Xsl sort

This question is related to xsl rathar than java/xml. I would really appreciate if someone could help. The xml document I use is something like this below.
<Performance>
     <SortInfo>
          <Column>CompletedMeasurement/VinCount</Column>
          <Order>TRUE</Order>
     </SortInfo>
     <PerformanceVO>
          <CompletedMeasurement>
                 <VinCount>35881</VinCount>
          </CompletedMeasurement>
     </PerformanceVO>
     <PerformanceVO>
     <CompletedMeasurement>
     <VinCount>45077</VinCount>
                      </CompletedMeasurement>
     </PerformanceVO>
</Performance>I have a requirement to use xsl:sort on the xsl to sort this xml document. The sort does not work if use an xsl variable. This is my xsl variable declaration:
<xsl:variable name="sortColumn" select="/Performance/SortInfo/Column"/>
This sort works (note that I have NOT used any xsl variable
<xsl:for-each select="PerformanceVO">
<xsl:sort data-type="number" select="CompletedMeasurement/VinCount" order="ascending"/>
</xsl:for-each>
This sort DOES NOT work (used xsl variable in xsl:sort)
<xsl:for-each select="PerformanceVO">
<xsl:sort data-type="number" select="$sortColumn" order="ascending"/>
</xsl:for-each>

This question is related to xsl rathar than java/xml.Then it would be better if you asked it in a place where experts in XSL answer questions. Check out the Mulberry mail-list.
This sort DOES NOT work (used xsl variable in xsl:sort)
<xsl:for-each select="PerformanceVO">
<xsl:sort data-type="number" select="$sortColumn"
order="ascending"/>
</xsl:for-each> I am pretty sure that must be an XSLT FAQ. I haven't looked but I bet you will find it in Dave Pawson's FAQs.

Similar Messages

Functions in xsl:sort

Is it possible to use extension functions in the select attribute of xsl:sort? This way I could sort the nodes based on a computed value.
e.g. <xsl:sort select="myns:myextfunc(ELEM1, ELEM2)">
I tried using a standard XSL function like
<xsl:sort select="concat(ELEM1, ELEM2)"/> and it seems to work but when I try to use an extension function it fails.
TIA
null

With XML Parser 9.0.1 Production, this works for me:
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:ext="http://www.oracle.com/XSL/Transform/java/Ext">
<xsl:output method="text"/>
<xsl:template match="/">
<xsl:for-each select="/Employees/Emp">

<xsl:sort data-type="number" order="descending" select="ext:max(Salary,Bonus)"
xmlns:ext="http://www.oracle.com/XSL/Transform/java/Ext"/>
<xsl:value-of select="Name"/>
</xsl:for-each>
</xsl:template>
</xsl:stylesheet>
public class Ext
public static double max( double n, double m ) {
return n > m ? n : m;
<Employees>
<Emp>
<Name>B</Name>
<Salary>10</Salary>
<Bonus>12</Bonus>
</Emp>
<Emp>
<Name>C</Name>
<Salary>25</Salary>
<Bonus>12</Bonus>
</Emp>
<Emp>
<Name>A</Name>
<Salary>10</Salary>
<Bonus>2</Bonus>
</Emp>
</Employees>If I make sure Ext.class is in my CLASSPATH, and transform the above XML by the above stylesheet, I get:
CBA
which is the reverse-numerically-sorted list of max() calculations.

Xsl:sort attribute value templates for select and order

I've tried the xdk_java_9_0_1_0_0.tar.gz update today and for that the reported bugs are still there : one urgent (quite a stopper for us) is this one
<xsl:template match="/resultlist">
<xsl:variable name="sortorder">descending</xsl:variable>
<xsl:variable name="sortfield">name</xsl:variable>
<xsl:for-each select="result">
<xsl:sort select="./{$sortfield}" order="{$sortorder}"/>
<xsl:value-of select="./index"/>,<xsl:value-of select="./name"/>
</xsl:for-each>
</xsl:template>
still it does not replace {$sortfield} and {$sortorder}.

This is known bug 1798018 in the Oracle XSLT processor.
The only workaround I can think of is to use <xsl:choose> to optionally use different <xsl:for-each> blocks, one with order="ascending" and one with order="descending" as literal values.

XSL:SORT not supported in current XSL-transformator

In iCS 5.1 the use of the xsl element sort (<xsl:sort>) in an xsl stylesheet does not sort the set of processed nodes (<xsl:for-each ...>) as we would expect it, but simply copies the xsl element <XSL:SORT> to the output stream at every location, where it is defined in the stylesheet (at the start of the output of each processed node) . This behavior is not according the W3C XSLT 1.0 specification. Are there any solutions available?
Thanks in advance,
Jakob.

