Search Results

Search found 1068 results on 43 pages for 'xsl fo'.

Page 14/43 | < Previous Page | 10 11 12 13 14 15 16 17 18 19 20 21  | Next Page >

  • Methods to see result fo a code change faster

    - by Can't Tell
    This question came to me when developing using Eclipse. I use JBoss Application Server and use hot code replacement. But this option requires that the 'build automatically' option to be enabled. This makes Eclipse build the workspace automatically (periodically or when a file is saved?) and for a large code base this takes too much time and processing which makes the machine freeze for a while. Also sometimes an error message is shown saying that hot code replacement failed. The question that I have is: is there a better way to see the result of a code change? Currently I have the following two suggestions: Have unit tests - this will allow to run a single test and see the result of a code change. ( But for a JavaEE application that uses EJBs is it easy to setup unit tests?) Use OSGi - which allows to add jars to the running system without bringing down the JVM. Any ideas on above suggestions or any other suggestion or a framework that allows to do this is welcome.

    Read the article

  • Exclude notes based on attribute wildcard in XSL node selection

    - by C A
    Using cruisecontrol for continuous integration, I have some annoyances with Weblogic Ant tasks and how they think that server debug information are warnings rather than debug, so are shown in my build report emails. The XML output from cruise is similar to: <cruisecontrol> <build> <target name="compile-xxx"> <task name="xxx" /> </target> <target name="xxx.weblogic"> <task name="wldeploy"> <message priority="warn">Message which isn't really a warning"</message> </task> </target> </build> </cruisecontrol> In the cruisecontrol XSL template the current selection for the task list is: <xsl:variable name="tasklist" select="/cruisecontrol/build//target/task"/> What I would like is something which selects the tasklist in the same way, but doesn't include any target nodes which have the attribute name="*weblogic" where * is a wildcard. I have tried <xsl:variable name="tasklist" select="/cruisecontrol/build//target[@name!='*weblogic']/task"/> but this doesn't seem to have worked. I'm not an expert with XSLT, and just want to get this fixed so I can carry on the real development of the project. Any help is much appreciated.

    Read the article

  • Exclude nodes based on attribute wildcard in XSL node selection

    - by C A
    Using cruisecontrol for continuous integration, I have some annoyances with Weblogic Ant tasks and how they think that server debug information are warnings rather than debug, so are shown in my build report emails. The XML output from cruise is similar to: <cruisecontrol> <build> <target name="compile-xxx"> <task name="xxx" /> </target> <target name="xxx.weblogic"> <task name="wldeploy"> <message priority="warn">Message which isn't really a warning"</message> </task> </target> </build> </cruisecontrol> In the cruisecontrol XSL template the current selection for the task list is: <xsl:variable name="tasklist" select="/cruisecontrol/build//target/task"/> What I would like is something which selects the tasklist in the same way, but doesn't include any target nodes which have the attribute name="*weblogic" where * is a wildcard. I have tried <xsl:variable name="tasklist" select="/cruisecontrol/build//target[@name!='*weblogic']/task"/> but this doesn't seem to have worked. I'm not an expert with XSLT, and just want to get this fixed so I can carry on the real development of the project. Any help is much appreciated.

    Read the article

  • test="" on a boolean always returns true

    - 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.

    Read the article

  • Writing JSON with XSLT

    - by JP
    Hi, I'm trying to write XSLT to transform a specific web page to JSON. The following code demonstrates how Ruby would do this conversion, but the XSLT doesn't generate valid JSON (there's one too many commas inside the array) - anyone know how to write XSLT to generate valid JSON? require 'rubygems' require 'nokogiri' require 'open-uri' doc = Nokogiri::HTML(open('http://bbc.co.uk/radio1/playlist')) xslt = Nokogiri::XSLT(DATA.read) puts out = xslt.transform(doc) # Now follows the XSLT __END__ <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns="http://www.w3.org/1999/xhtml"> <xsl:output method="text" encoding="UTF-8" media-type="text/plain"/> <xsl:template match="/"> [ <xsl:for-each select="//*[@id='playlist_a']//div[@class='artists_and_songs']//ul[@class='clearme']"> {'artist':'<xsl:value-of select="li[@class='artist']" />','track':'<xsl:value-of select="li[@class='song']" />'}, </xsl:for-each> ] </xsl:template> </xsl:stylesheet>

