xslt sorting from two different X paths

sen1953 · August 5th, 2013, 12:46 PM

Code:

<?xml version="1.0" encoding="UTF-8"?>
<accountList>
    <previousAccount>
        <account>
            <lastName>NASH</lastName>
            <accountStatus>REMOVED</accountStatus>
            <accNo>08</accNo>
        </account>
        <account>
            <lastName>ADOGA</lastName>
            <accountStatus>REMOVED</accountStatus>
            <accNo>01</accNo>
        </account>
        <account>
            <lastName>LUCAS</lastName>
            <accountStatus>HOLD</accountStatus>
            <accNo>09</accNo>
        </account>
        <account>
            <lastName>DONALD</lastName>
            <accountStatus>HOLD</accountStatus>
            <accNo>21</accNo>
        </account>
        <account>
            <accountStatus>HOLD</accountStatus>
            <lastName>LONDON</lastName>
            <accNo>24</accNo>
        </account>
    </previousAccount>
    <account>
        <Title>Mr</Title>
        <firstName>RICHARD</firstName>
        <lastName>JOHN</lastName>
        <city>london</city>
        <accNo>02</accNo>
    </account>
    <account>
        <Title>Mr</Title>
        <firstName>xxx</firstName>
        <lastName>JOHN</lastName>
        <city>London</city>
        <accNo>17</accNo>
    </account>
    <account>
        <Title>Mr</Title>
        <firstName>HEWIT</firstName>
        <lastName>JOHN</lastName>
        <city>LONDON</city>
        <accNo>20</accNo>
    </account>
    <account>
        <Title>Mr</Title>
        <firstName>xxx</firstName>
        <lastName>JOHN</lastName>
        <city>LONDON</city>
        <accNo>21</accNo>
    </account>
    <account>
        <Title>Mr</Title>
        <firstName>KEVIN</firstName>
        <lastName>PETE</lastName>
        <city>LONDON</city>
        <accNo>07</accNo>
    </account>
</accountList>
xslt code

    <xsl:stylesheet version="2.0"
        xsi:schemaLocation="http://www.w3.org/1999/XSL/Transform"
        xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
        xmlns:fn="http://www.w3.org/2005/xpath-functions"
        xmlns:xs="http://www.w3.org/2001/XMLSchema"
        xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
    <xsl:output method="text"/>
    <xsl:variable name="newline">
        <xsl:text>
        </xsl:text>
    </xsl:variable>
    <xsl:template match="/">
        <xsl:for-each-group select="accountList/account" group-starting-with="*[firstName != 'xxx' or lastName != preceding-sibling::*[1]/lastName]">
            <xsl:sort select="accNo" order="ascending"/>

            <xsl:value-of select="accNo"/>
            <xsl:text>         </xsl:text>
            <xsl:value-of select="lastName"/>
            <xsl:value-of select="$newline"/>
        </xsl:for-each-group>
        <xsl:value-of select="$newline"/>
        <xsl:for-each select="accountList/previousAccount/account">
            <xsl:sort select="accNo" order="ascending"/>
            <xsl:value-of select="accNo"/>
            <xsl:text> </xsl:text>
            <xsl:value-of select="accountStatus"/>
            <xsl:text> </xsl:text>
            <xsl:value-of select="lastName"/>
            <xsl:value-of select="$newline"/>
        </xsl:for-each>
    </xsl:template>
</xsl:stylesheet>

In my requirement ,I got to sort accNo which are different nodes.I think I'll do much better explaining this in code rather than words, accountList/previousAccount/account/accNo accountList/account/accNo

I used descendant,which is not working fine for my requirement.Here I have used two for-each for two nodes and sorted separtedly.But this is not a required output. Please help me to sort out this please find my output below

Actual Output

Code:

02         JOHN
        07         PETE
        20         JOHN

        01 REMOVED ADOGA
        08 REMOVED NASH
        09 HOLD LUCAS
        21 HOLD DONALD
        24 HOLD LONDON

Expecting output

Code:

