Quick comment, Harsh himself pointed this out recently.
In your Muenchian grouping condition:
generate-id() = generate-id(key('x', @key)[1])
The [1] is actually redundant, as generate-id() will only take the first
node returned. So you can use this:
generate-id() = generate-id(key('x', @key))
~ Scott
-----Original Message-----
From: Mukul Gandhi [mailto:gandhi(_dot_)mukul(_at_)gmail(_dot_)com]
Sent: Tuesday, October 09, 2007 11:25 AM
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: Re: [xsl] Add numbers
Hi Harsh,
Though the question has already been answered, but here's my attempt:
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
version="1.0">
<xsl:output method="xml" indent="yes" />
<xsl:key name="x" match="a" use="@key" />
<xsl:template match="/first">
<op>
<xsl:for-each select="second/a[generate-id() =
generate-id(key('x', @key)[1])]">
<xsl:element name="{(_at_)key}">
<val><xsl:value-of select="sum(key('x', @key)/@val)" /></val>
</xsl:element>
</xsl:for-each>
</op>
</xsl:template>
</xsl:stylesheet>
Ken's solution is quite generic, because it doesn't depend on element
name, and can work with deep hierarchies.
On 10/9/07, Chaudhary, Harsh <HCHAUDHA(_at_)amfam(_dot_)com> wrote:
Hi,
I have an XML:
<?xml version="1.0" encoding="UTF-8"?>
<first>
<second>
<a val="4" key="one">b</a>
<a val="2" key="two">b</a>
</second>
<second>
<a val="3" key="one">c</a>
</second>
</first>
I need to group together the nodes which have the same key and then I
need to add the attribute "val" in all such cases.
So, The output I need is:
<op>
<one>
<val>7</val>
</one>
<two>
<val>2</val>
</two>
</op>
How do I go about doing this? I am using Xalan and XSLT 1.0.
I have used the Meunichian method to group the nodes with same keys
together. But I don't know how to proceed from there. I tried using
the
sum() function but it won't work for me on account that even if I run
a
for-each loop over the set of nodes with the same key, it just prints
out the value of "val" for each a element. I think I need to be in the
parent element for this to work.
Thanks,
Harsh.
--
Regards,
Mukul Gandhi
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