Reputation: 27
My XML:
<root>
<Name>Test</Name>
<DOB>10-11-1989</DOB>
<root>
I need to get current time date format with milliseconds by using XSLT. I am using XSLT version-1.0..
After Trasformation my XML should look like below:
<root>
<Name>Test</Name>
<DOB>10-11-1989</DOB>
<currentDate>2014-05-21-01.25.32.000000</currentDate>
<root>
I found solutions for version-2.0 but, that doesn't help me.
Upvotes: 1
Views: 5472
Reputation: 7173
Here is an example, using Java, that works with Xalan and is default available in the ORACLE JRE.
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:java="http://xml.apache.org/xslt/java"
exclude-result-prefixes="java">
<xsl:output indent="yes"/>
<xsl:strip-space elements="*"/>
<xsl:template match="/root">
<xsl:copy>
<xsl:copy-of select="@*|node()"/>
<dateTimeStamp>
<xsl:value-of select="java:format(java:java.text.SimpleDateFormat.new('yyyy-MM-dd-HH.mm.ss.SSSSSS a'), java:java.util.Date.new())" />
</dateTimeStamp>
</xsl:copy>
</xsl:template>
</xsl:stylesheet>
From http://www.heber.it/?p=1053.
If you want to have 12-hr format, simply change
<xsl:value-of select="java:format(java:java.text.SimpleDateFormat.new('yyyy-MM-dd-HH.mm.ss.SSSSSS'), java:java.util.Date.new())" />
to
<xsl:value-of select="java:format(java:java.text.SimpleDateFormat.new('yyyy-MM-dd-hh.mm.ss.SSSSSS a'), java:java.util.Date.new())" />
it outputs:
<root>
<Test>tested</Test>
<dateTimeStamp>2014-05-23-01.57.04.000441 PM</dateTimeStamp>
</root>
Upvotes: 0
Reputation: 116992
XSLT 1.0 alone does not provide any means to get the current date or time.
If your processor supports it, you can use the date-time() EXSLT extension function - however, I don't think this will include milliseconds.
Alternatively, you could pass the timestamp as a parameter to the stylesheet during runtime.
Upvotes: 1