Re: [xsl] dateTime conversion

Subject: Re: [xsl] dateTime conversion
From: sudheshna iyer <sudheshnaiyer@xxxxxxxxx>
Date: Sun, 3 Oct 2010 16:35:22 -0700 (PDT)
Michael,

I am using xslt 1.0. I guess I need to find more from requirements.
My xsd has xs:datetime variable but I am receiving "2010-10-03T18:15:20-0400"
from my client and hence the validation is failing.  In xmlspy, if I auto
generate the xml from xsd, sample value for xs:datetime is in format
"2001-12-17T09:30:47Z". So I thought I might have to convert
"2010-10-03T18:15:20-0400" to format "2001-12-17T09:30:47Z". 

But the number
in the TO format is just a sample. I am not sure what is the efficient way to
convert date that is in format "2010-10-03T18:15:20-0400" to format
"2001-12-17T09:30:47Z"


--- On Sun, 10/3/10, Michael Kay <mike@xxxxxxxxxxxx>
wrote:

> From: Michael Kay <mike@xxxxxxxxxxxx>
> Subject: Re: [xsl] dateTime
conversion
> To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
> Date: Sunday, October 3,
2010, 7:07 PM
> 
> > Please ignore the portion that is before the timezone
>
and convert
> > T18:15:20-0400 to T09:30:47Z
> > 
> > Numbers are just the
samples. But if I take a closure
> look at them, -0400 needs to be replaced
with "Z"
> > 
> 
> Do you want to convert 12:00:00-04:00 to 08:00:00Z (the
>
same time instant in a different time zone) or to
> 12:00:00Z?
> 
> And are
you using XSLT 1.0 or 2.0?
> 
> Note: your date-time values are almost in ISO
8601 format,
> except for the missing colon between hours and minutes in
> the
timezone offset. Was this difference deliberate?
> 
> Michael Kay
> Saxonica
>
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