    Read the article

  • Retrieving XML node from a path specified in an attribute value of another node

    - by Olivier PAYEN
    From this XML source : <?xml version="1.0" encoding="utf-8" ?> <ROOT> <STRUCT> <COL order="1" nodeName="FOO/BAR" colName="Foo Bar" /> <COL order="2" nodeName="FIZZ" colName="Fizz" /> </STRUCT> <DATASET> <DATA> <FIZZ>testFizz</FIZZ> <FOO> <BAR>testBar</BAR> <LIB>testLib</LIB> </FOO> </DATA> <DATA> <FIZZ>testFizz2</FIZZ> <FOO> <BAR>testBar2</BAR> <LIB>testLib2</LIB> </FOO> </DATA> </DATASET> </ROOT> I want to generate this HTML : <html> <head> <title>Test</title> </head> <body> <table border="1"> <tr> <td>Foo Bar</td> <td>Fizz</td> </tr> <tr> <td>testBar</td> <td>testFizz</td> </tr> <tr> <td>testBar2</td> <td>testFizz2</td> </tr> </table> </body> </html> Here is the XSLT I currently have : <?xml version="1.0" encoding="utf-8"?> <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:msxsl="urn:schemas-microsoft-com:xslt" exclude-result-prefixes="msxsl"> <xsl:output method="html" indent="yes"/> <xsl:template match="/ROOT"> <html> <head> <title>Test</title> </head> <body> <table border="1"> <tr> <!--Generate the table header--> <xsl:apply-templates select="STRUCT/COL"> <xsl:sort data-type="number" select="@order"/> </xsl:apply-templates> </tr> <xsl:apply-templates select="DATASET/DATA" /> </table> </body> </html> </xsl:template> <xsl:template match="COL"> <!--Template for generating the table header--> <td> <xsl:value-of select="@colName"/> </td> </xsl:template> <xsl:template match="DATA"> <xsl:variable name="pos" select="position()" /> <tr> <xsl:for-each select="/ROOT/STRUCT/COL"> <xsl:sort data-type="number" select="@order"/> <xsl:variable name="elementName" select="@nodeName" /> <td> <xsl:value-of select="/ROOT/DATASET/DATA[$pos]/*[name() = $elementName]" /> </td> </xsl:for-each> </tr> </xsl:template> </xsl:stylesheet> It almost works, the problem I have is to retrieve the correct DATA node from the path specified in the "nodeName" attribute value of the STRUCT block.

    Read the article

  • Is this XSLT correct for the XML file which I have developed?

