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  • How do I de-duplicate a list of nodes in XSLT - and return the last node encountered?

    - by Broam
    I've seen lots of "de-duplicate this xml" questions but everyone wants the first node or the nodes are identical. I have a bit of a bigger puzzle. I have a list of articles in XML, a relevant snippet is shown: <item><key>Article1</key><stamp>100</stamp></item> <item><key>Article1</key><stamp>130</stamp></item> <item><key>Article2</key><stamp>800</stamp></item> <item><key>Article1</key><stamp>180</stamp></item> <item><key>Article3</key><stamp>900</stamp></item> <item><key>Article3</key><stamp>950</stamp></item> <item><key>Article4</key><stamp>990</stamp></item> <item><key>Article5</key><stamp>999</stamp></item> I'd like a list of nodes where the keys are unique and where the last instance is returned, not the first: Stamp (integer) is always increasing for elements of a particular key. Ideally I'd like "largest stamp" but they're always in order so the shortcut is ok. Desired result: (Order doesn't really matter.) <item><key>Article2</key><stamp>800</stamp></item> <item><key>Article1</key><stamp>180</stamp></item> <item><key>Article3</key><stamp>950</stamp></item> <item><key>Article4</key><stamp>990</stamp></item> <item><key>Article5</key><stamp>999</stamp></item> I'm somewhat confused on how to get this list. Any ideas? I'm using the Saxon processor if it matters.

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  • How to force javax xslt transformer to encode entities in utf-8?

    - by calavera.info
    I'm working on filter that should transform an output with some stylesheet. Important sections of code looks like this: PrintWriter out = response.getWriter(); ... StringReader sr = new StringReader(content); Source xmlSource = new StreamSource(sr, requestSystemId); transformer.setOutputProperty(OutputKeys.ENCODING, "UTF-8"); transformer.setParameter("encoding", "UTF-8"); //same result when using ByteArrayOutputStream xo = new java.io.ByteArrayOutputStream(); StringWriter xo = new StringWriter(); StreamResult result = new StreamResult(xo); transformer.transform(xmlSource, result); out.write(xo.toString()); The problem is that national characters are encoded as html entities and not by using UTF. Is there any way to force transformer to use UTF-8 instead of entities?

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  • Can anyone help me out in writing a xslt-fo for this xml file?

    - by atrueguy
    Currencies By Country Australia Australian Dollar Austria Schilling Belgium Belgium Franc Canada Canadian Dollar England Pound Fiji Fijian Dollar France Franc Germany DMark Hong Kong Hong Kong Dollar Italy Lira Japan Yen Netherlands Guilder Switzerland SFranc USA Dollar I started to write a xsl-fo to format the above xml in to a table, but I am really struggling with the flow of the tags, can any one help me out in writing a xsl-fo for this xml file? Is it possible for anyone to suggest me material for staring with xsl-fo, so that I can code my own xsl-fo., because the tags and syntax are very difficult to understand.

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  • Why don't all XSLT templates get executed at once?

    - by Matt W
    Refer to: http://stackoverflow.com/questions/1613454/retrieving-one-element-and-filter-by-another In the above linked posting, I don't understand why the two template blocks don't get executed upon in the incoming XML at the same time. As far I can can see, the XSL risks the second template executing for every Document element whether it is called with apply-templates or not. Could someone explain this to my stifled brain, please? Thanks, Matt.

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  • XSLT - make xsl:analyze-string return string instead of sequence of strings?

