How to format a date using XSLT?

Question

Trying to convert date to specific format but getting error:

<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns="http://www.openapplications.org/oagis/9"
xmlns:xs="http://www.w3.org/2001/XMLSchema">

<xsl:variable name="origDate" select="row[@num='6']/cell[@num='2']"/>
<PromisedDeliveryDateTime><xsl:value-of select="format-date(xs:date($origDate), '[Y] [M] [D]')" /></PromisedDeliveryDateTime>

But I get the error:

Failed to transform XQMessage part, index=0 (Invalid date "30.11.2011"
(Non-numeric component)

Michael Kay · Accepted Answer · 2012-09-10 08:04:14Z

3

The xs:date() constructor function expects a string in ISO date format: 2011-11-30. To manipulate dates in other formats you first need to convert them to ISO format, which you can do for example as

string-join(reverse(tokenize($in, '.')), '-')

answered Sep 10, 2012 at 8:04

Michael Kay

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2 Comments

VextoR Over a year ago

thanks! But still something wrong:

<xsl:variable name="origDate" select="row[@num='6']/cell[@num='2']"/> <xsl:variable name="origDate" select="string-join(reverse(tokenize($origDate, '.')), '-')"/> <PromisedDeliveryDateTime><xsl:value-of select="format-date(xs:date($origDate), '[Y] [M] [D]')" /></PromisedDeliveryDateTime>

Failed to transform XQMessage part, index=0 (Invalid date "----------" (Year is less than four digits)

Sean B. Durkin Over a year ago

@VextoR No-one is going to read XML in the comment feed. Update your question and also mention what XSLT processor your are using.

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