    - by atrueguy
    This is my XML file. I need to develop a xslt for this. <?xml version="1.0" encoding="ISO-8859-1"?> <!--<!DOCTYPE svg PUBLIC "-//W3C//DTD SVG 1.1//EN" "http://www.w3.org/Graphics/SVG/1.1/DTD/svg11.dtd">--> <!-- Generator: Arbortext IsoDraw 7.0 --> <svg width="100%" height="100%" viewBox="0 0 214.819 278.002"> <g id="Standard_x0020_layer"/> <g id="Catalog"> <line stroke-width="0.353" stroke-linecap="butt" x1="5.839" y1="262.185" x2="209.039" y2="262.185"/> <text transform="matrix(0.984 0 0 0.93 183.515 265.271)" stroke="none" fill="#000000" font-family="'Helvetica'" font-size="3.174">© 2009 k Co.</text> <text transform="matrix(0.994 0 0 0.93 7.235 265.3)" stroke="none" fill="#000000" font-family="'Helvetica'" font-size="3.174">087156-8-</text> <text transform="matrix(0.995 0 0 0.93 21.708 265.357)" stroke="none" fill="#000000" font-family="'Helvetica'" font-size="3.174" font-weight="bold">AB</text> <path stroke-width="0.088" stroke-linecap="butt" stroke-dasharray="2.822 1.058" d="M162.037 107.578L174.439 100.417L180.698 104.03"/> <g id="AUTOID_20445" class="52971"> <line stroke-width="0.088" stroke-linecap="butt" x1="68.859" y1="43.621" x2="65.643" y2="45.399"/> <text transform="matrix(0.944 0 0 0.93 69.165 43.356)" stroke="none" fill="#000000" font-family="'Helvetica'" font-size="2.775" font-weight="bold">52971</text> </g> </g> </svg> I have developed a XSLT for this in this way, but I am failing to produce the desired output can any one help me in this. <?xml version="1.0" encoding="ISO-8859-1"?> <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:svg="http://www.w3.org/2000/svg"> <xsl:template match="/"> <fo:root xmlns:fo="http://www.w3.org/1999/XSL/Format"> <fo:layout-master-set> <fo:simple-page-master master-name="simple" page-height="11in" page-width="8.5in"> <fo:region-body margin="0.7in" margin-top="1.15in" margin-left=".8in"/> <fo:region-before extent="1.5in"/> <fo:region-after extent="1.5in"/> <fo:region-start extent="1.5in"/> <fo:region-end extent="1.5in"/> </fo:simple-page-master> </fo:layout-master-set> <fo:page-sequence master-reference="simple"> <fo:flow flow-name="xsl-region-body"> <fo:block> <fo:instream-foreign-object> <svg:svg xmlns:svg="http://www.w3.org/2000/svg" height="100%" width="100%" viewBox="0 0 214.819 278.002"> <xsl:for-each select="svg/g"> <svg:g style="stroke:none;fill:#000000;"> <svg:path> <xsl:variable name="s"> <xsl:value-of select="translate(@d,' ','')"/> </xsl:variable> <xsl:attribute name="d"><xsl:value-of select="translate($s,',',' ')"/></xsl:attribute> </svg:path> </svg:g> </xsl:for-each> <xsl:for-each select="svg/g"> <svg:line x1 = "{$x1}" y1 = "{$y1}" x2 = "{$x2}" y2 = "{$y2}" style = "stroke-width: 0.088; stroke: black;"/> <line xmlns="http://www.w3.org/2000/svg" x1="{$x1}" y1="{$y1}" x2="{$x2}" y2="{$y2}" stroke-width="0.088" stroke="black" fill="#000000" /> </xsl:for-each> </svg:svg> </fo:instream-foreign-object> </fo:block> </fo:flow> </fo:page-sequence> </fo:root> </xsl:template> </xsl:stylesheet> Please help me in this

    Read the article

  • 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 .

    Read the article

  • Loading cross domain XML with Javascript using a hybrid iframe-proxy/xsl/jsonp concept?

    - by Josef
    On our site www.foo.com we want to download and use http://feeds.foo.com/feed.xml with Javascript. We'll obviously use Access-Control but for browsers that don't support it we are considering the following as a fallback: On www.foo.com, we set document.domain, provide a callback function and load the feed into a (hidden) iframe: document.domain = 'foo.com'; function receive_data(data) { // process data }; var proxy = document.createElement('iframe'); proxy.src = 'http://feeds.foo.com/feed.xml'; document.body.appendChild(proxy); On feeds.foo.com, add an XSL to feed.xml and use it to transform the feed into an html document that also sets document.domain and calls the callback function in its parent with the feed data as json: <?xml version="1.0"?> <xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0"> <xsl:template match="ROOT"> <html><body> <script type="text/javascript"> document.domain = 'foo.com'; parent.receive_data([<xsl:apply-templates/>]); </script> </body></html> </xsl:template> <!-- templates that transform data into json objects go here --> </xsl:stylesheet> Is there a better way to load XML from feeds.foo.com and what are the ramifications of this iframe-proxy/xslt/jsonp trick? (..and in what cases will it fail?) Remarks This does not work in Safari & Chrome but since both support Access-Control it's fine. We want little or no change to feeds.foo.com We are aware of (but not interested in) server-side proxy solutions update: wrote about it

    Read the article

  • How can I highlight the corresponding titles of clicked links?

    - by danielle
    I'm writing an XSL file that contains a side-nav menu with a list of links. When the user clicks on one of the links, the page jumps to the corresponding table of information for that link. How can I make it so that when the link is clicked, the title of that table (not the link itself) is highlighted? It should also un-highlight if another link is clicked. Here is the menu of links: <div onclick = "highlight(this);" onblur = "undoHighlight(this);"> <a href = "#{generate-id(.)}"> <xsl:value-of select = "."/> (<xsl:value-of select = "count(../n1:entry)"/>) </a> </div> This is the javascript for the highlight/undoHighlight functions: function highlight(link) { undoHighlight(link) link.style.background = "red"; } function undoHighlight(link) { link.style.background = "white"; } Any help would be appreciated. Thanks in advance!

