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Search found 13 results on 1 pages for 'xlst'.

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  • How- XLST Transformation

    - by Yuan Ray
    Just wanted to ask on how to get the author names in the given xml sample below and put an attribut of eq="yes". EQ means Equal Contributors. This is the XML. <ArticleFootnote Type="Misc"> <Para>John Doe and Jane Doe are equal contributors.</Para> </ArticleFootnote> This should be the output in other form of XML. <AuthorGroups> <Authors eq="yes">John Doe</Authors> <Authors eq="yes">Jane Doe</Authors> </AuthorGroups> Assuming that JOhn Doe and Jane Doe are already defined in the list of authors but after the transformation, author tag should have the attribute eq="yes". Please help as I don't know much writing in xlst. Thanks in advance.

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  • How to sort an XML file by date in XLST

    - by AdRock
    I am trying to sort by date and get an error message about the stylesheet can't be loaded I found an answer on how others have suggested but it doesn't work for me Here is where it is supposed to sort. The commented out line is where the sort should occur <xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0"> <xsl:template name="hoo" match="/"> <html> <head> <title>Registered Festival Organisers and Festivals</title> <link rel="stylesheet" type="text/css" href="userfestival.css" /> </head> <body> <h1>Registered Festival Organisers and Festivals</h1> <xsl:for-each select="folktask/member"> <xsl:if test="user/account/userlevel='3'"> <!--<xsl:sort select="concat(substring(festival/event/datefrom,1,4),substring(festival/event/datefrom, 6,2),substring(festival/event/datefrom, 9,2))" data-type="number" order="ascending"/>--> Sample node from XML <festival id="1"> <event> <eventname>Oxford Folk Festival</eventname> <url>http://www.oxfordfolkfestival.com/</url> <datefrom>2010-04-07</datefrom> <dateto>2010-04-09</dateto> <location>Oxford</location> <eventpostcode>OX1 9BE</eventpostcode> <coords> <lat>51.735640</lat> <lng>-1.276136</lng> </coords> </event> </festival>

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  • Remove Empty Attributes from XML

    - by er4z0r
    Hi, I have a buggy xml that contains empty attributes and I have a parser that coughs on empty attributes. I have no control over the generation of the xml nor over the parser that coughs on empty attrs. So what I want to do is a pre-processing step that simply removes all empty attributes. I have managed to find the empty attribus, but now I don't know how to remove them: XPathFactory xpf = XPathFactory.newInstance(); XPath xpath = xpf.newXPath(); XPathExpression expr = xpath.compile("//@*"); Object result = expr.evaluate(d, XPathConstants.NODESET); if (result != null) { NodeList nodes = (NodeList) result; for(int node=0;node<nodes.getLength();node++) { Node n = nodes.item(node); if(isEmpty(n.getTextContent())) { this.log.warn("Found empty attribute declaration "+n.toString()); NamedNodeMap parentAttrs = n.getParentNode().getAttributes(); parentAttrs.removeNamedItem(n.getNodeName()); } } } This code gives me a NPE when accessing n.getParentNode().getAttributes(). But how can I remove the empty attribute from an element, when I cannot access the element?

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  • CruiseControl.Net - not able to see Project Statistics

    - by Anders Juul
    Hi all, I've made a reinstallation of the buildserver and can no longer see the standard graphs of project statistics. The error message shown is "Missing/Invalid statistics reports. Please check if you have enabled the Statistics Publisher, and statistics have been collected atleast once after that." To the best of my knowledge, the ccnet.config file has not been changed in this respect and by inspection it is verified that I have a Statistics / statisticsList-section for the project. Furthermore, the values appear in the Artifacts\statistics.csv and Artifacts\report.xml files. My guess would then be StatisticsGraph.xslt, which I have copied fresh from distribution to both Server\xlst and WebDashboard\xslt (why are they located in both places, by the way!?). Rebuild and check - still same error message. Any hints to how to debug this would be appreciated!

