I am trying to sort a XML nodeset by one of two fields, conditional on the value in another field.
<xsl:for-each select="CampusCourseDeliveryItem">
<xsl:sort select="OffCampus"/>
<xsl:sort select="OrganisationName" />
<xsl:sort select="OffCampusLocation"/>
<!-- code to display node goes here -->
What I want is, if OffCampus='Y', use OffCampusLocation as the sort key, otherwise use OrganisationName.
Example data:
<CampusCourseDelivery>
<CampusCourseDeliveryItem>
<OrganisationName>Chicago</OrganisationName>
<OffCampus>N</OffCampus>
<OffCampusLocation></OffCampusLocation>
</CampusCourseDeliveryItem>
<CampusCourseDeliveryItem>
<OrganisationName>London</OrganisationName>
<OffCampus>Y</OffCampus>
<OffCampusLocation>Detroit</OffCampusLocation>
</CampusCourseDeliveryItem>
<CampusCourseDeliveryItem>
<OrganisationName>Seattle</OrganisationName>
<OffCampus>Y</OffCampus>
<OffCampusLocation>Berlin</OffCampusLocation>
</CampusCourseDeliveryItem>
<CampusCourseDeliveryItem>
<OrganisationName>Adelaide</OrganisationName>
<OffCampus>N</OffCampus>
<OffCampusLocation>Ignore this value</OffCampusLocation>
</CampusCourseDeliveryItem>
</CampusCourseDelivery>
Expected sort order:
Adelaide
Berlin
Chicago
Detroit
Sort by the following expression:
concat(
substring(OffCampusLocation, 1, string-length(OffCampusLocation) * (OffCampus='Y')),
substring(OrganisationName, 1, string-length(OrganisationName) * (OffCampus='N'))
)
Related
I'm new to xslt 2.0, I would like to set the value to a variable in for-each loop only once (means if the value set, I want to come out of the loop).
For now it keep iterating for all the users. I just want to come out of the loop once the value set (immediately after my first attemp). I'm not sure how to break if the value set once.
Can you please help me on the below code ?
XSLT Code:
<xsl:variable name="v_first_name">
<xsl:for-each select="$emailList/emails/child::*">
<xsl:variable name="mailid" select="id" />
<xsl:for-each select="$userList/users/child::*">
<xsl:if test="emailid = $mailid">
<xsl:if test="firstname eq 'Antony'">
<xsl:value-of select="firstname" />
</xsl:if>
</xsl:if>
</xsl:for-each>
</xsl:for-each>
</xsl:variable>
<xsl:if test="$v_first_name != ''">
<first_name>
<xsl:value-of select="$v_first_name" />
</first_name>
</xsl:if>
XML O/p:
<first_name>AntonyAntonyAntonyAntony</first_name>
Expected XML O/P:
<first_name>Antony</first_name>
Note1: Please note that I'm using xslt 2.0 and my lists can have duplicates (So Antony can come twice, but I want only once (or unique)).
Note2: I also tried with position(), but couldn't find it work as the condition () can match at any position.
Thanks in advance.
Start with XPath and simply select the nodes you are looking for instead of considering for-each a "loop". If you select e.g. $userList/users/*[emailid = $emailList/emails/*/id] you select child elements from users which have a matching emailid in $emailList/emails/*.
I am not sure which sense it makes to hard code a first name value and then output it but of course you can select e.g. $userList/users/*[emailid = $emailList/emails/*/id and firstname = 'Antony']/lastname. That gives you a sequence of element nodes, if you want the first use a positional predicate e.g. depending on the structure of your input $userList/users/*[emailid = $emailList/emails/*/id and firstname = 'Antony'][1]/lastname or, of all selected elements ($userList/users/*[emailid = $emailList/emails/*/id and firstname = 'Antony']/lastname)[1].