    Read the article

  • Add a Carriage Return to the Output of an XSL Transformation

    - by dsrekab
    I am trying to use XSLT to convert an XML document to a text file and the text of the document looks fine. However, I need to add a carriage return after the end of each line (NOT A CRLF) and I seem to be failing in every attempt. I have tried adding just a CR at the end of the line like this: <xsl:text>&#xD;</xsl:text> I have tried changing my media-type to string, I have tried to add the disable-output-escaping attribute to the text element, but it always adds a CRLF. This is on a Windows OS and I know that Windows uses CRLF for a new line, but I would have thought you could override that if you said to specifically use only the CR or the LF (e.g. VB.net's VBCR or VBLF). Does anyone know if it is possible to only output a CR with XSLT? Thanks in advance.

    Read the article

  • Using xsl param (if exists) to replcae attribute value

    - by Assaf
    I would like an xsl that replaces the value attribute of the data elements only if the relevant param names are passed. Input <applicationVariables applicationServer="tomcat"> <data name="HOST" value="localhost"/> <data name="PORT" value="8080"/> <data name="SIZE" value="1000"/> </applicationVariables> So for example if passing in a param HOST1=myHost and PORT=9080 the output should be: <applicationVariables applicationServer="tomcat"> <data name="HOST" value="myHost"/> <data name="PORT" value="9080"/> <data name="SIZE" value="1000"/> </applicationVariables> Note that HOST and PORT where replaced but SIZE was not replaced because there was no parameter with name SIZE Since the list of data elements is long (and may change), i would like a generic way of doing this with xsl

    Read the article

  • XSLT; parse escaped text to a node-set and extract subelements

    - by Tom W
    Hello SO; I've been fighting with this problem all day and am just about at my wit's end. I have an XML file in which certain portions of data are stored as escaped text but are themselves well-formed XML. I want to convert the whole hierarchy in this text node to a node-set and extract the data therein. No combination of variables and functions I can think of works. The way I'd expect it to work would be: <xsl:variable name="a" select="InnerXML"> <xsl:for-each select="exsl:node-set($a)/*"> 'do something </xsl:for-each> The input element InnerXML contains text of the form <root><element a>text</element a><element b><element c/><element d>text</element d></element b></root> but that doesn't really matter. I just want to navigate the xml like a normal node-set. Where am I going wrong?

    Read the article

  • XSLT, JSTL e JSF

    - by Paulo S.
    I have a xml file which I want to transform in a jsf code page. To do that I've created a xsl file. xml: <?xml version='1.0' encoding='ISO-8859-1'?> <questionario xmlns:xsi='http://www.w3.org/2001/XMLSchema-instance' xsi:noNamespaceSchemaLocation='Schema2.xsd'> <componente nome='input'> <id>input1</id> </componente> <componente nome='input'> <id>input2</id> </componente> </questionario> code: <%@ taglib uri="http://java.sun.com/jstl/core_rt" prefix="c" %> <%@ taglib uri="http://java.sun.com/jsp/jstl/xml" prefix="x" %> <c:set var="xml" value="${questionarioXSLBean.xml}"/> <c:set var="xsl"> <xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="2.0" xmlns:f="http://java.sun.com/jsf/core" xmlns:h="http://java.sun.com/jsf/html" exclude-result-prefixes="f h"> <xsl:template match="/"> <xsl:for-each select="questionario/componente"> <xsl:if test="attribute::nome = 'input'"> <xsl:variable name="id"> <xsl:value-of select="id" /> </xsl:variable> <h:inputText id="{$id}"/> </xsl:if> </xsl:for-each> </xsl:template> </xsl:stylesheet> </c:set> <x:transform xml="${xml}" xslt="${xsl}" /> The problem is that nothing is shown in my screen because the generated code for <h:inputText id="input1"/> is <h:inputText id="input_1" xmlns:h="http://java.sun.com/jsf/html"/> how can I replace the xmlns:h="http://java.sun.com/jsf/html" or suppress it. Thanks! Update: Let me clarify what I want to do. I want to generate a jsf page dynamically depending on the attributes of a xml file, for instance, 2 input texts, 3 check boxes, etc. To make the transformation to jsf I thought in two approaches, one using jstl and another using xslt. The problem with the former is that I couldn't integrate jstl with jsf code (to set jsf components attributes using jstl variables) and with the last approach, I'm facing the problem described above.I wouldn't like to create components in java (UIComponents). Any suggestions?