This question is related to xsl rathar than java/xml.Then it would be better if you asked it in a place where experts in XSL answer questions. Check out the Mulberry mail-list.
This sort DOES NOT work (used xsl variable in xsl:sort)
<xsl:for-each select="PerformanceVO">
<xsl:sort data-type="number" select="$sortColumn"
order="ascending"/>
</xsl:for-each> I am pretty sure that must be an XSLT FAQ. I haven't looked but I bet you will find it in Dave Pawson's FAQs.

Xsl:sort order attribute value template

I am trying the use the following in XDB 10.1.0.2
<xsl:sort order="{$myOrder} ...
It appears that the order attribute does not allow an attribute value template.
Any suggestions would be appreciated.
Anthony

This is known bug 1798018 in the Oracle XSLT processor.
The only workaround I can think of is to use <xsl:choose> to optionally use different <xsl:for-each> blocks, one with order="ascending" and one with order="descending" as literal values.

Xsl:sort or sorting

Hi all,
Ineed to use a sorting function in one of the transformations in a BPEL project. The version of BPEL and Jdev i am using is 10.1.3.4
Is it possible to use xsl:sort? If no, is there any alternative to perform sorting on the xml?
Regards
Silas

It should work as it is supported by xslt 2.0. Please refer to Error: "xsl:sort" XSL Element Currently not Supported in the transform

Xsl:sort order parameter

Does the oracle xml parser support this syntax (works with Xalan)
<xsl:sort order="{$order}" select="sortnode" />
I need to parameterize the order="ascending/descending". I am sucessfully passing in my parameter (I can output it with <xsl:value-of select="$order"/>), but it doesn't seem to have any effect on the actual sorting.

This is Bug 1798018. It was supposed to be fixed in 9.0.2C, but
I just retested it and the problem still persists.
The order and data-type attributes of <xsl:sort> don't
behave properly as Attribute Value Templates.

XSL - Sort and Group?

Hello,
Anyone know if there is a way to do something like this?
Original XML:
<item>
<word>word1</word><num>10</num><num2>9</num2>
</item>
<item>
<word>word2</word><num>5</num><num2>3</num2>
</item>
<item>
<word>word1</word><num>7</num><num2>6</num2>
</item>
I want to do something like group on <word> and sort accending on sum(<num>), min(<num2>) and end up with this:
<item>
<word>word1</word><num>17</num><num2>6</num2>
</item>
<item>
<word>word2</word><num>5</num><num2>3</num2>
</item>
Is this possible?
thanks,
chad.

Here's a start. See Chapter 9 of my book for a complete explanation of how to exploit XSLT keys like database functional indexes to make grouping operations like this be fast in XSLT. It takes a bit to wrap your mind around why and how it works and the book has figures that help illustrate this.

<list>
<item>
<word>word1</word><num>10</num><num2>9</num2>
</item>
<item>
<word>word2</word><num>5</num><num2>3</num2>
</item>
<item>
<word>word1</word><num>7</num><num2>6</num2>
</item>
</list>

<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">
<xsl:key name="g" match="item" use="word"/>
<xsl:output indent="yes"/>
<xsl:template match="/">
<list>
<xsl:for-each select="list/item[generate-id(.)=generate-id(key('g',word)[1])]">
<xsl:sort data-type="number" select="sum(key('g',word)/num)"/>
<xsl:variable name="items-with-same-word" select="key('g',word)"/>
<item>
<word><xsl:value-of select="word"/></word>
<num><xsl:value-of select="sum($items-with-same-word/num)"/></num>
<num2><xsl:value-of select="sum($items-with-same-word/num2)"/></num2>
</item>
</xsl:for-each>
</list>
</xsl:template>
</xsl:stylesheet>This produces the output:
<list>
<item>
<word>word2</word>
<num>5</num>
<num2>3</num2>
</item>
<item>
<word>word1</word>
<num>17</num>
<num2>15</num2>
</item>
</list>