    - by tohuwawohu
    Is it possible to make xsl:analyze-string return one string instead of a sequence of strings? Background: I'd like to use xsl:analyze-string in a xsl:function that should encapsulate the pattern matching. Ideally, the function should return an xs:string to be used as sort criteria in an xsl:sort element. At the moment, i have to apply string-join() on every result of the function call since xsl:analyze-string returns a sequence of strings, and xsl:sort doesn't accept such a sequence as sort criteria. See line 24 of the stylesheet: <?xml version="1.0" encoding="UTF-8"?> <xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:my="www.my-personal-namespa.ce" xmlns:xs="http://www.w3.org/2001/XMLSchema"> <xsl:output indent="yes" method="xml" /> <xsl:function name="my:sortierung" > <xsl:param name="inputstring" as="xs:string"/> <xsl:analyze-string select="$inputstring" regex="[0-9]+"> <xsl:matching-substring> <xsl:value-of select="format-number(number(.), '00000')" /> </xsl:matching-substring> <xsl:non-matching-substring> <xsl:value-of select="." /> </xsl:non-matching-substring> </xsl:analyze-string> </xsl:function> <xsl:template match="/input"> <result> <xsl:apply-templates select="value" > <xsl:sort select="string-join((my:sortierung(.)), ' ')" /> </xsl:apply-templates> </result> </xsl:template> <xsl:template match="value"> <xsl:copy-of select="." /> </xsl:template> </xsl:stylesheet> with this input: <?xml version="1.0" encoding="UTF-8"?> <input> <value>A 1 b 120</value> <value>A 1 b 1</value> <value>A 1 b 2</value> <value>A 1 b 1a</value> </input> In my example, is there a way to modify the xsl:function to return a xs:string instead of a sequence?

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  • XSLT: How do I trigger a template when there is no input file?

    - by Ben Blank
    I'm creating a template which produces output based on a single string, passed via parameter, and does not use an input XML document. xsltproc seems to happily run with a single parameter specifying the stylesheet, but I don't see a way to trigger a template without an input file (no parameter to xsltproc to run a named template, for example). I'd like to be able to run: xsltproc --stringparam bar baz foo.xsl But I'm currently having to run, with the "main" template matching "/": echo '<xml/>' | xsltproc --stringparam bar baz foo.xsl - How can I get this to work? I'm sure I've seen other templates in the past which were meant to be run without an input document, but I don't remember how they worked or where to find them again. :-)

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  • How to apply or chain multiple matching templates in XSLT?

    - by Ignatius
    I am working on a stylesheet employing many templates with match attributes: <xsl:template match="//one" priority="0.7"> <xsl:param name="input" select="."/> <xsl:value-of select="util:uppercase($input)"/> <xsl:next-match /> </xsl:template> <xsl:template match="/stuff/one"> <xsl:param name="input" select="."/> <xsl:value-of select="util:add-period($input)"/> </xsl:template> <xsl:function name="util:uppercase"> <xsl:param name="input"/> <xsl:value-of select="upper-case($input)"/> </xsl:function> <xsl:function name="util:add-period"> <xsl:param name="input"/> <xsl:value-of select="concat($input,'.')"/> </xsl:function> What I would like to do is be able to 'chain' the two functions above, so that an input of 'string' would be rendered in the output as 'STRING.' (with the period.) I would like to do this in such a way that doesn't require knowledge of other templates in any other template. So, for instance, I would like to be able to add a "util:add-colon" method without having to open up the hood and monkey with the existing templates. I was playing around with the <xsl:next-match/> instruction to accomplish this. Adding it to the first template above does of course invoke both util:uppercase and util:add-period, but the output is an aggregation of each template output (i.e. 'STRINGstring.') It seems like there should be an elegant way to chain any number of templates together using something like <xsl:next-match/>, but have the output of each template feed the input of the next one in the chain. Am I overlooking something obvious?

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  • compact XSLT code to drop N number of tags if all are null.