    Read the article

  • XML cross-browser support

    - by 1anthony1
    I need help getting the file to run in Firefox: I have tried adapting scripts so that my file runs in both IE and Firefox but so far it still only works in IE. (The file can be tested at http://www.eyle.org/crosstest.html - simply type the word Mike in the text box using IE (doesn't work in Firefox).The HTML document is: <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Untitled Document</title> <script type="text/javascript"> var xmlDoc; //loads xml using either IE or firefox function loadXmlDoc() { //test for IE if(window.ActiveXObject) { xmlDoc = new ActiveXObject("Microsoft.XMLDOM"); xmlDoc.async = false; xmlDoc.load("books2.xml"); } //test for Firefox else if(document.implementation && document.implementation.createDocument) { xmlDoc = document.implementation.createDocument("","",null); xmlDoc.load("books2.xml"); } //if neither else {document.write("xml file did not load");} } //window.onload = loadXmlDoc(); var subject; //getDetails adds value of txtField to var subject in outputgroup(subject) function getDetails() { //either this or window.onload = loadXmlDoc is needed loadXmlDoc(); var subject = document.getElementById("txtField1").value; function outputgroup(subject) { var xslt = new ActiveXObject("Msxml2.XSLTemplate"); var xslDoc = new ActiveXObject("Msxml2.FreeThreadedDOMDocument"); var xslProc; xslDoc.async = false; xslDoc.resolveExternals = false; xslDoc.load("contains3books.xsl"); xslt.stylesheet = xslDoc; xslProc = xslt.createProcessor(); xslProc.input = xmlDoc; xslProc.addParameter("subj", subject); xslProc.transform(); document.write(xslProc.output); } outputgroup(subject); } </script> </head> <body> <input type="text" id="txtField1"> <input type="submit" onClick="getDetails(); return false"> </body> </html> The file includes books2.xml and contains3books.xsl (I have put the code for these files at ...ww.eyle.org/books2.xml ...ww.eyle.org/contains3books.xsl) (NB: replace ...ww. with http: // www)

    Read the article

  • XSLT Pagination

    - by dbomb101
    I have created a xslt document which formats an xml document, but I would like the results from the xslt sheet to be paginated. here is the orginal xlst document <?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" version="1.0" encoding="UTF-8" indent="yes"/> <xsl:template match="/"> <xsl:for-each select="musicInformation/musicdetails"> <label for="artistname{position()}" id="artistnameLabel{position()}">Artist Name:</label> <span id ="artistname{position()}"><xsl:value-of select="artistname" /></span> <br/> <label for="recordname{position()}" id="recordnameLabel{position()}">Record Name:</label> <span id ="recordname{position()}"><xsl:value-of select="recordname" /></span> <br/> <label for="recordtype{position()}" id="recordtypeLabel{position()}">Record Type:</label> <span id ="recordtype{position()}"><xsl:value-of select="recordtype" /></span> <br/> <label for="format{position()}" id="formatLabel{position()}">Format:</label> <span id ="format{position()}"><xsl:value-of select="format" /></span> <br/> <a href="xmlDetail.php?mid={@m_id}" >See Details</a> <br/><br/> </xsl:for-each> </xsl:template> </xsl:stylesheet>

    Read the article

  • Trying to use an Xslt for an xml in asp.net

    - by Josemalive
    Hello, i have the following xslt sheet: <?xml version="1.0" encoding="UTF-8" ?> <xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0"> <xsl:variable name="nhits" select="Answer[@nhits]"></xsl:variable> <xsl:output method="html" indent="yes"/> <xsl:template match="/"> <div> <xsl:call-template name="resultsnumbertemplate"/> </div> </xsl:template> <xsl:template name="resultsnumbertemplate"> <xsl:value-of select="$nhits"/> matches found </xsl:template> </xsl:stylesheet> And this is the xml that im trying to mix with the previous xslt: <Answer xmlns="exa:com.exalead.search.v10" context="n%3Dsl-ocu%26q%3Dlavadoras" last="9" estimated="false" nmatches="219" nslices="0" nhits="219" start="0"> <time> <Time interrupted="false" overall="32348" parse="0" spell="0" exec="1241" synthesis="15302" cats="14061" kwds="14061"> <sliceTimes>15272 </sliceTimes> </Time> </time> </Answer> Im using a xslcompiledtransform and that's working fine: XslCompiledTransform transformer = new XslCompiledTransform(); transformer.Load(HttpContext.Current.Server.MapPath("xslt\\" + requestvariables["xslsheet"].ToString())); transformer.Transform(xmlreader, null, writer); My problems comes when im trying to put into a variable the "nhits" attribute value placed on the Answer element, but i'm not rendering anything using my xslt sheet. Do you know what could be the cause? Could be the xmlns attribute in my xml file? Thanks in advance. Best Regards. Jose

