In XSLT replace value - xslt

Question

I have this XML document :Now I want to replace LineNo so that the output will be line no will 1 ,2 . I have tried some thing like this.
<xsl:value-of select="replace( '000010',1)"/>
<Rder>
<Order>
<OrderNo>458</OrderNo>
<LineNo>000010</LineNo>
<SerialNO>96</SerialNO>
<VNo>543</VNo>
</Order>
<Order>
<OrderNo>458</OrderNo>
<LineNo>000020</LineNo>
<SerialNO>32</SerialNO>
<VNo>543</VNo>
</Order>
</Rder>
I want to replace the value of LineNo= 000010 ,000020 by 1,2 in XSLT below one i have tried.
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">
<xsl:output method="xml" indent="yes" />
<xsl:key name="orders" match="Order" use="OrderNo" />
<xsl:template match="/*">
<SalesOrders>
<xsl:for-each select="Rder/Order[generate-id() = generate-id(key('orders', OrderNo)[1])]">
<Order VNo="{VNo}" OrderNo="{OrderNo}">
<OrderLines>
<xsl:apply-templates select="key('orders', OrderNo)" />
</OrderLines>
</Order>
</xsl:for-each>
</SalesOrders>
</xsl:template>
<xsl:template match="Order">
<OrderLine LineNo="{LineNo}" SerialNO="{SerialNO}"/>
</xsl:template>
</xsl:stylesheet>
Actually I getting those lineno details in same format i have tried couple cases its doesn't giving that expected format.
Any help would be appreciated.

Answer 1

Why don't you do simply:
<xsl:template match="Order">
<OrderLine LineNo="{position()}" SerialNO="{SerialNO}"/>
</xsl:template>
or:
<xsl:template match="Order">
<OrderLine LineNo="{number(LineNo) div 10}" SerialNO="{SerialNO}"/>
</xsl:template>