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

[Joerg dot Heinicke at gmx dot de]

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>&#xa;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

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