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XSLT: Get node where one certain value is present

- by Kim Andersen

Hi there I have the following XML: <data> <page id="1118"> <itms> <values> <value>1104</value> </values> </itms> </page> <page id="1177"> <itms> <values> <value>1273</value> <value>1215</value> </values> </itms> </page> </data> I need to get the @id from the < page , where a certain value is present in one of the < value -tags. The id that need to be in the < value is kept in this: $itm/@id. This means that if my $itm/@id is equal to 1273, I need to get 1177 returned. I'm not quite sure how to achieve this. Actually I could have XML that looks like this as well: <data> <page id="1118"> <itms> <values> <value>1104</value> </values> </itms> </page> <page id="1177"> <itms> <values> <value>1273</value> <value>1215</value> </values> </itms> </page> <page id="1352"> <itms> <values> <value>1242</value> <value>1273</value> </values> </itms> </page> </data> If that's the case, I need the latest id, so this means that if the $itm/@id matches values in more < page 's, then I need to grab the value from the latest page. I the above case that would be 1352. Hope this makes sense to you guys. And by the way, I work with Umbraco CMS if that does any difference. Best Regards, Kim

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XSLT: use parameters in xls:sort attributes (dynamic sorting)

- by fireeyedboy

How do I apply a parameter to a select and order attribute in a xsl:sort element? I'ld like to do this dynamic with PHP with something like this: $xsl = new XSLTProcessor(); $xslDoc = new DOMDocument(); $xslDoc->load( $this->_xslFilePath ); $xsl->importStyleSheet( $xslDoc ); $xsl->setParameter( '', 'sortBy', 'viewCount' ); $xsl->setParameter( '', 'order', 'descending' ); But I'ld first have to now how to get this to work. I tried the following, but it gives me a 'compilation error' : 'invalid value $order for order'. $sortBy doesn't seem to do anything either: <?xml version="1.0" encoding="utf-8"?> <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <xsl:output method="xml" indent="yes"/> <xsl:param name="sortBy" select="viewCount"/> <xsl:param name="order" select="descending"/> <xsl:template match="/"> <media> <xsl:for-each select="media/medium"> <xsl:sort select="$sortBy" order="$order"/> // <someoutput> </xsl:for-each> </media> </xsl:template> </xsl:stylesheet>

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xslt or php for rendering output

- by Arsenal

I'm creating something where users can upload an xml and data get's imported to the database. Now I'm building some kind of a preview page where users will get to see how their input will look once it's stored. What would be the fastest (in execution time), using XSL to transform the xml to a html page, or using php to render the output? My guess is XSL is far more suitable (+ faster) for this (and by using a DTD there's no need for code written validation, right?)

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Leave entity intact in XML + XSLT

- by Kuroki Kaze

I transform XML to (sort of) HTML with XSL stylesheets (using Apache Xalan). In XML there can be entities like —, which must be left as is. In beginning of XML file I have a doctype which references these entities. What should I do for entity to be left unchanged? <!DOCTYPE article [ <!ENTITY mdash "—"> ]> gives me SAXParseException: Recursive entity expansion, 'mdash' when encountering &mdash in XML text.

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Exception durin processing XSLT transformation!

- by Artic

I'm usin such code to generate contents file. try { StreamResult result = new StreamResult(); TransformerFactory tf = TransformerFactory.newInstance(); for (String item: groups){ item = item.replaceAll(" ", "-").toLowerCase(); result.setOutputStream(new FileOutputStream(path+item+".html")); Templates templ = tf.newTemplates(xsltSource); Transformer transf = templ.newTransformer(); transf.clearParameters(); transf.setParameter("group", item); transf.transform(xmlSource, result); } } catch (TransformerConfigurationException e) { throw new SinkException(e.getMessage()); } catch (TransformerException e) { throw new SinkException(e.getMessage()); } But on second iteration I have an exception ERROR: javax.xml.transform.TransformerException: com.sun.org.apache.xml.internal.utils.WrappedRuntimeException: Read error Cann't understand what is the reason?

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Adding a variable href attribute to a hyperlink using XSLT

- by Pete

I need to create links using dynamic values for both the URL and the anchor text. I have column called URL, which contains URL's, i.e., http://stackoverflow.com, and I have a column called Title, which contains the text for the anchor, i.e., This Great Site. I figure maybe all I need to do is nest this tags, along with a little HTML to get my desired result: <a href="> <xsl:value-of select="@URL">"> <xsl:value-of select="@Title"/> </xsl:value-of> </a> In my head, this would render: <a href=" http://stackoverflow.com"> This Great Site </a> I know there's more to it, but I haven't been able to clearly understand what more I need. I think this issue has been addressed before, but not in a way I could understand it. I'd be more than happy to improve the question's title to help noobs like myself find this item, and hopefully an answer. Any help is greatly appreciated.

