0
這裏是我的XSLT:風格標籤不渲染
<?xml version="1.0" encoding="ISO-8859-1"?>
<xsl:stylesheet version="1.1" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="html" encoding="UTF-8" omit-xml-declaration="yes" doctype-system="about:legacy-compat" />
<xsl:template match="/">
<html>
<xsl:call-template name="header">
<xsl:with-param name="title">Strip Club List - Top 100
<xsl:choose>
<xsl:when test="/*/general/viewmethod='numcomments'">
Highest Number of Comments
</xsl:when>
<xsl:when test="/*/general/viewmethod='numreviews'">
Highest Number of Reviews
</xsl:when>
<xsl:when test="/*/general/viewmethod='highestreviews'">
Highest Rating
</xsl:when>
<xsl:when test="/*/general/viewmethod='numlikes'">
Highest Number of Likes
</xsl:when>
<xsl:when test="/*/general/viewmethod='numdislikes'">
Highest Number of Dislikes
</xsl:when>
<xsl:when test="/*/general/viewmethod='numfollowers'">
Highest Number of Followers
</xsl:when>
</xsl:choose>
<xsl:value-of select="/*/locations/name" /></xsl:with-param>
<xsl:with-param name="stylesheets">fonts.css,core.css,state.css,top.css</xsl:with-param>
<xsl:with-param name="scripts">core.js,state.js</xsl:with-param>
<xsl:choose>
<xsl:when test="/*/general/viewmethod='numcomments'">
<style>
.linked_location .rating {
right: 110px;
}
</style>
</xsl:when>
<xsl:when test="/*/general/viewmethod='numreviews'">
<style>
.linked_location .rating {
right: 100px;
}
</style>
</xsl:when>
</xsl:choose>
</xsl:call-template>
<body>
<div id="body"></div>
</body>
</html>
</xsl:template>
</xsl:stylesheet>
它在HTML文檔的頭部被渲染。問題是它沒有渲染到頁面上。我相信這與在XSLT document
中使用style tags
有關。
XML:
<root>
<general>
<viewmethod>numreviews</viewmethod>
</general>
</root>
我也只是嘗試:
<xsl:choose>
<xsl:when test="/*/general/viewmethod='numcomments'">
<style>
.linked_location .rating {
right: 110px;
}
</style>
</xsl:when>
<xsl:when test="/*/general/viewmethod='numreviews'">
<style>
.linked_location .rating {
right: 100px;
}
</style>
</xsl:when>
<xsl:when test="/*/general/viewmethod='highestreviews'">
<style>
.linked_location .rating {
right: 100px;
}
</style>
</xsl:when>
<xsl:when test="/*/general/viewmethod='numlikes'">
<style>
.linked_location .rating {
right: 90px;
}
</style>
</xsl:when>
<xsl:when test="/*/general/viewmethod='numdislikes'">
<style>
.linked_location .rating {
right: 90px;
}
</style>
</xsl:when>
<xsl:when test="/*/general/viewmethod='numfollowers'">
<style>
.linked_location .rating {
right: 90px;
}
</style>
</xsl:when>
</xsl:choose>
這裏是我的header template
<xsl:template name="header">
<xsl:param name="title" />
<xsl:param name="keywords" />
<xsl:param name="description" />
<xsl:param name="stylesheets" />
<xsl:param name="scripts" />
<xsl:param name="emp" />
<head>
<xsl:if test="string-length($keywords) > 0"><meta name="keywords" content="{$keywords}" /></xsl:if>
<xsl:if test="string-length($description) > 0"><meta name="description" content="{$description}" /></xsl:if>
<xsl:if test="string-length($stylesheets) > 0"><xsl:call-template name="headercss"><xsl:with-param name="stylesheets" select="$stylesheets" /></xsl:call-template></xsl:if>
<xsl:if test="string-length($scripts) > 0"><xsl:call-template name="headerjs"><xsl:with-param name="scripts" select="$scripts" /></xsl:call-template></xsl:if>
<script src="http://domain/www/delivery/i.php?id=13&blockcampaign=1&target=_blank"></script>
<xsl:if test="$emp > 0">
<link rel="stylesheet" type="text/css" href="/emp/style.css" />
<script src="/emp/gl.js"></script>
<script src="/emp/tween.js"></script>
<script src="/emp/emp.js"></script>
</xsl:if>
<title><xsl:value-of select="$title" /></title>
</head>
</xsl:template>
感謝
http://stackoverflow.com/help/mcve –
@ michael.hor257k請看更新的問題 – jkushner
你真的不明白的問題不能轉載只有該片段去? –