    Read the article

  • Actually XSLT Lookup (Store variables during loop and use in it another template)

    - by krisvandenbergh
    Is there a way to store a variable/param during a for-each loop in a sort of array, and use it in another template, namely <xsl:template match="Foundation.Core.Classifier.feature">. All the classname values that appear during the for-each should be stored. How would you implement that in XSLT? Here's my current code. <xsl:for-each select="Foundation.Core.Class"> <xsl:for-each select="Foundation.Core.ModelElement.name"> <xsl:param name="classname"> <xsl:value-of select="Foundation.Core.ModelElement.name"/> </xsl:param> </xsl:for-each> <xsl:apply-templates select="Foundation.Core.Classifier.feature" /> </xsl:for-each> Here's the template in which the classname parameters should be used. <xsl:template match="Foundation.Core.Classifier.feature"> <xsl:for-each select="Foundation.Core.Attribute"> <owl:DatatypeProperty rdf:ID="{Foundation.Core.ModelElement.name}"> <rdfs:domain rdf:resource="$classname" /> </owl:DatatypeProperty> </xsl:for-each> </xsl:template> The input file can be found at http://krisvandenbergh.be/uml_pricing.xml

    Read the article

  • Access data of a XSL file from a JSF...

    - by Asela
    Hi all, I'm having 2 simple XML & XSL files as follows. form_1.xml <?xml version="1.0" encoding="windows-1252"?> <?xml-stylesheet type="text/xsl" href="form_1.xsl"?> <myform> </myform> form_1.xsl <?xml version="1.0" encoding="windows-1252"?> <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <xsl:template match="myform"> <html> <body> <form> <div align="center"> <h2>My first form in XSL</h2> <table cellspacing="1" cellpadding="1"> <tr> <td>First name : </td> <td> <input type="text"></input> </td> </tr> <tr> <td>Last name : </td> <td> <input type="text"></input> </td> </tr> <tr> <td>Address : </td> <td> <input type="text"></input> </td> </tr> </table> </div> </form> </body> </html> </xsl:template> </xsl:stylesheet> Now I have a JSF file where I have embeded the form_1.xml file inside an iFrame. Inside my JSF, I have submit & reset buttons as follows. myJsf.xhtml <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml" xmlns:h="http://java.sun.com/jsf/html" xmlns:f="http://java.sun.com/jsf/core" xmlns:ui="http://java.sun.com/jsf/facelets" xmlns:utils="http://java.sun.com/jsf/composite/utils"> <h:head> <title>::: The form filler application :::</title> <link href="./css/styles.css" rel="stylesheet" type="text/css" /> </h:head> <h:body> <div align="center"> <table cellspacing="2" cellpadding="2"> <tr> <th class="title">&nbsp; My form filler &nbsp;</th> </tr> </table> <br /> <fieldset><legend>Fill appropriate data in the following form</legend> <h:form> <div align="center"> <table> <tr> <td colspan="2"><iframe src="form_1.xml" frameborder="0" width="500px" height="500px"></iframe></td> </tr> <tr></tr> <tr> <td align="right"><h:commandButton value="Save data" action="#{myManagedBean.printValuesEnteredInTheForm}" /></td> <td align="left"><h:commandButton type="reset" value="Clear" /></td> </tr> </table> </div> </h:form></fieldset> </div> </h:body> </html> Now my question is that upon clicking the Submit button in my JSF, how do I access the values which I have entered in the XSL file? Any help is greatly appreciated. Thanks in advance. Reagrds, Asela.

    Read the article

  • Is it possible to navigate to the parent node of a matched node during XSLT processing?

