Re: Mapping from two sources

Thank you for your answers. 

But the solution returns only elements from
Doc1, I need element of Doc2 if
doc1.OLN = doc2.OLN
========
Problem

I need
to have two sources:

input1 and input2. 

input1:
<?xml version="1.0"
encoding="ISO-8859-1"?>
<Order>
	<OrderLine>
			<OLN>1</OLN>
<Fname>aa</Fname>
	</OrderLine>
	<OrderLine>
			<OLN>2</OLN>
<Fname>bb</Fname>
	</OrderLine>	
</Order>


input2:
<?xml version="1.0"
encoding="ISO-8859-1"?>
<POOrder>
	<POOrderLine>
			<OLN>1</OLN>
<ID>123</ID>
			<LName>aa</LName>
	</POOrderLine>
	<POOrderLine>
<OLN>2</OLN>
			<ID>324</ID>
			<LName>bb</LName>
	</POOrderLine>
<POOrderLine>
			<OLN>3</OLN>
			<ID>456</ID>
			<LName>bb</LName>
</POOrderLine>	
</POOrder>

I need the output from both sources combined.
Please note that first two elements are coming from input1 and thrid element
is from input2. What is the 
optimal way of doing this?

<?xml version="1.0"
encoding="ISO-8859-1"?>
<OrderResponse>
	<Oline>
		<OLN>1</OLN>
<Fname>aa</Fname>
		<ID>123</ID>
	</Oline>
	<Oline>
		<OLN>2</OLN>
<Fname>bb</Fname>
		<ID>324</ID>
	</Oline>
</OrderResponse>
=====

Solution
proposed:

<xsl:for-each-group select="$doc1//OrderLine, $doc2//POOrderline"
group-by="OLN">
<Oline>
<OLN><xsl:value-of
select="current-grouping-key()"></OLN>
<xsl:copy-of
select="current-group()/(FName, ID)"/>
</Oline>
</xsl:for-each-group>


--- On
Sat, 10/2/10, Mukul Gandhi <gandhi.mukul@xxxxxxxxx> wrote:

> From: Mukul
Gandhi <gandhi.mukul@xxxxxxxxx>
> Subject: Re:  Mapping from two sources
> To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
> Date: Saturday, October 2, 2010, 1:07
PM
> On Sat, Oct 2, 2010 at 4:29 PM,
> sudheshna iyer
<sudheshnaiyer@xxxxxxxxx>
> wrote:
> > does xsl:for-each-group acts like inner
join?
> 
> Well I think, for-each-group cannot be treated as inner
> join
(which is
> a relational algebra abstraction, as Mike also says) -- we
> are
not
> joining any two sets with for-each-group. The input to
> for-each-group
> is just a sequence, with definition of grouping key (for
> e.g, the
> value
of group-by attribute).
> 
> I would imagine for-each-group as a kind of
mapping
> function,
> something similar to for-each but with a kind of layer
of
> grouping on
> top of that.
> 
> 
> 
> -- 
> Regards,
> Mukul Gandhi
> 
>
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