    - by infant programmer
    This is my input xml: <root> <node1/> <node2/> <node3/> <node4/> <othertags/> </root> The output must be: <root> <othertags/> </root> if any of the 4 nodes isn't null then none of the tags must be dropped. example: <root> <node1/> <node2/> <node3/> <node4>sample_text</node4> <othertags/> </root> Then the output must be same as input xml. <root> <node1/> <node2/> <node3/> <node4>sample_text</node4> <othertags/> </root> This is the XSL code I have designed :: <xsl:template match="@*|node()"> <xsl:copy> <xsl:apply-templates select="@*|node()"/> </xsl:copy> </xsl:template> <xsl:template match="/root/node1[.='' and ../node2/.='' and ../node3/.='' and ../node4/.=''] |/root/node2[.='' and ../node1/.='' and ../node3/.='' and ../node4/.=''] |/root/node3[.='' and ../node1/.='' and ../node2/.='' and ../node4/.=''] |/root/node4[.='' and ../node1/.='' and ../node2/.='' and ../node3/.='']"/> As you can see the code requires more effort and becomes more bulky as the number of nodes increase. Is there any alternative way to overcome this bottleneck?

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  • Writing a template for XSLT Copy-Of to transform a message with differences. How?

    - by Emanuel Schuster
    I have an xml message that is in the older schema (xsd) format. My new schema is exactly the same but I embedded an element inside the older one. For example : My old schema had an element : <exclude> MyRestriction </exclude> but my new schema is like this : <exclude> <restriction> MyRestriction </restriction> </exclude> and the entire message is the same as before. Last time I used to do a copy-of but now I need to have a template that copy-of everything but move the value of the exclude to the restriction tag. Anyone can help me please ? Thanks

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  • How can I dynamically set the default namespace declaration of an XSLT transformation's output XML?

    - by Paralife
    I can do it, but not for the default namespace, using the <xsl:namespace>. If I try to do it for the default namespace: <xsl:namespace name="" select"myUri"/> it never works. It demands that I explicitly define the namespace of the element to be able to use the above null prefix declaration. The reason I want this is because I have a task to transform an input XML file to another output xml. The output XML has many elements and i dont want to have to explicitly set the namespace for every element. Thats why I want to set the default and never bother again. But the default must be computed from some data in the source XML. It does not change during the whole transformation, but it is dependent on input XML data. Any solution? EDIT 1: To sup up: I want to create a namespace dynamically and set it to be the default namespace of the output xml document. The uri of the namespace is derived from some data in the input XML. If I use <xsl:namespace> in my root output element, I cannot create a default namespace for it, only a prefixed one. And even with the prefixed one, it does not propagate to children.

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  • 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?

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  • XSLT : I need to parse the xml with same element name with sequence of order to map in to another xml with different name

    - by Karuna
    As the below source XML Value/string element value has to be replace with target element value, Could some please help me out how to create the XSL to transform from source xml into target xml .Please. Source XML: <PricingResultsV6> <subItems> <SubItem> <profiles> <ProfileValues> <values> <strings>800210</strings> <strings>THC</strings> <strings>10.0</strings> <strings>20.0</strings> <strings>30.0</strings> <strings>40.0</strings> <strings>550.0</strings> <strings>640.0</strings> </values> </ProfileValues> </rofiles> </SubItem> </subItems> </PricingResultsV6> Target XML : <CalculationOutput> <PolicyNumber> 800210 </PolicyNumber> <CommissionFactorMultiplier> THC </CommissionFactorMultiplier> <PremiumValue>10.0</PremiumValue> <SalesmanCommissionValue>20.0</SalesmanCommissionValue> <ManagerCommissionValue>30.0</ManagerCommissionValue> <GL_COR> 550.0</GL_COR> <GL_OPO>640.0</GL_OPO> </CalculationOutput>

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  • How to display workflow related tasks in the item display page where the workflow is currently running on in SharePoint2013

