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Re: [xsl] How to sort based on the number of child elements?

2022-05-09 12:58:24
 I cannot use xsl:sort for this, right?

Why not?

See my reply using <xsl:sort> to solve the problem.

On Mon, May 9, 2022 at 10:31 AM Roger L Costello costello(_at_)mitre(_dot_)org <
xsl-list-service(_at_)lists(_dot_)mulberrytech(_dot_)com> wrote:

Hi Folks,

I have an XML Schema that contains a xs:choice. I want to sort the
branches of the choice, in longest-to-shortest order.

Here is a xs:choice with two branches:

<xs:choice>
    <xs:element name="MilitaryDayTime">
        <xs:complexType>
            <xs:sequence>
                <xs:element name="Day" type="xs:string"/>
                <xs:element name="HourTime" type="xs:string"/>
                <xs:element name="MinuteTime" type="xs:string"/>
                <xs:element name="TimeZone" type="xs:string"/>
            </xs:sequence>
        </xs:complexType>
    </xs:element>
    <xs:element name="DateTimeGroup">
        <xs:complexType>
            <xs:sequence>
                <xs:element name="Day" type="xs:string"/>
                <xs:element name="HourTime" type="xs:string"/>
                <xs:element name="MinuteTime" type="xs:string"/>
                <xs:element name="TimeZone" type="xs:string"/>
                <xs:element name="MonthName" type="xs:string"/>
                <xs:element name="Year" type="xs:string"/>
            </xs:sequence>
        </xs:complexType>
    </xs:element>
</xs:choice>

The first branch is an element with 4 child elements. The second branch is
an element with 6 child elements. So sorting the branches
longest-to-shortest will result in reversing the order of the branches.

I cannot use xsl:sort for this, right?

Is there an easy solution to this task?

/Roger




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