUNT</th>
<th>CLASS</th>
</tr>
<xsl:for-each select="xs:element[@name='student']">
<xsl:sort select="class"/>
<xsl:variable name="pos" select="position() mod 2"/>
<xsl:if test="$pos=1">
<tr bgcolor="#dddddd">
<td>
<xsl:value-of select="xs:firstname"/>
</td>
<td>
<xsl:value-of select="xs:lastname"/>
</td>
<td>
<xsl:value-of select="xs:ulid"/>
</td>
<td>
<xsl:value-of select="xs:account"/>
</td>
<td>
<xsl:value-of select="xs:class"/>
</td>
</tr>
</xsl:if>
<xsl:if test="$pos=0">
<tr>
<td>
<xsl:value-of select="xs:firstname"/>
</td>
<td>
<xsl:value-of select="xs:lastname"/>
</td>
<td>
<xsl:value-of select="xs:ulid"/>
</td>
<td>
<xsl:value-of select="xs:account"/>
</td>
<td>
<xsl:value-of select="xs:class"/>
</td>
</tr>
</xsl:if>
</xsl:for-each>
</table>
</body>
</html>
</xsl:template>
</xsl:stylesheet>
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XML.ORG.CN讨论区 - XML技术 → 『 XSL/XSLT/XSL-FO/CSS 』 → [求助]新人求助如何用xslt 去显示有schema 的 xml 请各位教教小第!!!
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2007/9/1 11:40:00
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