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  • Ignoring 'A' and 'The' when sorting with XSLT

    - by ChrisV
    I would like to have a list sorted ignoring any initial definite/indefinite articles 'the' and 'a'. For instance: The Comedy of Errors Hamlet A Midsummer Night's Dream Twelfth Night The Winter's Tale I think perhaps in XSLT 2.0 this could be achieved along the lines of: <xsl:template match="/"> <xsl:for-each select="play"/> <xsl:sort select="if (starts-with(title, 'A ')) then substring(title, 2) else if (starts-with(title, 'The ')) then substring(title, 4) else title"/> <p><xsl:value-of select="title"/></p> </xsl:for-each> </xsl:template> However, I want to use in-browser processing, so have to use XSLT 1.0. Is there any way to achieve this in XLST 1.0?

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  • How to get a non-XML output using JDOM XSLTransformer?

    - by Neil McF
    Hello, I have an XML file which I'd like to parse into a non-XML (text) file based on a XLST file. The code in both seem correct, and it works when testing manually, but I'm having a problem doing this programatically. I'm using JDOM's XSLTransformer class to apply the XSLT to the XML and it returns it in the format of a JDOM Document. The problem here is that I can't seem to access anything in the Document as it is not a proper XML file and I get a "java.lang.IllegalStateException: Root element not set" error. Is there a better way within Java to obtain a non-XML file as a result of XSLT?

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  • Firefox: how to apply XSLT from plugin to opened XML document? (how to replace document with another

    - by aloispaulin
    Hi! I've developed a plug-in for Firefox, that can read and manipulate the content of the currently opened document. I would like to apply a XLST to the document in case it is XML. I have no problem to read the XML document and apply a XSLT to it in memory, however, I have no idea of how to replace the existing document with the newly created one. Thus, I see two possible scenarios: a) I replace the document with the result of the transformation b) I apply the XSLT directly to the XML All my attempts to realize one of both possible solutions have failed... Hoping the community can provide help! Many tnx in advance! Alois

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  • 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>

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  • Java object graph -> xml when direction of object association needs to be reversed.

    - by Sigmoidal
    An application I have been working on has objects with a relationship similar to below. In the real application both objects are JPA entities. class Underlying{} class Thing { private Underlying underlying; public Underlying getUnderlying() { return underlying; } public void setUnderlying(final Underlying underlying) { this.underlying = underlying; } } There is a requirement in the application to create xml of the form: <template> <underlying> <thing/> <thing/> <thing/> </underlying> </template> So we have a situation where the object graph expresses the relationship between Thing and Underlying in the opposite direction to how it's expressed in the xml. I expect to use JAXB to create the xml but ideally I don't want to have to create a new object hierarchy to reflect the associations in the xml. Is there any way to create xml of the form required from the entities in their current form (through the use of xml annotations or something)? I don't have any experience using JAXB but from the limited research I've done it doesn't seem like it's possible to reverse the direction of association in any straightforward way. Any help/advice would be greatly appreciated. One other option that has been suggested is to use XLST to transform the xml into the correct format. I have done no research on this topic as yet but I'll add to the question when I have some more info. Thanks, Matt.

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  • XSLT 1.0 grouping to reformat element defined by date into element defined by task

    - by Daniel
    Hi folks, I have a tricky XSLT transformation and I'd like your advise My xml is formatted as below: <Person> <name>John</name> <date>June12</date> <workTime taskID=1>34</workTime> <workTime taskID=2>12</workTime> </Person> <Person> <name>John</name> <date>June12</date> <workTime taskID=1>21</workTime> <workTime taskID=2>11</workTime> </Person> The output xml should be: <Person> <name>John</name> <taskID>1</taskID> <workTime> <date>June12</date> <time>34</time> </worTime> <workTime> <date>June13</date> <time>21</time> </worTime> </Person> <Person> <name>John</name> <taskID>2</taskID> <workTime> <date>June12</date> <time>12</time> </worTime> <workTime> <date>June13</date> <time>11</time> </worTime> </Person> Essentially, as an input, a "Person" object gathers all the task/workTime for a specific date. As an output, I want the "Person" object to gather the date/workTime for a specific task. I need to use XLST 1.0. I've been trying to use grouping with key but get very puzzled. Appreciate your help. Daniel

