如何在xslt中访问变量? [英] how to access variable in xslt?
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问题描述
能否请您告诉我如何在xslt中访问变量?我在js
中创建变量.我想打印它的值.所以我在xsl:variable中分配了变量值
当我尝试打印变量值时,它给了我string
而不是求值
这是我的代码
http://xsltransform.net/pPJ8LWb
could you please tell me how to access variable in xslt ? .I make variable in js
.I want to print it's value .So I assign the variable value in xsl:variable
when I try to print variable value it gives me string
instead of evaluating
here is my code
http://xsltransform.net/pPJ8LWb
<?xml version="1.0" encoding="UTF-8" ?>
<xsl:transform xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="2.0">
<xsl:output method="html" doctype-public="XSLT-compat" omit-xml-declaration="yes" encoding="UTF-8" indent="yes" />
<xsl:template match="/">
<hmtl>
<head>
<title>New Version!</title>
</head>
<script>
console.log('starte');
var start = 5;
'<xsl:variable name="start">{start}</xsl:variable>'
console.log('<xsl:value-of select="$start"/>')
</script>
</hmtl>
</xsl:template>
</xsl:transform>
预期输出 :: 5 任何更新
Expected output :: 5 any update
推荐答案
好吧,我建议您这样做:
Well, i can suggest you do like this:
演示- https://jsfiddle.net/skyr9999/behf2h7t/
<?xml version="1.0" encoding="UTF-8" ?>
<xsl:transform xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="2.0">
<xsl:output method="html" doctype-public="XSLT-compat" omit-xml-declaration="yes" encoding="UTF-8" indent="yes" />
<xsl:template match="/">
<hmtl>
<head>
<title>New Version!</title>
</head>
<script>
console.log('starte');
var start = 5;
document.write('<xsl:variable name="start">'+start+'</xsl:variable>');
console.log("<xsl:value-of select='"+start+"'/>")
</script>
</hmtl>
</xsl:template>
</xsl:transform>
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