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Re: [xsl] Sorting on two booleans

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Ah! That was it!
number()

It didn't occur to me that xslt 1.0 you couldn't sort on a boolean.
Thank you so much!
Joelle

Michael Kay wrote:

Ah sorry, I missed that subtlety. So the first sort key is @active,
descending, and the second is (active = published), false<true:

<xsl:sort select="number(@active = @published)" data-type="number"/>

(or in XSLT 2.0 you can sort booleans directly, with false<true, giving

<xsl:sort select="@active eq @published" />

Alternatively, you can probably do it with a single sort key:

<xsl:for-each select="project">
<xsl:sort select="substring('dcab', 2*@active + @published + 1, 1)"/>

But I haven't tested it and it's a bit obscure for practical use.

Michael Kay
http://www.saxonica.com/

-----Original Message-----
From: Joelle Tegwen [mailto:tegwe002@xxxxxxx] Sent: 26 February 2009 17:55
To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
Subject: Re: [xsl] Sorting on two booleans

That gives me
active, not published
active, published
not active, not published
not active, published

But I need the last two reversed.
not active, published
not active, not published

Thanks
Joelle

Michael Kay wrote:

I've got a document that looks like this:
<projects>
    <project active="1" published="1">stuff</project>
    <project active="1" published="0">stuff</project>
    <project active="0" published="1">stuff</project>
    <project active="0" published="0">stuff</project> </projects

And I need to be able to sort them (on the fly - it's part of a sortable table in html) in this order and the reverse:
active, not published
active, published
not active, published
not active, not published.

So your major sort key is @active (true<false) and your

minor sort key

is @published (false<true).

That's

<xsl:for-each select...>
<xsl:sort select="@active" data-type="number" order="descending"/>
<xsl:sort select="@published" data-type="published" order="ascending"/>

For the reverse order, flip all the order="" attributes.

Michael Kay
http://www.saxonica.com/

I've tried all kinds of ways to sort them to get this to happen including brute force:
<xsl:for-each select="project">
<xsl:sort select="@active=1 and

@published=0"

order="ascending"
data-type="number"/>
<xsl:sort select="@active=1 and

@published=1"

order="ascending"
data-type="number"/>
<xsl:sort select="@active=0 and

@published=1"

order="ascending"
data-type="number"/>
<xsl:sort select="@active=0 and

@published=0"

order="ascending"
                            data-type="number"/>
                        <xsl:copy-of select="."/>
                    </xsl:for-each>

Nothing works. It always comes out with at least one in the wrong spot.

Could someone point me in the right direction? Help me understand what I'm missing about how sort works?

Thanks much in advance
Joelle

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