Subject: Re: [xsl] Date-Format: How can i use this?? From: Joerg.Heinicke@xxxxxx Date: Wed, 30 Jan 2002 12:06:36 +0100 (MET) |
Hi Asim, you can use the substring()-function: <xsl:template match="OBX"> <xsl:text>Datum: </xsl:text> <xsl:value-of select="substring(@ObservationDT,9,2)"/> <xsl:text>.</xsl:text> <xsl:value-of select="substring(@ObservationDT,6,2)"/> <xsl:text>.</xsl:text> <xsl:value-of select="substring(@ObservationDT,1,2)"/> <xsl:text>
Uhrzeit: </xsl:text> <xsl:value-of select="substring(@ObservationDT,12,5)"/> </xsl:template> Maybe you have to change the exact numbers, but that's the way. Your second question: <xsl:value-of select="translate(@Unit_ID, 'U', 'L')"/> Regards, Joerg > Hi, > > i have this XML: > > <tdPAVSelection> > <OBX ObservationDT="2001-11-04T08:00:00" Observation_ID="2157-6" > ValueType_ID="NM" Value="105" Unit_ID="U/L" AbnormFlag_ID="H" > ResultState_ID="F" FootnoteID="29"/> > </tdPAVSelection> > > But i need this Result for ObservationDT: > > Datum : 04.11.2001 > Uhrzeit : 08:00 > > My second question: > > I have in Tag Unit_ID = "U/L". But i want Replace U with ?. So i have > ?/L instead of U/L. > > Thanks for Help for a newbie > > Asim -- GMX - Die Kommunikationsplattform im Internet. http://www.gmx.net XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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