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  • xslt help - my transform is not rendering correctly

    - by Hcabnettek
    Hi All, I'm trying to apply an xlst transformation using the following file. This is very basic, but I wanted to build off of this when I get it working correctly. <?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:template match="/"> <div> <h2>Station Inventory</h2> <hr/> <xsl:apply-templates/> </div> </xsl:template> Here is some xml I'm using for the source. <StationInventoryList xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="http://www.dummy-tmdd-address"> <StationInventory> <station-id>9940</station-id> <station-name>Zone 9940-SEB</station-name> <station-travel-direction>SEB</station-travel-direction> <detector-list> <detector> <detector-id>2910</detector-id> <detector-name>1999 West Smith Exit SEB</detector-name> </detector> <detector> <detector-id>9205</detector-id> <detector-name>CR-155 Exit SEB</detector-name> </detector> <detector> <detector-id>9710</detector-id> <detector-name>Pt of View SEB</detector-name> </detector> </detector-list> </StationInventory> </StationInventoryList> Any ideas what I'm doing wrong? The simple intent here is to make a list of station, then make a list of detectors at a station. This is a small piece of the XML. It would have multiple StationInventory elements. I'm using the data as the source for an asp:xml control and the xslt file as the transformsource. var service = new InternalService(); var result = service.StationInventory(); invXml.DocumentContent = result; invXml.TransformSource = "StationInventory.xslt"; invXml.DataBind(); Any tips are of course appreciated. Have a terrific weekend. Cheers, ~ck

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  • Issue with undefined namespace

    - by SoBeK
    "xmlns="VL01" Seems to be causing the style-sheet to fail(works fine if removed), do no know how to address it the style-sheet. I feel like this is basic XLST 101 but I am having a hard time wrapping my brain around it. Any assisting would be greatly appreciated. Cheers XML <?xml version="1.0" encoding="utf-8"?> <Report xsi:schemaLocation="VL01 http://site.com/ReportServer?%2FVMS%20Reports%2FVL01&amp;rs%3ACommand=Render&amp;rs%3AFormat=XML&amp;rs%3ASessionID=lk44ff55z5q3ck3b5pfuxo45&amp;rc%3ASchema=True" Name="VL01" textbox41="VL01 - Checklist Report&#xD;&#xA;" textbox1946=" 2) Target Element&#xD;&#xA;     Target Element List&#xD;&#xA;        Windows 7&#xD;&#xA;3) Report Options&#xD;&#xA; xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="VL01"> <list2> <Item_Collection> <Item /> </Item_Collection> </list2> <list1> <list1_Details_Group_Collection> <list1_Details_Group Key="V0001070" EffectiveDate="04 Mar 1998 16:03:47:000" LongName2="Name..."/> </list1_Details_Group_Collection> </list1> </Report> XSL <?xml version="1.0" encoding="UTF-8"?> <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"> <xsl:output method="xml" version="4.0" indent="yes"/> <xsl:variable name="var-checklist_name"> <xsl:value-of select="substring-after(substring-before(translate(Report/@textbox1946, '&#xD;&#xA;', ''),'3)'),'Target&#x00A0;Element&#x00A0;List')"/> </xsl:variable><xsl:template match="/"> <html> <body> <table> <xsl:apply-templates select="Report/list1/list1_Details_Group_Collection"/> </table> </body> </html> </xsl:template> <xsl:template match="Report/list1/list1_Details_Group_Collection"> <xsl:for-each select="list1_Details_Group"> <Import_List> <Checklist_Name> <xsl:value-of select="normalize-space(translate($var-checklist_name, '&#x00A0;', ' '))"/> </Checklist_Name> <Vuln_ID><xsl:value-of select="number(substring-after(@VulKey,'V'))"/></Vuln_ID> <Short_Name><xsl:value-of select="substring(@LongName2,1,255)"/></Short_Name> <Release_Date><xsl:value-of select="substring(@EffectiveDate,1,11)"/></Release_Date> </Import_List> </xsl:for-each> </xsl:template> </xsl:stylesheet>

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