01 REMOVED ADOGA    
02         JOHN
07         PETE
08 REMOVED NASH
09 HOLD     LUCAS
20         JOHN
21 HOLD DONALD
24 HOLD LONDON 
accNo will be in sorte

Output is in sorted format.

sen1953 · August 21st, 2013, 09:08 AM

Please give some idea , whether its is possible or not? I am unable to find solution for this . Can we sort one element which is from two different template.?

samjudson · August 21st, 2013, 12:25 PM

Try starting with something like this:

Code:

<xsl:for-each select="accountList/account | accountList/previousAccount/account">
            <xsl:sort select="accNo" order="ascending"/>
  <!-- do something in here with each item -->
</xsl:for>

sen1953 · August 29th, 2013, 11:57 AM

Thanks SAM,,Already I tried this approach ,again i check xpath and come back again..Thanks lot for your reply

sen1953 · August 30th, 2013, 10:45 AM

Given solution is not working for me..I am getting error like this when use two select operation in for-each for sorting

Error
A sequence of more than one item is not allowed as the @select attribute of xsl:sort; SystemID: file

Please help me SAM..

samjudson · August 30th, 2013, 10:58 AM

I didn't have a sequence in the select of the xsl:sort - I had a sequence in the select of the for-each.

I've tested this using the following XML and it works fine:

Code:

<?xml version="1.0" encoding="UTF-8"?>
<root>
	  <A id="1"/>
	  <B id="2"/>
	  <A id="2"/>
	  <B id="1"/>
	  <A id="4"/>
	  <B id="1"/>
</root>

XSLT:

Code:

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
                xmlns:xs="http://www.w3.org/2001/XMLSchema"
                xmlns:saxon="http://saxon.sf.net/"
                version="1.0"
                exclude-result-prefixes="xs saxon">

   <xsl:template match="/">
	     <xsl:for-each select="root/A | root/B">
		       <xsl:sort select="@id" data-type="number"/>
		       <xsl:copy-of select="."/>
		    </xsl:for-each>
   </xsl:template>
    
</xsl:stylesheet>

Output:

Code:

<A id="1"/><B id="1"/><B id="1"/><B id="2"/><A id="2"/><A id="4"/>

sen1953 · August 30th, 2013, 12:53 PM

I run and modified the input xml and run it form my requirement.I didn't get the expected result.See this is my problem , This is my xml structure

<?xml version="1.0" encoding="UTF-8"?>
<root>
<E>
<F>
<A>8</A>
<A>2</A>
<A>7</A>
<A>10</A>
<A>3</A>
<A>7</A>
</F>
</E>
<d>
<A>5</A>
<A>1</A>
<A>9</A>
</d>

</root>

I tried this way...

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:xs="http://www.w3.org/2001/XMLSchema"
xmlns:saxon="http://saxon.sf.net/"
version="1.0"
exclude-result-prefixes="xs saxon">

<xsl:template match="/">
<xsl:for-each select="root/E/F | root/D">
<xsl:sort select="A"data-type="number"order="ascending"/>

<xsl:copy-of select="."/>
</xsl:for-each>
</xsl:template>

</xsl:stylesheet>

My wrong output

<?xml version="1.0" encoding="UTF-8"?>
<F>
<A>8</A>
<A>2</A>
<A>7</A>
<A>10</A>
<A>3</A>
<A>7</A>

</F>

The problem is missing out this element

This is my requirement input xml..Please advice me...

sen1953 · September 2nd, 2013, 09:49 AM

I am looking for your reply ? Give me any suggestion pls

mhkay · September 2nd, 2013, 10:12 AM

You are trying to sort the F elements based on the value of their child A elements. But they have more than one child A element, so you get an error, because the sort key must select a single value for each item in the sequence being sorted.

samjudson · September 2nd, 2013, 11:58 AM

That and the fact that XML is case sensitive, so root/D will not find 'd' elements.