Xsl:sort not sorting

I tried the example of John Butler.
Source:
<?xml version="1.0" encoding="US-ASCII"?><SortTest>
<SortKey>Allergens</SortKey> <SortKey>Muscle, Skeletal</SortKey> <SortKey>Food Handling</SortKey> <SortKey>Muscle Contraction</SortKey> <SortKey>allergen Blo t 5</SortKey> <SortKey>Food</SortKey> <SortKey>Food Hypersensitivity</SortKey> <SortKey>Muscle</SortKey>
</SortTest>
and XSL:
<?xml version="1.0" encoding="US-ASCII"?><xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0" default-space="strip"> <xsl:output indent="no" /> <xsl:template match="/"> <xsl:apply-templates select="*"/> </xsl:template>
<xsl:template match="SortTest"> <xsl:element name="{name(.)}"> <xsl:apply-templates select="SortKey"> <xsl:sort order="ascending" data-type="text" case-order="upper-first" select="."/> </xsl:apply-templates>
<xsl:text>
</xsl:text>
</xsl:element> </xsl:template>
<xsl:template match="SortKey"> <xsl:text>
</xsl:text>
<xsl:copy-of select="."/> </xsl:template></xsl:stylesheet>
and get the (unsorted!) output
<SortTest>
<SortKey>Allergens</SortKey>
<SortKey>Muscle, Skeletal</SortKey>
<SortKey>Food Handling</SortKey>
<SortKey>Muscle Contraction</SortKey>
<SortKey>allergen Blo t 5</SortKey>
<SortKey>Food</SortKey>
<SortKey>Food Hypersensitivity</SortKey>
<SortKey>Muscle</SortKey>
</SortTest>
How do I get XSL:SORT to work?

Here is my results
D:\src\XMLMAP~1>type testsort2.xml
<stuff>
<item>boot</item>
<item>apple</item>
</stuff>
D:\src\XMLMAP~1>type testsort2.xsl
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:template match="/">
<xsl:apply-templates select="stuff/item">
<xsl:sort select="."/>
</xsl:apply-templates>
</xsl:template>
<xsl:template match="item">
<xsl:value-of select="."/>
<xsl:if test="position() != last()">,</xsl:if>
</xsl:template>
</xsl:stylesheet>
D:\src\XMLMAP~1>..\oraclexml\bin\xslsample testsort2.xml testsort2.xsl
XSL processor sample
Parsing XML file testsort2.xml
Parsing Stylesheet testsort2.xsl
Initializing the result context
XSL Processing
<root>bootapple</root>
D:\src\XMLMAP~1>

Custom Sort via the XSL and custom template?

I am trying to add a sort order to a CQWP. I would like a template style that does the sorting? I have to sort by a field called SortOrder, then I want it to be by file name - here is the code:
I am thinking to do a foreach but not able to get it working? Is there another easier way?

    <xsl:template name="PolicySortedBullets" match="Row[@Style='PolicySortedBullets']" mode="itemstyle">
        <xsl:variable name="SafeLinkUrl">
            <xsl:call-template name="OuterTemplate.GetSafeLink">
                <xsl:with-param name="UrlColumnName" select="'LinkUrl'"/>
            </xsl:call-template>
        </xsl:variable>
        <xsl:variable name="DisplayTitle">
            <xsl:call-template name="OuterTemplate.GetTitle">
                <xsl:with-param name="Title" select="@Title"/>
                <xsl:with-param name="UrlColumnName" select="'LinkUrl'"/>
            </xsl:call-template>
        </xsl:variable>
        <div class="item link-item bullet">
            <xsl:call-template name="OuterTemplate.CallPresenceStatusIconTemplate"/>
   <a href="{$SafeLinkUrl}" title="{@LinkToolTip}" target="_blank">
              <xsl:if test="$ItemsHaveStreams = 'True'">
                <xsl:attribute name="onclick">
                  <xsl:value-of select="@OnClickForWebRendering"/>
                </xsl:attribute>
              </xsl:if>
              <xsl:if test="$ItemsHaveStreams != 'True' and @OpenInNewWindow = 'True'">
                <xsl:attribute name="onclick">
                  <xsl:value-of disable-output-escaping="yes" select="$OnClickTargetAttribute"/>
                </xsl:attribute>
              </xsl:if>
              <xsl:value-of select="$DisplayTitle"/>
            </a>
        </div>
    </xsl:template>


http://stackoverflow.com/questions/5415383/xslt-sortingboth-ascending-and-descending
http://sharepoint.stackexchange.com/questions/81167/content-query-web-part-group-by-sort
You can use the XSLT <xsl:sort> element to customize your sort order.
https://sites.google.com/site/icsharepoint2010/xslt/sorting-in-xslt
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">
<xsl:output method="text"/>
<xsl:template match="employees">
<xsl:apply-templates>
<xsl:sort select="salary"/>
</xsl:apply-templates>
</xsl:template>
<xsl:template match="employee">
Last: <xsl:apply-templates select="last"/>
First: <xsl:apply-templates select="first"/>
Salary: <xsl:apply-templates select="salary"/>
Hire Date: <xsl:apply-templates select="@hireDate"/>
<xsl:text>
</xsl:text>
</xsl:template>
</xsl:stylesheet>