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XSLT: How to get XML

- by Jyotsna Sonawane

I have a XSL that transforms one format of XML into another. In input XML I have a node with following value - which is actually a XML string if we replace < with < (less than) for e.g. <Paragraph><Title><!CDATA[Pour les nuits du 2012-10-01 - 2012-10-30]]></Title><Text><![CDATA[TAXES INCLUSES.]]></Text></Paragraph> I want to have the content of otherInfo as a XML node in output XML. if I do , I do not get it as a XML node - it is output just as text. How can I make XSL output the content of otherInfo as XML node ?

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Preserve certain html tags during XSLT

- by Bilzac

I have looked up solutions on stackflow, but none of them seem to work for me. Here is my question. Lets say I have the following text : Source: <description> $300$ <br/> $250 <br/> $200! <br/> <p> Yes, that is right! <br/> You can own a ps3 for only $200 </p> </description> Output: <newprice> $300$ <br/> $250 <br/> $200! <br/> Yes, that is right! <br/> You can own a ps3 for only $200 </newprice> I can't seem to find a way to do that.

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How to determine the inner context node within an XSLT select

- by Marcus Rickert

Hi there! I'm trying to determine a subset of nodes stored in the variable objs by using a user defined function cube:is_active: <xsl:variable name="active_cubes" select="$objs[cube:is_active(XXX) = 'TRUE']"/ Since the function does not allow a local context I'm required to pass it as a parameter (denoted by XXX). However, the usual suspects "." or "current()" do not work since they refer to context node of the surrounding block and NOT to the current element of objs which is evaluated. The only solution so far which DOES work is: XXX=SOME_CHILD_TAG/.. But this is really ugly since it depends on the existence of the child tag for the parent node to work correctly. Is there any other way? Thanks!

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default attribute called xmlns getting added to element - using xslt transformation

- by Amit

Hello All, I am using xsl to generate an o/p in xml. while creating element I have specified following when o/p is getting created - it is appending a default attribute called xmlns="" with the element... Any idea why ??? o/p-- any value here Thanks in advance...

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XSLT - Comparing preceding-sibling's elements with current's node element

- by siondream

Hello, I have this XML file: <recursos> <recurso url="http://w3c.com"> <descripcion>Consorcio W3C</descripcion> <tipo>externo</tipo> <idioma>ingles</idioma> <contenido>General</contenido> <unidad>Unidad 2</unidad> </recurso> <recurso url="http://html.com"> <descripcion>Especificación HTML</descripcion> <tipo>externo</tipo> <idioma>castellano</idioma> <contenido>HTML</contenido> <version>4.01</version> <unidad>Unidad 3</unidad> </recurso> </recursos> I want to compare one "recurso"'s preceding sibling element "unidad" with the "unidad" of the current "recurso" to check if they're different. I was trying: <xsl:if test="preceding-sibling::recurso[position()=1]::unidad != unidad"> </xsl:if> But I know it's horribly wrong :( I hope you could help me, thank you very much.

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XSLT: replace an integer by a string

- by binogure

I have a little problem. A node in my XML may contains and integer, and i have to replace this integer by a string. Each number match with a string. For example i have: Integer - String 1 - TODO 2 - IN PROGRESS 3 - DONE 4 - ERROR 5 - ABORTED Original XML: <root> <status>1</status> </root> Converted XML: <root> <status>TODO</status> </root> So i want replace 1 by "TODO", 2 by "IN PROGRESS" ... <?xml version="1.0" encoding="ISO-8859-1"?> <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <xsl:template match="/root/status"> <root> <status> <xsl:variable name="text" select="." /> <xsl:choose> <xsl:when test="contains($text, '1')"> <xsl:value-of select="'TODO'"/> </xsl:when> <xsl:otherwise> <xsl:value-of select="$text"/> </xsl:otherwise> </xsl:choose> </status></root> </xsl:template> </xsl:stylesheet> I'am asking if there is another way to do that.

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XSLT split text and preserve HTML tags