I'm trying to use preceding-sibling and following-sibling with a subset of records with a sort on them. The problem that the preceding / following brings back values from the original xml order:
<Salaries>
<Salary>
<Base>1000</Base>
<CreatedDate xmlns:d7p1="http://schemas.datacontract.org/2004/07/System">
<d7p1:DateTime>2016-01-09T14:38:54.8440764Z</d7p1:DateTime>
<d7p1:OffsetMinutes>0</d7p1:OffsetMinutes>
</CreatedDate>
</Salary>
<Salary>
<Base>2000</Base>
<CreatedDate xmlns:d7p1="http://schemas.datacontract.org/2004/07/System">
<d7p1:DateTime>2015-01-09T14:38:54.8440764Z</d7p1:DateTime>
<d7p1:OffsetMinutes>0</d7p1:OffsetMinutes>
</CreatedDate>
</Salary>
<Salary>
<Base>3000</Base>
<CreatedDate xmlns:d7p1="http://schemas.datacontract.org/2004/07/System">
<d7p1:DateTime>2017-01-09T14:38:54.8440764Z</d7p1:DateTime>
<d7p1:OffsetMinutes>0</d7p1:OffsetMinutes>
</CreatedDate>
</Salary>
</Salaries>
When I use a sort under a for-each (Salaries/Salary) with a c# function to add offset minutes into a date and convert to a long number 201701010000 for example(to make manipulation in xslt easier).
<xsl:sort select="number(cs:Convertdatetolong(cs:AddOffsetMinutes(substring(p:CreatedDate/d5p1:DateTime,1,19),p:CreatedDate/d5p1:OffsetMinutes)))" order="ascending"/>
The sort works perfectly and I get the records out in the following order:
2000
1000
3000
The problem comes if I use preceding-sibling / preceding (and following). I would expect the first record (2000) to have no preceding record and the last record (3000) to have no following.
However when I use the preceding / following I get the previous record and the next record from the original XML:
2000 (preceding - 1000 / following - 3000)
1000 (preceding - / following - 2000)
3000 (preceding - 2000 / following - )
I would like to be able to compare against the previous record (in the sorted order) and the current record (in the sorted order):
2000 (preceding - / following - 1000)
1000 (preceding - 2000 / following 3000)
3000 (preceding - 1000 / following - )
I've tried preceding-sibling and preceding
<xsl:value-of select="preceding::p:Salary[1]/p:Base"/>
<xsl:value-of select="preceding-sibling::p:Salary[1]/p:Base"/>
<xsl:value-of select="preceding::p:Salary[position()=1]/p:Base"/>
(the salary is in a different namespace (p)
Is this actually possible or do I have to use variables to save the previous record's data to compare against?
Any ideas gratefully received. I'm using xslt 1.0
Although XSLT/XPath often talks of a "sequence of nodes", it's actually more accurate to think of it as a "sequence of node references" - because, for example, the same node can appear more than once in the sequence. When you sort a sequence of node references, you don't change the individual nodes in any way, you only change the sequence. That means the nodes still exist in their original tree exactly where they were before, and their parents, siblings, and descendants are exactly as they were before.
What you want is not the preceding and following siblings of the node, but the nodes that come before and after it in the sorted sequence, which is a quite different thing.
One way to do this is to construct a new tree containing copies of the original nodes, which you get, for example, if you do
<xsl:variable name="x">
<xsl:for-each ...>
<xsl:sort ...>
<xsl:copy-of select="."/>
The sibling relationships of the copied nodes will then reflect the sorted order. There's the minor problem that in XSLT 1.0, $x is a result tree fragment so you have to convert it to a node-set using the exslt:node-set() function.
In fact in XSLT 1.0 that's probably the only way of doing it, because the XSLT 1.0 data model only has node sets, not sequences, which means there is no way of capturing and processing a sequence of nodes in anything other than document order. The 2.0 model has much more flexibility and power. Upgrade if you can - XSLT 1.0 is approaching 20 years old.