    - by ybbest
    In one of the project, I need to display workflow related tasks in the item display page where the workflow is currently running on. To achieve this, I’d like to add the tasks list view web part and using the connected web part to achieve this.(ID=workflowitemid) However, to make it work I need to unhide the workflowitemid field in the task list, as it is hidden field and also cantogglehidden field is set to false. I need to use reflection to change the cantogglehidden field to true as it only has getter in the API and then I am able to unhide the field. You can download the script here. However, it is not ideal (make your environment not supported by Microsoft) to display tasks this way. Another way to display the related task is to use SharePoint designer solution with List view web part and data source. Here are the steps. 1. Create a new list display form as below 2. Edit the custom display form in advanced mode. 3. Find the PlaceHolderMain contentplace hoder and insert the DataView by choosing the associated workflow tasks list as below 4. Go to the List View Tools >> OPTIONS 5. Create a Parameter called workflowitemId Parameter which retrieve the value from the ID querystring as below 6. Create a filter based on UIVersion = workflowitemId as below ,we are going to change the UIVersion to WorkflowItemId property later as WorkflowItemId is a hidden field and cannot be selected from the wizard. 7. Replace UIVersion with WorkflowItemId in the caml for the XsltListViewWebPart. From: TO 8. Go to the new custom display page at http://yourserver/Lists/aa/CustomDisplayPage.aspx?ID=414, you will see the associated tasks are showing in the page. References: http://office.microsoft.com/en-us/sharepoint-designer-help/watch-this-design-a-document-review-workflow-solution-HA010256417.aspx (Video 12 and 13)

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  • Multiple XML/XSLT files in PHP, transform one with XSLT and add others but process it first with PHP

    - by ipalaus
    I am processing XML files transformations with XSLT in PHP correctly. Actually I use this code: $xml = new DOMDocument; $xml->LoadXML($xml_contents); $xsl = new DOMDocument; $xsl->load($xsl_file); $proc = new XSLTProcesoor; $proc->importStyleSheet($xsl); echo $proc->transformToXml($xml); $xml_contents is the XML processed with PHP, this is done by including the XML file first and then assigning $xml_contents = ob_get_contents(); ob_end_clean();. This forces to process the PHP code on the XML, and it works perfectly. My problem is that I use more than one XML file and this XML files has PHP code on it that need to be processed AND have a XSLT file associated to process the data. Actually I'm including this files in XSLT with the next code: <!-- First I add the XML file --> <xsl:param name="menu" select="document('menu.xml')" /> <!-- Next I add the transformations for menu.xml file --> <xsl:include href="menu.xsl" /> <!-- Finally, I process it on the actual ("parent") XML --> <xsl:apply-templates select="$menu/menu" /> My questiion is how I can handle this. I need to add mutiple XML(+XSLT) files to my first XML file that will containt PHP so it needs to be processed. Thank you in advance!

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  • Why is xslt converter ignoring the content of <link>-tag

    - by Kristoffer Nolgren
    When I put forexample this in my xslt-stylesheet: <link><xsl:text>test</xsl:text></link> Or this: <link>test</link> I get the following result: <link xmlns=""></link> This however: <linkb>test</linkb> Render the following result: <linkb xmlns="">test</linkb> The rest of the xslt does not seem to make any difference, i've tried it in several different and empty xslt-stylesheets This problem appears in backend conversion (php) aswell as frontend-konversion in chrome browser (but not in Firefox) Example of error: dev.resihop.nu (right above the footer)

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  • Invalid URI: The Uri scheme is too long

    - by phenevo
    Hi, I have XML: Which is result of this part of query: SELECT Countries.FileSystemName as country ,Regions.DefaultName as region ,Provinces.DefaultName as province,cities.defaultname as city,cities.code as cityCode, IndividualFlagsWithForObjects.value as Status I have xslt: <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <xsl:output method="text" encoding="iso-8859-1"/> <xsl:param name="delim" select="string(',')" /> <xsl:param name="quote" select="string('&quot;')" /> <xsl:param name="break" select="string('&#xD;')" /> <xsl:template match="/"> <xsl:apply-templates select="results/countries" /> </xsl:template> <xsl:template match="countries"> <xsl:apply-templates /> <xsl:if test="following-sibling::*"> <xsl:value-of select="$break" /> </xsl:if> </xsl:template> <xsl:template match="*"> <!-- remove normalize-space() if you want keep white-space at it is --> <xsl:value-of select="concat($quote, normalize-space(.), $quote)" /> <xsl:if test="following-sibling::*"> <xsl:value-of select="$delim" /> </xsl:if> </xsl:template> <xsl:template match="text()" /> </xsl:stylesheet> And is part of code XmlReader reader = cmd.ExecuteXmlReader(); doc.LoadXml("<results></results>"); XmlNode newNode = doc.ReadNode(reader); while (newNode != null) { doc.DocumentElement.AppendChild(newNode); newNode = doc.ReadNode(reader); } doc.Save(@"c:\listOfCities.xml"); XslCompiledTransform XSLT = new XslCompiledTransform(); XsltSettings settings = new XsltSettings(); settings.EnableScript = true; XSLT.Load(@"c:\xsltfile1.xslt", settings, new XmlUrlResolver()); XSLT.Transform(doc.OuterXml,@"c:\myCities.csv"); <-here I get error Why I get error. Is seems to be good .