- by Lycaon

I have an xml that has a description node: <config> <desc>A <b>first</b> sentence here. The second sentence with some link <a href="myurl">The link</a>. The <u>third</u> one.</desc> </config> I am trying to split the sentences using dot as separator but keeping in the same time in the HTML output the eventual HTML tags. What I have so far is a template that splits the description but the HTML tags are lost in the output due to the normalize-space and substring-before functions. My current template is given below: <xsl:template name="output-tokens"> <xsl:param name="sourceText" />  <xsl:variable name="newlist" select="concat(normalize-space($sourceText), ' ')" />  <xsl:choose> <xsl:when test ="contains($newlist, '.')">  <xsl:variable name="first" select="substring-before($newlist, '.')" />  <xsl:variable name="remaining" select="substring-after($newlist, '.')" />  <xsl:if test="normalize-space($first)!='.' and normalize-space($first)!=''"> <p><xsl:value-of select="normalize-space($first)" />.</p> </xsl:if>  <xsl:if test="$remaining"> <xsl:call-template name="output-tokens"> <xsl:with-param name="sourceText" select="$remaining" /> </xsl:call-template> </xsl:if> </xsl:when>  <xsl:otherwise> <p>  <xsl:if test="normalize-space($sourceText)!=''"> <xsl:value-of select="normalize-space($sourceText)" />. </xsl:if> </p> </xsl:otherwise> </xsl:choose> </xsl:template> and I am using it in the following manner: <xsl:template match="config"> <xsl:call-template name="output-tokens"> <xsl:with-param name="sourceText" select="desc" /> </xsl:call-template> </xsl:template> The expected output is: <p>A <b>first</b> sentence here.</p> <p>The second sentence with some link <a href="myurl">The link</a>.</p> <p>The <u>third</u> one.</p>

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XSLT Similar functionality to SQL "WHERE LIKE"

- by Fishcake

I have the following XML <Answer> <Label>FCVCMileage</Label> <Value>3258</Value> <Iteration>0</Iteration> <DataType>NUMBER</DataType> </Answer> And need to get the Value underneath the Mileage label. However the 4 letter prefix to Mileage could be any 1 of 8 different prefixes. I know I could find the value by testing for every combination using xsl:if or xsl:choose but is there a more elegant way that would work similar to the following SQL and also wouldn't need changes to code being made if other prefixes were added. WHERE label LIKE '%Mileage' NB. There will only ever be 1 label element containing the word Mileage Cheers!

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test="" on a boolean always returns true xslt

- by user70448

Why does <xsl:if test="<XPATH to boolean value here>"> ... </xsl:if> ALWAYS return true? Since boolean can be 0,1,"false" and "true" by definition, the ONLY way to test for a boolean value is to do string comparison against these. This can't be right.

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Navigating sorted XML data (XSLT)

- by Andrew Parisi

I have an XML file with company data in it, for 30 companies across 8 industries, on a portfolio page. The user has the option to sort this data by Industry, and this XML file will be added to constantly. This sorting is done in my XSL file using <xsl:choose>. Example: <xsl:when test="(invest[@investid='con'])"> <xsl:for-each select="$invest-port/portfolio/company[@industry='Communications']"> <xsl:sort select="name" /> <div class="invest-port-thumb"> <a> <xsl:attribute name="href"> <xsl:value-of select="link" /> </xsl:attribute> </a> </div> </xsl:for-each> </xsl:when> When navigating to an individual company's page, there are "previous" and "next" buttons at the bottom of the window. My issue is that I need these to dynamically link to the previous and next elements from within the XML data that has been sorted. Is this possible? Or is there an easier way to do this? (such as place each company in industry-divided XML files instead of one) Any insight would be greatly appreciated!

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How to sort by hexadecimals in xslt?

- by fielding

Hi, im trying to sort my transformed output by a element which contains a hex value. <xsl:sort select="Generation/Sirio/Code" data-type="number"/> Values are plain old Hex: 00 01 02 ... 0A ... FF but they are getting sorted like that: 0A FF 00 01 02, which indicates the sorting method fails as soon as there are character involved. How can i work around this? Thank you very much!

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xslt and xpath: match preceding comments

- by miasbeck

given this xml: <root> <list>  <item name="foo" /> <item name="bar" />  <item name="another foo" /> </list> </root> I'd like to use a XPath to select all item-nodes that have a comment immediately preceding them, that is I like to select the "foo" and "another foo" items, but not the "bar" item. I already fiddled about the preceding-sibling axis and the comment() function but to no avail.

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How can I make XSLT work in chrome?

- by Eric

I have an XML document here that is served with a corresponding XSL file. The transformation is left to be executed client-side, without JavaScript. This works fine in IE (shock horror), but in Google Chrome, just displays the document's text nodes. I know that it is possible to do client-side XSL in Chrome, as I have seen examples of it, but I am yet to be able to replicate this success myself What am I doing wrong?