    - by Darin
    I'm working with an OpenXML document, processing the main document part with some XSLT. I've selected a set of nodes via <xsl:template match="w:sdt"> </xsl:template> In most cases, I simply need to replace that matched node with something else, and that works fine. BUT, in some cases, I need to replace not the w:sdt node that matched, but the closest w:p ancestor node (ie the first paragraph node that contains the sdt node). The trick is that the condition used to decide one or the other is based on data derived from the attributes of the sdt node, so I can't use a typical xslt xpath filter. I'm trying to do something like this <xsl:template match="w:sdt"> <xsl:choose> <xsl:when test={first condition}> {apply whatever templating is necessary} </xsl:when> <xsl:when test={exception condition}> <!-- select the parent of the ancestor w:p nodes and apply the appropriate templates --> <xsl:apply-templates select="(ancestor::w:p)/.." mode="backout" /> </xsl:when> </xsl:choose> </xsl:template> <!-- by using "mode", only this template will be applied to those matching nodes from the apply-templates above --> <xsl:template match="node()" mode="backout"> {CUSTOM FORMAT the node appropriately} </xsl:template> This whole concept works, BUT no matter what I've tried, It always applies the formatting from the CUSTOM FORMAT template to the w:p node, NOT it's parent node. It's almost as if you can't reference a parent from a matching node. And maybe you can't, but I haven't found any docs that say you can't Any ideas?

    Read the article

  • How to fetch particular element from string after splitting it, using XSLT split() ?

    - by Vijay
    Hi All, I have one query regarding XSLT functions. I am using split function to fetch values from a string. I am using separator as "*". e.g. String : {"item1*item2*item3*item4"} split function I am using is as; <xsl:template name="SplitItemsCollection"> <xsl:param name="list" select="''"/> <xsl:param name="separator" select="'*'"/> <xsl:if test="not($list = '' or $separator = '')"> <xsl:variable name="head" select="substring-before(concat($list, $separator), $separator)"/> <xsl:variable name="tail" select="substring-after($list, $separator)"/> <xsl:value-of select="$head"/> <xsl:call-template name="split"> <xsl:with-param name="list" select="$tail"/> <xsl:with-param name="separator" select="$separator"/> </xsl:call-template> </xsl:if> </xsl:template> This function returns me all the items. Is there any way to fetch particular item from these separated items ? Say I want to fetch item3. Can we do it directly ? Thanks in advance.

    Read the article

  • Calling a .NET C# class from XSLT

    - by HanSolo
    If you've ever worked with XSLT, you'd know that it's pretty limited when it comes to its programming capabilities. Try writing a for loop in XSLT and you'd know what I mean. XSLT is not designed to be a programming language so you should never put too much programming logic in your XSLT. That code can be a pain to write and maintain and so it should be avoided at all costs. Keep your xslt simple and put any complex logic that your xslt transformation requires in a class. Here is how you can create a helper class and call that from your xslt. For example, this is my helper class:  public class XsltHelper     {         public string GetStringHash(string originalString)         {             return originalString.GetHashCode().ToString();         }     }   And this is my xslt file(notice the namespace declaration that references the helper class): <?xml version="1.0" encoding="UTF-8" ?> <xsl:stylesheet  xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0" xmlns:ext="http://MyNamespace">     <xsl:output method="text" indent="yes" omit-xml-declaration="yes"/>     <xsl:template  match="/">The hash code of "<xsl:value-of select="stringList/string1" />" is "<xsl:value-of select="ext:GetStringHash(stringList/string1)" />".     </xsl:template> </xsl:stylesheet>   Here is how you can include the helper class as part of the transformation: string xml = "<stringList><string1>test</string1></stringList>";             XmlDocument xmlDocument = new XmlDocument();             xmlDocument.LoadXml(xml);               XslCompiledTransform xslCompiledTransform = new XslCompiledTransform();             xslCompiledTransform.Load("XSLTFile1.xslt");               XsltArgumentList xsltArgs = new XsltArgumentList();                        xsltArgs.AddExtensionObject("http://MyNamespace", Activator.CreateInstance(typeof(XsltHelper)));               using (FileStream fileStream = new FileStream("TransformResults.txt", FileMode.OpenOrCreate, FileAccess.ReadWrite, FileShare.ReadWrite))             {                 // transform the xml and output to the output file ...                 xslCompiledTransform.Transform(xmlDocument, xsltArgs, fileStream);                            }

    Read the article

< Previous Page | 10 11 12 13 14 15 16 17 18 19 20 21  | Next Page >