Sorting XSLT output items

[Jure Sah]

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Hello,

I have an XSLT that checks several XML files (using document() ) and
outputs their combined contents. The items inside it have a date
argument according to which I would like to sort the result. How can
this be done?

That is... the usual sort will allow you to sort values from the input
file, but since my XSLT uses several other files in succession and
processes each of their items in a nested loop, I do not see where I
should put the sort tag to achieve the desired effect.

The only solution I see is to use an XSLT sort on the XSLT file output,
but that seems silly.

Thanks for any help in advance.

LP,
Jure
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[Martin Honnen]

Jure Sah wrote:

> I have an XSLT that checks several XML files (using document() ) and
> outputs their combined contents. The items inside it have a date
> argument according to which I would like to sort the result. How can
> this be done?

<xsl:apply-templates select="document('file1.xml')/foo/bar |
document('file2.xml')/foo/bar">
<xsl:sort select="dateElement"/>
</xsl:apply-templates/>

Note that only XSLT 2.0 has support for sorting dates in the W3C XML
schema date format. With XSLT 1.0 you will need to make sure you have
the date in a format like '2009-02-28' that allows string sorting.

--

Martin Honnen
http://JavaScript.FAQTs.com/

[Jure Sah]

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Martin Honnen pravi:

> <xsl:apply-templates select="document('file1.xml')/foo/bar |
> document('file2.xml')/foo/bar">
> <xsl:sort select="dateElement"/>
> </xsl:apply-templates/>

I can't use that, my files are defined in the XML file I am processing.

Any additional ideas?

LP,
Jure
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[Martin Honnen]

Jure Sah wrote:

> Martin Honnen pravi:
>> <xsl:apply-templates select="document('file1.xml')/foo/bar |
>> document('file2.xml')/foo/bar">
>> <xsl:sort select="dateElement"/>
>> </xsl:apply-templates/>
>
> I can't use that, my files are defined in the XML file I am processing.
>
> Any additional ideas?

Nevertheless you should be able to process the union of nodes in those
documents and sort them. But you keep us guessing, show us your XML
input(s) and the result you want to produce, then we can suggest a way
how to solve that.

[Jure Sah]

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Martin Honnen pravi:

>>> <xsl:apply-templates select="document('file1.xml')/foo/bar |
>>> document('file2.xml')/foo/bar">
>>> <xsl:sort select="dateElement"/>
>>> </xsl:apply-templates/>
>>
>> I can't use that, my files are defined in the XML file I am processing.
>>
>> Any additional ideas?
>
> Nevertheless you should be able to process the union of nodes in those
> documents and sort them. But you keep us guessing, show us your XML
> input(s) and the result you want to produce, then we can suggest a way
> how to solve that.

My code is available here:
http://dustwolf.ctrl-alt-del.si/downloads/XSLTaggregator_src.zip

As the name says, the code is a RSS Aggregator. Picks the URLs of the
RSS feeds to combine from an XML file and outputs the resulting feed.
Everything works fine, except the results are not sorted by date.

The code goes a little like this:
<xsl:for-each select="feed:rss2">
<xsl:variable name="myURL" select="." />
<xsl:for-each select="document($myURL)/rss/channel/item">

..that is the two nested "for-each"es.

Any ideas?

Thanks.. really. :)

LP,
Jure
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[David Carlisle]

Jure Sah wrote:

> The code goes a little like this:
> <xsl:for-each select="feed:rss2">
> <xsl:variable name="myURL" select="." />
> <xsl:for-each select="document($myURL)/rss/channel/item">
>
> ..that is the two nested "for-each"es.
>
> Any ideas?
>

just use one for-each then it's easier to sort over the whole collection.

<xsl:for-each select="document(feed:rss2)/rss/channel/item">
<xsl:sort select= something ...

--
http://dpcarlisle.blogspot.com

[Jure Sah]

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David Carlisle pravi:

> just use one for-each then it's easier to sort over the whole collection.
>
> <xsl:for-each select="document(feed:rss2)/rss/channel/item">
> <xsl:sort select= something ...

That worked great! Thanks a bunch!

LP,
Jure
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