This is the mail archive of the
xsl-list@mulberrytech.com
mailing list .
URGENT : Substitue value of xsl:variable into output !!!
- To: <XSL-List at mulberrytech dot com>
- Subject: URGENT : Substitue value of xsl:variable into output !!!
- From: "Mahesh Nanavate" <mahesh dot nanavate at ecapsol dot com>
- Date: Tue, 9 May 2000 21:44:47 +0530
- Reply-To: xsl-list at mulberrytech dot com
Hi Everybody,
I have one problem :
Consider the following XML snippet:
---
---
<ItemList>
<oneItem>
<ItemID>1212</ItemID>
<ItemName>XYZ</ItemName>
<Qty>1</Qty>
</oneItem>
</ItemList>
---
---
Consider the following XSL snippet:
---
---
<xsl:template match="oneItem">
<p><xsl:value-of select="ItemName" /><br/>
<xsl:variable name="inVar" select="concat('q',ItemID)"/>
<input name="$inVar" value="{Qty}" type="number"/> <br/>
<anchor>Change
<go
ref="http://10.10.10.94:7001/servlet?mySvc=myApp&myP1={ItemID}&myP2=
$inVar"></go>
</anchor>
</p>
</xsl:template>
---
---
My Problem is that I am not able to substitute the value of the
xsl:variable with name inVar
into 'name' attribute of the input element.
Also I require that the myP2 be assigned a value of concat('$',value of
inVar).
The depicted code is not substituting the value of inVar instead it's
giving the string as it is.
Any help in this regard is most appreciated ,
Thanks in advance,
Mahesh.
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list