asked on

Read data backwards

Hi,

I want to display the following XML using XSLT:

<root>
<no_of_entries>4</no_of_entries>
<data>
<number>1</number>
</data>
<data>
<number>2</number>
</data>
<data>
<number>3</number>
</data>
<data>
<number>4</number>
</data>
</root>

For this I would use:

<xsl:for-each select="data">
<xsl:value-of select="number" />
</xsl:for-each>

and have the data printed "1 2 3 4".

I want to have this data read backwards "4 3 2 1". It is not possible to change the XML.

Please advice
SirTam

ASKER CERTIFIED SOLUTION

Gertone (Geert Bormans)

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sirtam

ASKER

Thank you. I will try this and get back to you when tested.

SirTam

sirtam

ASKER

I noticed that all of the attributes was needed. I used the following sort:

<xsl:sort select="number" order="descending"/>

I now also see that my the data always will be numbered from 1 and up with 1 as min and 4 as max.

Gertone (Geert Bormans)

nope, you need all attributes in this case
<xsl:sort select="number" order="descending"/>
is equal to
<xsl:sort select="position()" order="descending" data-type="text" />
(text is the default, not number)

XSLT1 processors will cast the position() to a string and sort stringwise
If you would have more than 10 items, 10 will be sorted in between 1 and 2
and that is NOT the effect you want, please stick to the original answer

sirtam

ASKER

You do have a point here.

In this case there wont be numbers up to 10, so I wouldn't have noticed it. But if the system should be changed this could have made an awkward bug,

Thanks again