Question

XML Sort on attribute using xsl:sort

Asked by: Bagur

Hi,
I need to sort an xml document by an attribute. I am unable to do this by using the attached xsl stylesheet. Please help.

I need to sort the below document (dataField Nodea) by it's Table_Id attribute

As I said the above xslt does not work for sorting on Table_Id attribute, although the same thing works for Node_Id attribute if I just replace "DataField/@Table_Id" with "DataField/@Node_Id" in xslt. I am completely baffled by it's behavior. I am using VB6 (I think that should not matter).

I am very concerned as to why it works with one field and not the other.

Any help with some explaination is greatly appreciated.

Thanks in advance

My xml document
-----------------------------
<SubCompetency Node_Id = "3" Node_Name = "SubComp1">
    <DataField
        Table_Id = "14"
        Node_Id = "13"
        Node_Name = "Data1"
        ReqToCmp = "1"
        Def_Val = "NULL">
        <HtmField Node_Id = "16" Node_Name = "Field1">
            <HtmGrpField
                Node_Id = "79"
                Node_Name = "Grp1"
                Node_Val = "5"/>
            <HtmGrpField
                Node_Id = "80"
                Node_Name = "Grp2"
                Node_Val = "4"/>
            <HtmGrpField
                Node_Id = "81"
                Node_Name = "Grp3"
                Node_Val = "3"/>
            <HtmGrpField
                Node_Id = "82"
                Node_Name = "Grp4"
                Node_Val = "2"/>
            <HtmGrpField
                Node_Id = "83"
                Node_Name = "Grp5"
                Node_Val = "1"/>
            <HtmGrpField
                Node_Id = "84"
                Node_Name = "Grp6"
                Node_Val = "0"/>
        </HtmField>
    </DataField>
    <DataField
        Table_Id = "8"
        Node_Id = "25"
        Node_Name = "Data2"
        ReqToCmp = "1"
        Def_Val = "NULL">
        <HtmField Node_Id = "28" Node_Name = "Field2">
            <HtmGrpField
                Node_Id = "151"
                Node_Name = "Grp1"
                Node_Val = "5"/>
            <HtmGrpField
                Node_Id = "152"
                Node_Name = "Grp1"
                Node_Val = "4"/>
            <HtmGrpField
                Node_Id = "153"
                Node_Name = "Grp1"
                Node_Val = "3"/>
            <HtmGrpField
                Node_Id = "154"
                Node_Name = "Grp4"
                Node_Val = "2"/>
            <HtmGrpField
                Node_Id = "155"
                Node_Name = "Grp5"
                Node_Val = "1"/>
            <HtmGrpField
                Node_Id = "156"
                Node_Name = "Grp6"
                Node_Val = "0"/>
        </HtmField>
    </DataField>
    <DataField
        Table_Id = "3"
        Node_Id = "26"
        Node_Name = "Data3"
        ReqToCmp = "1"
        Def_Val = "NULL">
        <HtmField Node_Id = "29" Node_Name = "Field3">
            <HtmGrpField
                Node_Id = "157"
                Node_Name = "Grp1"
                Node_Val = "5"/>
            <HtmGrpField
                Node_Id = "158"
                Node_Name = "Grp2"
                Node_Val = "4"/>
            <HtmGrpField
                Node_Id = "159"
                Node_Name = "Grp3"
                Node_Val = "3"/>
            <HtmGrpField
                Node_Id = "160"
                Node_Name = "Grp4"
                Node_Val = "2"/>
            <HtmGrpField
                Node_Id = "161"
                Node_Name = "Grp5"
                Node_Val = "1"/>
            <HtmGrpField
                Node_Id = "162"
                Node_Name = "Grp6"
                Node_Val = "0"/>
        </HtmField>
    </DataField>
</SubCompetency>
------------------------------------------------------
 
My Xsl stylesheet
---------------------------------------------------------
 
<?xml version="1.0" encoding="UTF-8"?>"
    <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">"
    <xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes"/>"
    <xsl:template match="SubCompetency|@*">"
    <xsl:copy><xsl:apply-templates select="node()|@*">"
    <xsl:sort select="DataField/@Table_Id" data-type="number" order="descending"/>"
    </xsl:apply-templates></xsl:copy></xsl:template>"
    <xsl:template match="node()|@*"><xsl:copy><xsl:apply-templates select="node()|@*"/>"
    </xsl:copy></xsl:template></xsl:stylesheet>"
--------------------------------------------------------------

                                  
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Asked On
2009-07-04 at 19:25:36ID24544420
Tags

XM attribute Sort

,

xsl:sort

Topics

Extensible Stylesheet Language Transformation (XSLT)

,

Extensible HTML (XHTML)

,

XPath

Participating Experts
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Points
500
Comments
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Answers

 

by: GertonePosted on 2009-07-05 at 04:35:14ID: 24779667

I think that it should not work with @Node_Id either
You are nesting one element too deep
<xsl:sort select="DataField/@Table_Id"
should be
<xsl:sort select="@Table_Id"
I think that nodes are incidently preorderd on Node_Id, because it should not have worked

There are some issues with ambiguous templates for @*,
I rearranged you stylesheet a bit, see below

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
    <xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes"/>
    <xsl:template match="SubCompetency">
        <xsl:copy>
            <xsl:copy-of select="@*"/>
            <xsl:apply-templates select="node()">
                <xsl:sort select="@Table_Id" data-type="number" order="descending"/>
            </xsl:apply-templates>
        </xsl:copy>
    </xsl:template>
    <xsl:template match="node()">
        <xsl:copy>
            <xsl:copy-of select="@*"/>
            <xsl:apply-templates select="node()"/>
        </xsl:copy>
    </xsl:template>
</xsl:stylesheet>
                                              
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by: BagurPosted on 2009-07-05 at 14:29:46ID: 24781371

Hi Gertone,

The solution worked, gr8.., but I still need to understand 2 more things

a) The @Node_Id was not ordered incidentally, I made sure of that and the sort works for @Node_Id

b) Why do you say that it is ordered one element deep? I am naive to this, so please explain how. I thought since we are matching subCompetency node, the DataField/@Table_Id falls directly underneath subCompetency Node and it should work fine. Anyway, the @Table_Id is an attribute of DataField, so what's wrong with mentioning DataField/@Table_Id? I know it does not work like that, but please explain how/why.

Thanks.

 

by: GertonePosted on 2009-07-05 at 14:36:36ID: 24781379

You are right that the template is on subCompentency,
in there you state
            <xsl:apply-templates select="node()">
                <xsl:sort select="@Table_Id" data-type="number" order="descending"/>
            </xsl:apply-templates>
The node() in this case is the child elements (so one level deep), including DataField
So you are pushing the DataField nodes to the templates and those DataField nodes you want to sort.
The sort select XPath starts from the nodes mentioned in the select on the apply-templates
So, what you really want is sorting on the @Table_id of the DataField
... you want to sort
DataField on
subCompetency/DataField/@Table_id
and not on
subCompetency/DataField/DataField/@Table_id
that was your error.
You lost sight of the context awareness of the XPath

 

by: GertonePosted on 2009-07-05 at 14:37:37ID: 24781385

about your a)
 DataField/@Node_Id can not work,
likely you had put
 @Node_Id

 

by: BagurPosted on 2009-07-05 at 14:40:53ID: 31599861

Thank you very much for the detailed explaination!

 

by: GertonePosted on 2009-07-05 at 14:41:26ID: 24781398

welcome

20120131-EE-VQP-002

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