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xslt to show most number of copies in catalog.xml file

- by SANJAY RAO

In this catalog.xml file i have two books who have the same inventory i.e 20 . I want to write a xsl file that will display the most number of copies of a book in a catalog .if there are two or more books of the same inventory then they have to be displayed . <?xml version="1.0"?> <?xml-stylesheet type="text/xsl" href="catalog3.xsl"?> <!DOCTYPE catalog SYSTEM "catalog.dtd"> <catalog> <Book> <sku>12345</sku> <title>Beauty Secrets</title> <condition>New</condition> <current_inventory>20</current_inventory> <price>99.99</price> </Book> <Book> <sku>54321</sku> <title>Picturescapes</title> <current_inventory>20</current_inventory> <condition>New</condition> <price>50.00</price> </Book> <Book> <sku>33333</sku> <title>Tourist Perspectives</title> <condition>New</condition> <current_inventory>0</current_inventory> <price>75.00</price> </Book> <Book> <sku>10001</sku> <title>Fire in the Sky</title> <condition>Used</condition> <current_inventory>0</current_inventory> <price>10.00</price> </Book> below is my catalog3.xsl file which is able to display only one out of the two books <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <xsl:variable name="max"/> <xsl:template match="/"> <html> <body> <h2>Titles of Books for which Most Copies are Available</h2> <table border="2"> <tr bgcolor="#9acd32"> <th>Title</th> <th>No of Copies</th> </tr> <xsl:apply-templates/> </table> </body> </html> </xsl:template> <xsl:template match="catalog"> <xsl:for-each select="Book"> <xsl:sort select="current_inventory" data-type="number" order="descending"/> <tr> <xsl:if test="position()= 1"> <p><xsl:value-of select="$max = "/></p> <td><xsl:value-of select="title"/></td> <td><xsl:value-of select="current_inventory"/></td> </xsl:if> </tr> could anybody correct me to achieve my goal of displaying all the copies having the same maximum inventory in the catalog . Thanks .

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Convert scientific notation into double with XSLT(FO)

- by Mork0075

Hello, the XML i would like to transform contains something like 7.3248378E7, which is not supported natively by the XSL-FO processor (ends up in NaN in the PDF file). Is there a nice way (which is the opposite of a dirty hack) to convert this number format? Thanks a lot.

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Exception during processing XSLT transformation!

- by Artic

I'm using this code to generate contents file. try { StreamResult result = new StreamResult(); TransformerFactory tf = TransformerFactory.newInstance(); Templates templ = tf.newTemplates(xsltSource); Transformer transf = templ.newTransformer(); for (String item: groups){ item = item.replaceAll(" ", "-").toLowerCase(); result.setOutputStream(new FileOutputStream(path+item+".html")); transf.clearParameters(); transf.setParameter("group", item); transf.transform(xmlSource, result); } } catch (TransformerConfigurationException e) { throw new SinkException(e.getMessage()); } catch (TransformerException e) { throw new SinkException(e.getMessage()); } But on second iteration I have an exception ERROR: javax.xml.transform.TransformerException: com.sun.org.apache.xml.internal.utils.WrappedRuntimeException: Read error Cann't understand what is the reason?

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Filter a CSV list in XSLT

- by Tom W

Hello SO; I'm trying to convert xml elements whose content is a csv list of identifiers, the exact form doesn't matter. I want to exclude every item that contains the character "/". The best I've come up with is translate(./someElement, ",*/*", "") but this has at least two problems; I'm pretty sure that xsl doesn't accept wildcards, and "," is an illegal character in xpaths. Is there a straightforward way to do this sort of thing?

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XSLT Type Checking

- by mo

Hi Folks Is it possible to check an elements ComplexType? i have this (simplified): complexType Record complexType Customer extension of Record complexType Person extension of Record <xsl:template match="/"> <records> <xsl:apply-templates /> </records> </xsl:template> <xsl:template match="!!! TYPECHECK FOR RECORD !!!" name="Record"> <record><xsl:value-of select="." /></record> </xsl:template> is it possible to check elementstype incl. inheritence? i dont know the elements name only that they are a subtype of Record. schema 1: complexType name="Customer" extension base="Record" element name="customers" element name="customer" type="Customer" schema 2: complexType name="Person" extension base="Record" element name="persons" element name="person" type="Person" schema ?: complexType name="UnknownType" extension base="Record" element name="unknowns" element name="unknown" type="UnknownType" xml 1: <customers> <customer /> <customer /> </customers> xml 2: <persons> <person /> <person /> </persons> xml ?: <?s> <? /> <? /> </?s> the xml input ist custom so i have to match by the type (i think)

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xslt: operations on new elements

- by user1495523

Could you please explain in case we could perform any operations on newly included elements using xsl? To explain using an example: if we have the following input file <?xml version="1.0" encoding="UTF-8"?> <top> <Results> <a>no</a> <b>10</b> <c>12</c> <d>9</d> </Results> <Results> <a>Yes</a> <b>8</b> <c>50</c> <d>12</d> </Results> </top> We need the final result as <?xml version="1.0" encoding="UTF-8"?> <top> <Results> <a>no</a> <b>10</b> <b_>10</b_> <c>12</c> <c_>12</c_> <d>9</d> <e_>11</e_> </Results> <Results> <a>Yes</a> <b>8</b> <b_>8</b_> <c>50</c> <c_>50</c_> <d>12</d> <e_>29</e_> </Results> </top> Where: b_ = b, c_ = c, & e_ = (b_ + c_)/2

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