Thanks to Michael for the answer. Posted here for completeness. Complicated because of the name spaces in use in the xml:
<!-- Puts the whole of the Salary Node into a variable-->
<xsl:variable name="SALARY" >
<xsl:copy-of select="p:Salaries" />
</xsl:variable>
<!-- Puts the the required key data into a node-set with the correct sort applied-->
<xsl:variable name="SAL">
<xsl:for-each select="msxsl:node-set($SALARY)//p:Salary">
<xsl:sort select="number(cs:Convertdatetolong(cs:AddOffsetMinutes(substring(p:CreatedDate/d5p1:DateTime,1,19),p:CreatedDate/d5p1:OffsetMinutes)))" order="ascending"/>
<xsl:copy-of select="." />
</xsl:for-each>
</xsl:variable>
<!-- Quick Output-->
<xsl:for-each select="msxsl:node-set($SAL)//p:Salary">
<xsl:text>Sa:</xsl:text>
<xsl:value-of select="position()" />
<xsl:text>Preceding:</xsl:text>
<xsl:value-of select="preceding-sibling::p:Salary[1]/p:Base"/>
<xsl:value-of select="$newline" />
<xsl:text>Current:</xsl:text>
<xsl:value-of select="p:Base"/>
<xsl:value-of select="$newline" />
<xsl:text>Following:</xsl:text>
<xsl:value-of select="following-sibling::p:Salary[1]/p:Base"/>
<xsl:value-of select="$newline"/>
</xsl:for-each>
The preceding-sibling axis gets the preceding siblings of the context node in document order.
To refer to the preceding siblings of a node after sorting, you will need to store the sorted nodes in a variable first - and, in XSLT 1.0, convert the variable into a node-set.
I have the following xml. What I want to do with my XSL is sort the output on the total value of the elements productDesignRating, productPriceRating and productPerfromanceRating. So far no luck in trying to get this done. Any help will be appreciated i need to be able to do this in xsl 1 so no xsl2 functions.
<DocumentElement xmlns="DotNetNuke/UserDefinedTable">
<QueryResults>
<productCategory>cat1</productCategory>
<productTitle>product1</productTitle>
<productImage><img alt="productImage" title="productImage" src="/skinconversion/Portals/12/babynokiko.jpg" /></productImage>
<productDesignRating>3</productDesignRating>
<productPriceRating>4</productPriceRating>
<productPerformanceRating>4</productPerformanceRating>
<productPrice>10</productPrice>
<productSummary>description</productSummary>
<productUrl>http://www.2dnn.com</productUrl>
</QueryResults>
<QueryResults>
<productCategory>cat2</productCategory>
<productTitle>product2</productTitle>
<productImage><img alt="productImage" title="productImage" src="/skinconversion/Portals/12/babynokiko.jpg" /></productImage>
<productDesignRating>3</productDesignRating>
<productPriceRating>3</productPriceRating>
<productPerformanceRating>3</productPerformanceRating>
<productPrice>10</productPrice>
<productSummary>description</productSummary>
<productUrl>http://www.2dnn.com</productUrl>
</QueryResults>
<QueryResults>
<productCategory>cat3</productCategory>
<productTitle>product3</productTitle>
<productImage><img alt="productImage" title="productImage" src="/skinconversion/Portals/12/babynokiko.jpg" /></productImage>
<productDesignRating>1</productDesignRating>
<productPriceRating>2</productPriceRating>
<productPerformanceRating>3</productPerformanceRating>
<productPrice>56</productPrice>
<productSummary>description</productSummary>
<productUrl>http://www.2dnn.com</productUrl>
</QueryResults>
</DocumentElement>
Try this:
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:my="DotNetNuke/UserDefinedTable">
<xsl:template match="node()">
<xsl:copy>
<xsl:apply-templates select="node()">
<xsl:sort select="my:productDesignRating +
my:productPriceRating +
my:productPerformanceRating"
data-type="number"/>
</xsl:apply-templates>
</xsl:copy>
</xsl:template>
</xsl:stylesheet>
This is just a simplified identity template (not processing any attributes), but applying a numerical sort on the sum of the values of the 3 specified child elements – if they are present.
