我有一个XML代码,应该在HTML5表格中显示XSL样式表,但我不确定我是否正朝着正确的方向前进
XSD代码是否会影响样式表是否在HTML5表中显示XML内容或者是否没有相关性
<?xml version = "1.0"?>
<?xml-stylesheet type = "text/xsl" href = "nutrition.xsl"?>
<nutrition:items xmlns:nutrition = "http://www.grandmascookies.com/nutrition">
<product name = "Grandma White's Cookies">
<item>
<servingsize>
<amount> 1 </amount>
<unit> package </unit>
</servingsize>
<calories>
<amount> 260 </amount>
<unit> calories </unit>
</calories>
<caloriesFat>
<amount> 100 </amount>
<unit> grams </unit>
</caloriesFat>
<gramsFat>
<amount> 11 </amount>
<unit> grams </unit>
</gramsFat>
</item>
</product>
</nutrition:item>
<xsl:stylesheet version = "1.0"
xmlns:xsl = "http://www.w3.org/1999/XSL/Transform">
<xsl:output method = "html" doctype-system = "about:legacy-compat"/>
<xsl:template match = "/">
<html xmlns = "http://www.w3.org/1999/xhtml">
<head>
<meta charset = "utf-8">
<link rel = "stylesheet" type = "text/css" href = "style.css"/>
<title> Cookies </title>
</head>
<body>
<table border = "1" syle = "background-color: blue">
<thead>
<tr>
<th> Calories</th>
<th> Fat - Calories</th>
<th> Fat - Grams </th>
</tr>
</thead>
</table>
</body>
</html>
</xsl:template>
</xsl:stylesheet>
答案 0 :(得分:0)
XSD是无关紧要的,除非你想验证xml输入 - 完全从XSL中删除。
我认为你对XSL有误解。 HTML5表格不显示XSL,它正在生成HTML(或任何您想要的其他内容)。
您的样式表与根节点(“/”,在您的情况下为营养:项目)匹配,并在每次找到时生成代码(即一次)。现在,您希望在正确的位置匹配表格中某一行的每个项目:
</thead>
<xsl:for-each select="./product/item">
<!-- Generate html table row -->
<tr>
<!-- Calories (relative to current node) -->
<td><xsl:value-of select="calories/amount"/></td>
<!-- Repeat for other fields. Probably want the name somewhere, too -->
</tr>
</xsl:for-each>
</table>
您还可以使用&lt; xsl:apply-templates ...&gt;和&lt; xsl:template-match select =“calories / amount”&gt;