XSL transform sort problem in 1.5

I am having problems applying an XSL transform to some XML data I have under the 1.5 JRE.
Under 1.4.2 (and in several tools and browsers), the transform works correctly. In 1.5,
<xsl:sort...> directives don't seem work correctly. Here's a simplified version of my
code. Assume that the generateXMLReport outputs an XML document to a stream.
public void generateHTMLReport( OutputStream out, String strXslSheet ){
     File f = new File( strXslSheet );
     if( f.exists() ){
          ByteArrayOutputStream bos = new ByteArrayOutputStream();
          // Generate the XML data we will work from
          generateXMLReport( bos );
          bos.flush();
          // Convert the XML to HTML by applying an XSL transform
          ByteArrayInputStream bis = new ByteArrayInputStream( bos.toByteArray() );
          TransformerFactory tFactory = TransformerFactory.newInstance();
          Transformer transformer = tFactory.newTransformer( new StreamSource( strXslSheet ) );
          transformer.transform( new StreamSource( bis), new StreamResult( out ) );
}Here's an XSL template:
<xsl:template name="totalTable">
     <xsl:param name="matchSet"/>
     <xsl:variable name="maxValue">
     <xsl:for-each select="$matchSet">
     <xsl:sort select="total" data-type="number" order="descending"/>
     <xsl:if test="position() = 1">
          <xsl:value-of select="number( total )" />
         </xsl:if>
         </xsl:for-each>
         </xsl:variable>
</xsl:template>If I remove the sort directive, I get the first value in the list, as I would expect.
Other 'for-each' loops behave similarly. The problem in 1.5. When I run against 1.4.2,
the transformation works fine.

Show us the code please. Not the WHOLE code, just an example that demonstrates what you're seeing. I've never seen the problem you're encountering, but I rather doubt that it's a failing in the api. More than likely it's a semantic of how you're doing the listener but that's a guess on my part. Let's see what you have and go from there.
PS.

Sorting xml document with xsl

How can I sort below xml on empNo using xsl, I want an output to be in xml format.
<employeeList>
<employee empNo=1000>
<Info >
<fname></fname>
<lname></lname>
</Info>
</employee>
<employee empNo=1001>
<Info >
<fname></fname>
<lname></lname>
</Info>
</employee>
</employeeList>
Can any one tell me the xsl for above.......
Thanks in advance

Hi,
Your xml is not well-formed. Attributes have to be enclosed in double quotes.
You can use the following 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"></xsl:output>
          <xsl:template match="/employeeList">
          <employeeList>
          <xsl:apply-templates select="employee">
          <xsl:sort select="@empNo"/>
          </xsl:apply-templates>
          </employeeList>
          </xsl:template>
     <xsl:template match="employee">
     <xsl:copy-of select="."/>
     </xsl:template>
</xsl:stylesheet>

Multiple order sort in xsl

Hi all,
Does anyone know how I would go about doing a multiple order sort through my xsl? For example, sort by name first, then sort those results by date, thanks in advance.

okay, that would work, but i think i have a problem, i have an xml that looks like this:
<?xml version="1.0" encoding="UTF-8" ?>
<report>
- <column_headers>
<heading>Heading1</heading>
<heading >Heading2</heading>
</column_headers>
- <results>
- <row>
<attr1>value1</attr1>
<attr2>value2</attr2>
</row>
- <row>
<attr1>value3</attr1>
<attr2>value4</attr2>
</row>
- <row>
<attr1>consulting agreement</attr1>
<attr2>yesterday</attr2>
</row>
</results>
- <sort default="attr1">
- <order-by id="attr2">
<clause>attr2</clause>
<clause>attr1</clause>
</order-by>
</sort>
</report>
I need to sort the row nodes by attr2 first, then attr1, and i have something like this in my xsl:
<xsl:if test="$sortField != 'default'">
<xsl:for-each select="/report/sort/order-by">
<xsl:variable name="idField" select="@id"/>
<xsl:if test="$idField=$sortField">
<xsl:for-each select="/report/results/row">
<xsl:sort select="*[name()=$newSort]"/> --> i need the sort statements here for both attr2 and attr1, but how do i get them out of the xml whenI don't know beforehand how many clause nodes there will be, or what they will contain
<tr>
<xsl:for-each select="*">
<td class="report">
<xsl:value-of select="."/>
</td>
</xsl:for-each>
</tr>
</xsl:for-each>