Specifically note the use of the data-type attribute, allowing to specify that the sorting base should be numbers, so the order here is 6,9,11 (for strings, which is default, it would be 11,6,9)…
[To reverse the order, simply add the order="descending" attribute]
I have an XML document as follows:
<objects>
<object uid="0" />
<object uid="1" />
<object uid="2" />
</objects>
I can select multiple elements using the following query:
doc.xpath("//object[#uid=2 or #uid=0 or #uid=1]")
But this returns the elements in the same order they're declared in the XML document (uid=0, uid=1, uid=2) and I want the results in the same order as I perform the XPath query (uid=2, uid=0, uid=1).
I'm unsure if this is possible with XPath alone, and have looked into XSLT sorting, but I haven't found an example that explains how I could achieve this.
I'm working in Ruby with the Nokogiri library.
There is no way in XPath 1.0 to specify the order of the selected nodes.
XPath 2.0 allows a sequence of nodes with any specific order:
//object[#uid=2], //object[#uid=1]
evaluates to a sequence in which all object items with #uid=2 precede all object items with #uid=1
If one doesn't have anXPath 2.0 engine available, it is still possible to use XSLT in order to output nodes in any desired order.
In this specific case the sequence of the following XSLT instructions:
<xsl:copy-of select="//object[#uid=2]"/>
<xsl:copy-of select="//object[#uid=1]"/>
produces the desired output:
<object uid="2" /><object uid="1" />
I am assuming you are using XPath 1.0. The W3C spec says:
The primary syntactic construct in XPath is the expression. An expression matches the production Expr. An expression is evaluated to yield an object, which has one of the following four basic types:
* node-set (an unordered collection of nodes without duplicates)
* boolean (true or false)
* number (a floating-point number)
* string (a sequence of UCS characters)
So I don't think you can re-order simply using XPath. (The rest of the spec defines document order and reverse document order, so if the latter does what you want you can get it using the appropriate axis (e.g. preceding).
In XSLT you can use <xsl:sort> using the name() of the attribute. The XSLT FAQ is very good and you should find an answer there.
An XSLT example:
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:param name="pSequence" select="'2 1'"/>
<xsl:template match="objects">
<xsl:for-each select="object[contains(concat(' ',$pSequence,' '),
concat(' ',#uid,' '))]">
<xsl:sort select="substring-before(concat(' ',$pSequence,' '),
concat(' ',#uid,' '))"/>
<xsl:copy-of select="."/>
</xsl:for-each>
</xsl:template>
</xsl:stylesheet>
Output:
<object uid="2" /><object uid="1" />
I don't think there is a way to do it in xpath but if you wish to switch to XSLT you can use the xsl:sort tag:
<xsl:for-each select="//object[#uid=1 or #uid=2]">
<xsl:sort: select="#uid" data-type="number" />
{insert new logic here}
</xsl:for-each>
more complete info here:
http://www.w3schools.com/xsl/el_sort.asp
This is how I'd do it in Nokogiri:
require 'nokogiri'
xml = '<objects><object uid="0" /><object uid="1" /><object uid="2" /></objects>'
doc = Nokogiri::XML(xml)
objects_by_uid = doc.search('//object[#uid="2" or #uid="1"]').sort_by { |n| n['uid'].to_i }.reverse
puts objects_by_uid
Running that outputs:
<object uid="2"/>
<object uid="1"/>
An alternative to the search would be:
objects_by_uid = doc.search('//object[#uid="2" or #uid="1"]').sort { |a,b| b['uid'].to_i <=> a['uid'].to_i }
if you don't like using sort_by with the reverse.
XPath is useful for locating and retrieving the nodes but often the filtering we want to do gets too convoluted in the accessor so I let the language do it, whether it's Ruby, Perl or Python. Where I put the filtering logic is based on how big the XML data set is and whether there are a lot of different uid values I'll want to grab. Sometimes letting the XPath engine do the heavy lifting makes sense, other times its easier to let XPath grab all the object nodes and filter in the calling language.
