RE: [xsl] Date-Format: How can i use this??

["Andrew Welch" <andrew@xxxxxxxxxxxxxxxxxxxxxxx>]

Hi,

To replace to U with ? try:

select="translate(@Unit_ID,'U','?')"

and to change the date format try:

<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,1,4)"/>

etc...

hth's

cheers

andrew

===


-----Original Message-----
From: owner-xsl-list@xxxxxxxxxxxxxxxxxxxxxx
[mailto:owner-xsl-list@xxxxxxxxxxxxxxxxxxxxxx]On Behalf Of Asim Tozlu
Sent: Wednesday, January 30, 2002 10:05 AM
To: XSL-List@xxxxxxxxxxxxxxxxxxxxxx
Subject: [xsl] Date-Format: How can i use this??


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



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