XSLT mapping sorting against an output field

Hi, all
I am testing my XSLT mapping. Here is the output message type defintion:
I want to sort f2 in my xslt mapping. Here is my mapping:
<?xml version="1.0" encoding="UTF-8" ?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:ns0="http://fileconversion" xmlns:ns1="http://fileconversion">
    <xsl:template match="/">
      <ns1:TESTOUTPUT_MT>
        <Msg>
        <xsl:for-each select="ns0:TESTDATA_MT/Trans/Detail">
        <xsl:sort select="Transaction/Data/f2" />
          <Transaction>
            <Data>
              <f1>
                <xsl:value-of select="PERNR" />
              </f1>
              <f2>
                <xsl:value-of select="POSITION" />
              </f2>
              <f3>
                <xsl:value-of select="TEL" />
              </f3>
              <f4>
                <xsl:value-of select="concat(concat(STREET,' '),CITY)" />
              </f4>
            </Data>
          </Transaction>
            </xsl:for-each>
        </Msg>
      </ns1:TESTOUTPUT_MT>
    </xsl:template>
</xsl:stylesheet>
The sorting did not execute on field "f2" and the data is in the same order as the input.
Note that I can do <xsl:sort select="POSITION " /> instead and it will work. But I am trying to test on sorting against output field as this is the requirement in the real interface.
What did I do wrong? Please advise.
Thanks,
Jonathan.

Hi Jon - You can sort the source values. If you use "sort select="Transaction/Data/f2" i don't think it has got some effect.
You can use <xsl:sort select="POSITION " /> Any issues with that?
Just to add - when you map Position with F2... how does the sorting will be different when you execute it on F2 or Position?
If you are doing some calculation for F2 then should handle the Sort function accordingly..

Sort in Cross-Tab report

I have a cross-tab report that lists dates as headers and employees as rows. The data at every cell shows the qty (it is like timesheet report). The totals are shown for every row and column. The report layout looks something like this:
----------- <date> <date> <date> .... <Total>
<name> <qty> <qty> <qty> <qty>
<name> <qty> <qty> <qty> <qty>
<name> <qty> <qty> <qty> <qty>
<Total> <qty> <qty> <qty> <qty>
The report definition looks like this (I numbered the columns to make it easier to write about them):
01: <?horizontal-break-table:1?>
02: <?for-each-group@column:EMPLOYEE_TIME;./TRANSACTION_DATE?><?sort:TRANSACTION_DATE;'ascending';data-type='date'?>
03: <?TRANSACTION_DATE?>
04: <?end for-each-group?>
05: <?for-each-group:EMPLOYEE_TIME;./EMPLOYEE_NAME?><?sort:EMPLOYEE_NAME?><?variable@incontext:G1;current-group()?>
06: <?EMPLOYEE_NAME?>
07: <?for-each-group@cell://EMPLOYEE_TIME;./TRANSACTION_DATE?>
08: <?sum ($G1[(./TRANSACTION_DATE=current()/TRANSACTION_DATE)]/TRANSACTION_QTY)?>
09: <?end for-each-group?>
10: <?sum ($G1/TRANSACTION_QTY)?>
11: <?end for-each-group?>
12: <?variable@incontext:T;.//EMPLOYEE_TIME?>
13: <?for-each-group@cell://EMPLOYEE_TIME;./TRANSACTION_DATE?>
14: <?sum (current-group()/TRANSACTION_QTY)?>
15: <?end for-each-group?>
16: <?sum ($T/TRANSACTION_QTY)?>
The XML data is not sorted in any way and the report sorts it by employee and date. As you see at the line 02 I applied the sort by the transaction date. The sort works as expected, however the actual QTY are not showing at correct cells. If I remove sort it works fine but the dates are in random order. I suspect it is because I sort transaction date at header level and the QTY is not aware about it. I tried to apply the sort at line 07 but it does not help.
Does anyone has any idea how to get this report to display data correctly?

I have a SR which has associated bug 6131270.
Here is the latest from my SR:
UPDATE
=======
Hi Tomas,
The bug has been updated. Development has recommended
"You also need the patch from bug 4332399 to solve the java.util.EmptyStackException. "
Please review above bug if you have access.
Please consider applying Patch 4332399 - XSL:SORT FAILED IN A 2 LEVEL XSL:FOR-EACH-GROUP CASE
Backport is included in patch 5840358 for EBS customers.

Xsl sort

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