I am currently working on a project that uses the Windows event log. I am using wevtutil to get the results from the event logs. I know that wevtutil supports xpath queries, but since I'm new to xpath I don't know that I can achieve what I'm trying to do.
In SQL, what I would be doing is something like this:
SELECT log.*, COUNT(1) numHits
FROM Application log
GROUP BY Source, Task, Level, Description
ORDER BY numHits DESC
LIMIT 10
Is it possible to do such a thing using xpath?
Edit: Here is a sample Event:
<Event xmlns='http://schemas.microsoft.com/win/2004/08/events/event'>
<System>
<Provider Name='MSSQL$SQLEXPRESS' />
<EventID Qualifiers='16384'>17403</EventID>
<Level>4</Level>
<Task>2</Task>
<Keywords>0x80000000000000</Keywords>
<TimeCreated SystemTime='2010-10-20T20:06:18.000Z' />
<EventRecordID>9448</EventRecordID>
<Channel>Application</Channel>
<Computer>SHAZTOP</Computer>
<Security />
</System>
<EventData>
<Data>73094</Data>
<Binary>
FB4300000A000000130000005300480041005A0054004F0050005C00530051004C004500580050005200450053005300000000000000</Binary>
</EventData>
</Event>
XPath 1.0 has four data types: string, number, boolean and node set.
The only XPath ordering criteria is document order (in the given axis direction). That is how you can limit any result node set as #Dimitre and #Welbog have sugested with fn:position().
But, there is no specification that an XPath engine must provide a node set result in any given order. So, you can't sort nor grouping in XPath 1.0. You can select the firsts of each group, but not efficiently. As example:
//Event[not(System/Level = preceding::Level) or
not(System/Task = preceding::Task)]
XPath 2.0 has the sequence data type. A sequence has the exclicit order of construction. So, you can group. As example:
for $event (//Event)[index-of(//Event/System/concat(Level,'++',Task),
System/concat(Level,'++',Task))[1]]
result //Event[System/Level = $event/System/Level]
[System/Task = $event/System/Task]
But, because XPath 2.0 has not built-in sorting nor recursion mechanism (you could provide an extension function...) you can't sort.
For that you need a language with built-in sorting or a way to express its algorithm. Both XSLT (1.0 or 2.0) and XQuery have these features.
In SQL, what I would be doing is
something like this:
SELECT log.*, COUNT(1) numHits
FROM Application log
GROUP BY Source, Task, Level, Description
ORDER BY numHits DESC
LIMIT 10
Is it possible to do such a thing
using xpath?
In case no sorting is necessary, one can get the first $n nodes selected by any XPath expression by:
(ExpressionSelectingNodeSet)[not(position() > $n)]
where $n can be substituted by a specific number
If there is a requirement that the nodes be sorted on one or more sort-keys, then this is not possible pure XPath, but one can easily perform such tasks with XSLT, using the <xsl:sort> instruction and the XPath position() function:
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output omit-xml-declaration="yes" indent="yes"/>
<xsl:template match="/*">
<nums>
<xsl:for-each select="num">
<xsl:sort data-type="number" order="descending"/>
<xsl:if test="not(position() > 5)">
<xsl:copy-of select="."/>
</xsl:if>
</xsl:for-each>
</nums>
</xsl:template>
</xsl:stylesheet>
When this transformation is applied on the following XML document:
<nums>
<num>01</num>
<num>02</num>
<num>03</num>
<num>04</num>
<num>05</num>
<num>06</num>
<num>07</num>
<num>08</num>
<num>09</num>
<num>010</num>
</nums>
the correct result, containing only the top 5 numbers is produced:
<nums>
<num>010</num>
<num>09</num>
<num>08</num>
<num>07</num>
<num>06</num>
</nums>
You can use the position() function to limit the results you're getting:
/root/element[position()<=10]
For example, that would select the first ten element elements which are children of the root.
If your structure is more complicated, you can use the position element in different places. For example, if the element element can exist in more than one parent, but you want the first ten of them regardless of parent, you can do it this way:
(/root/parent1/element | /root/parent2/element)[position()<=10]