Subject: Re: [xsl] dateTime conversion From: Michael Kay <mike@xxxxxxxxxxxx> Date: Mon, 04 Oct 2010 09:29:58 +0100 |
Michael Kay Saxonica
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 timezoneand convertT18:15:20-0400 to T09:30:47Z
Numbers are just the samples. But if I take a closurelook 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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