我正在编写一个XSL转换程序来将XML转换为HTML。这是我的枪.xml:
<?xml version="1.0" encoding="utf-8"?>
<?xml-stylesheet type="text/xsl" href="guns.xslt"?>
<guns xsi:noNamespaceSchemaLocation="guns2.xsd"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
<gun>
<model>revolver</model>
<handy>1</handy>
<origin>Britain</origin>
<ttc>20mm</ttc>
</gun>
</guns>
这里是guns.xslt:
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:template match = "/">
<html>
<body>
<h2>Gun Collection</h2>
<table border = "1">
<tr bgcolor = "#9acd32">
<th>Model</th>
<th>Origin</th>
<th>TTC</th>
</tr>
<xsl:for-each select="guns/gun">
<tr>
<td><xsl:value-of select = "model"/></td>
<td><xsl:value-of select = "handy"/></td>
<td><xsl:value-of select = "origin"/></td>
<td><xsl:value-of select = "ttc"/></td>
</tr>
</xsl:for-each>
</table>
</body>
</html>
</xsl:template>
</xsl:stylesheet>
两个文件都位于同一个包中,但XPAth无法识别
"guns/gun"
块中的for-each
。我错过了什么? 最佳答案
您已在xml样式表中声明为“text/xsl”,请将文件扩展名从“guns.xslt”更改为“guns.xsl”,并将xml更改如下:
<?xml-stylesheet type="text/xsl" href="guns.xsl"?>
那么您的结果如下: