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Old April 3rd, 2007, 06:56 AM
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Default seconds-from-duration


I'm fairly new to XSLT and I'm trying to find an example on how to use seconds-from-duration. Basically I have a field in one XML document that looks like this :


and when I transform it, I want it to come out looking like this:


... but finding documentation on using seconds-from-duration that is fit for human use is proving to be utterly impossible. Is there a simple way to use this function?


Old April 3rd, 2007, 07:05 AM
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Here's an example:

  <xsl:value-of select="seconds-from-duration(xs:dayTimeDuration(preview_start_time))"/>

With your example that will give you 15 rather than 65. I don't know where you expect 65 to come from. If you want the total number of seconds (1 min 15 secs = 75 secs), you don't want seconds-from-duration (which will give you 15), you want to do

xs:dayTimeDuration(preview_start_time) idiv xs:dayTimeDuration('PT1S')

Remember that you need an XSLT 2.0 processor and that in the above examples you need to declare the xs namespace.

Michael Kay
Author, XSLT Programmer's Reference and XPath 2.0 Programmer's Reference
Old April 3rd, 2007, 08:51 AM
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Thanks very much for the helpful response. I'm sure I'll be back once I have a 2.0 processor installed.

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