Subject: URGENT : Substitue value of xsl:variable into output !!! From: "Mahesh Nanavate" <mahesh.nanavate@xxxxxxxxxxx> Date: Tue, 9 May 2000 22:03:27 +0530 |
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
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