xml文件具有以下结构
<?xml version="1.0" encoding="utf-8"?>
<?xml-stylesheet type="text/xsl" href="TestFile.xsl"?>
<RootElement>
<Date FileModified="7/2/2010 12:54:53 PM" />
<Child Name="A"/>
<Child Name="B"/>
<Child Name="C"/>
<Child Name="D"/>
<Child Name="E"/>
</RootElement>
我需要从文件中读取日期属性值并传递给
var d=new Date( date );
这是我的xsl文件
<?xml version="1.0" encoding="ISO-8859-1"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:template match="/">
<html>
<head></head>
<body>
<center>
<b>
<script type="text/javascript">
var d_names = new Array("Sunday", "Monday", "Tuesday","Wednesday", "Thursday", "Friday", "Saturday");
var m_names = new Array("January", "February", "March","April", "May", "June", "July", "August","September","October","November", "December");
var d = new Date(-----------); // here i need to get the date from the xsl file
var curr_day = d.getDay();
var curr_date = d.getDate();
var sup = "";
if (curr_date == 1 || curr_date == 21 || curr_date ==31)
{
sup = "st";
}
else if (curr_date == 2 || curr_date == 22)
{
sup = "nd";
}
else if (curr_date == 3 || curr_date == 23)
{
sup = "rd";
}
else
{
sup = "th";
}
var curr_month = d.getMonth();
var curr_year = d.getFullYear();
var curr_hour=d.getHours();
var curr_period="AM"
if(curr_hour>12)
{
curr_hour=curr_hour-12;
curr_period="PM"
}
document.write(d_names[curr_day]+ "     " +curr_date + "<sup>"+ sup + " </sup> " + m_names[curr_month] + "  " + curr_year+ "      "+curr_hour+" : "+d.getMinutes()+" : "+d.getSeconds()+" "+curr_period);
</script>
</b>
<xsl:choose>
<xsl:when test="//Child">
<br/>
<br/>
<br/>
<b>SampleTable</b>
<br/>
<br/>
<table border="1">
<tr bgcolor="RGB(0,0,127)" >
<th width="5">
<font color="white">S.No</font>
</th>
<th width="250">
<font color="white"> Name</font>
</th>
</tr>
<xsl:for-each select ="//Child">
<tr>
<td>
<xsl:number value="position()" format="01"/>
</td>
<td>
<xsl:value-of select="@Name"/>
</td>
</tr>
</xsl:for-each>
</table>
</xsl:when >
<xsl:otherwise ></xsl:otherwise>
</xsl:choose >
</center>
</body >
</html>
</xsl:template>
</xsl:stylesheet>
有没有办法做到这一点
答案 0 :(得分：1)
var d = new Date('<xsl:value-of select="RootElement/Date/@FileModified"/>')
答案 1 :(得分：0)
此外，您可以使用纯XSLT获得de desire结果，如下所示：
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:variable name="m_names">
<January/><February/><March/><April/><May/><June/><July/><August/><September/><October/><November/><December/>
</xsl:variable>
<xsl:variable name="d_names">
<Sunday/><Monday/><Tuesday/><Wednesday/><Thursday/><Friday/><Saturday/>
</xsl:variable>
<xsl:variable name="modules" select="'033614625035034025036146'"/>
<xsl:variable name="date" select="/*/*/@FileModified"/>
<xsl:variable name="day" select="substring-before($date,'/')"/>
<xsl:variable name="month" select="substring-before(substring-after($date,'/'),'/')"/>
<xsl:variable name="year" select="substring($date,7 + string-length($date) - 22,4)"/>
<xsl:variable name="calc-day" select="(($year - 1) mod 7 +
(floor(($year - 1) div 4) - 3 * floor((floor(($year - 1) div 100) + 1) div 4)) mod 7 +
substring($modules,
12 * ($year mod 4 = 0 and $year mod 100 != 0) + $month,
1) +
$day mod 7)
mod 7 + 1"/>
<xsl:template match="/">
<html>
<head></head>
<body>
<center>
<b>
<xsl:value-of select="concat(
name(document('')/*/xsl:variable[@name='d_names']/*[$calc-day]),
'     ',
$day
)"/>
<sup>
<xsl:choose>
<xsl:when test="$day = 1 or $day = 21">st</xsl:when>
<xsl:when test="$day = 2 or $day = 22">nd</xsl:when>
<xsl:when test="$day = 3 or $day = 23">rd</xsl:when>
<xsl:otherwise>th</xsl:otherwise>
</xsl:choose>
</sup>
<xsl:value-of select="concat(
'     ',
name(document('')/*/xsl:variable[@name='m_names']/*[number($month)]),
'  ',
$year,
'     ',
substring($date,12 + string-length($date) - 22,2),
' : ',
substring($date,15 + string-length($date) - 22,2),
' : ',
substring($date,18 + string-length($date) - 22,2),
' ',
substring($date,string-length($date)-1)
)"/>
</b>
<xsl:apply-templates/>
</center>
</body >
</html>
</xsl:template>
<xsl:template match="Child[1]">
<br/>
<br/>
<br/>
<b>SampleTable</b>
<br/>
<br/>
<table border="1">
<tr bgcolor="RGB(0,0,127)" >
<th width="5">
<font color="white">S.No</font>
</th>
<th width="250">
<font color="white"> Name</font>
</th>
</tr>
<xsl:apply-templates select="../Child" mode="child"/>
</table>
</xsl:template>
<xsl:template match="Child" mode="child">
<tr>
<td>
<xsl:number value="position()" format="01"/>
</td>
<td>
<xsl:value-of select="@Name"/>
</td>
</tr>
</xsl:template>
</xsl:stylesheet>
注意：如果您使用更好的日期格式，日期提取可能会更简洁。