Subject: Re: [xsl] dateTime conversion From: Michael Kay <mike@xxxxxxxxxxxx> Date: Mon, 04 Oct 2010 21:44:02 +0100 |
You may indeed be wrong, and so might I. He didn't explain the problem very clearly. I did try to ask whether he was after a timezone conversion, and the response was that he seemed to want to change the timezone to Z because that's the only way he knew how to make it valid against the schema.I might be wrong but the problem that sudheshna iyer is facing may not only be date time format, but convert the time to a UTC time zone.
Is there a function in XSLT to convert localtime to UTC time?
Michael Kay Saxonica
Ubi Nunes Topaz Solutions LTD.
-----Original Message----- From: Michael Kay [mailto:mike@xxxxxxxxxxxx] Sent: Monday, 4 October 2010 9:30 p.m. To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx Subject: Re: [xsl] dateTime conversion
If a client is sending you data that doesn't conform to the agreed schema, then you should really persuade your client to correct the program that is generating the data. If you relent, and accept invalid data, then the schema has failed in its purpose of defining the agreement between you and your client as to how data gets exchanged; if they ignore the schema now, they will continue to ignore it in the future, and your problems in repairing the data can only get worse.
However, if for political reasons you have to repair the data you are sent, the obvious repair is to insert the missing colon in the timezone so that -0400 becomes -04:00. This can be done simply using concat() and
substring().
Michael Kay Saxonica
On 04/10/2010 12:35 AM, sudheshna iyer wrote: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 whatis 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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