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  • Parameters not being passed into template when using the .Net transform classes

    - by Chris F
    I am using the .Net XslCompiledTranform to run some simple XSLT (see below for a simplified example). The example XSLT is meant to do simply show the value of the parameter that is passed in to the template. The output is what I expect it to be (i.e. <result xmlns:p1="http://www.doesnotexist.com"> <valueOfParamA>valueA</valueOfParamA> </result> when I use Saxon 9.0, but when I use XslCompiledTransform (XslTransform) in .net I get <result xmlns:p1="http://www.doesnotexist.com"> <valueOfParamA></valueOfParamA> </result> The problem is that that the parameter value of paramA is not being passed into the template when I use the .Net classes. I completely stumped as to why. when I step through in Visual Studio, the debugger says that the template will be called with paramA='valueA' but when execution switches to the template the value of paramA is blank. Can anyone explain why this is happening? Is this a bug in the MS implementation or (more likely) am I doing something that is forbidden in XSLT? Any help greatly appreciated. This is the XSLT that I am using <?xml version="1.0" encoding="utf-8"?> <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:extfn="http://exslt.org/common" exclude-result-prefixes="extfn" xmlns:p1="http://www.doesnotexist.com"> <!-- Replace msxml with xmlns:extfn="http://exslt.org/common" xmlns:extfn="urn:schemas-microsoft-com:xslt" --> <xsl:output method="xml" indent="yes"/> <xsl:template match="/"> <xsl:variable name="resultTreeFragment"> <p1:foo> </p1:foo> </xsl:variable> <xsl:variable name="nodeset" select="extfn:node-set($resultTreeFragment)"/> <result> <xsl:apply-templates select="$nodeset" mode="AParticularMode"> <xsl:with-param name="paramA" select="'valueA'"/> </xsl:apply-templates> </result> </xsl:template> <xsl:template match="@* | node()" mode="AParticularMode"> <xsl:param name="paramA"/> <valueOfParamA> <xsl:value-of select="$paramA"/> </valueOfParamA> </xsl:template> </xsl:stylesheet>

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  • Explicit script end tag always converted to self-closing

    - by Jonas
    I'm using xslt to transform xml to an aspx file. In the xslt, I have a script tag to include a jquery.js file. To get it to work with IE, the script tag must have an explicit closing tag. For some reason, this doesn't work with xslt below. <?xml version="1.0" encoding="utf-8"?> <xsl:stylesheet version="1.0" xmlns="http://www.w3.org/1999/xhtml" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:msxsl="urn:schemas-microsoft-com:xslt" exclude-result-prefixes="msxsl" xmlns:asp="remove"> <xsl:output method="html"/> <xsl:template match="/"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <title>TEST</title> <script type="text/javascript" src="jquery-1.2.6.js"></script> But if I change the script tag as shown below, it works. <script type="text/javascript" src="jquery-1.2.6.js"> // <![CDATA[ // ]]> </script> I thought that the <xsl:output method="html" /> would do the trick, but it doesn't seem to work? /Jonas

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  • How to feed an xml database with tags obtained thru html forms ?

    - by blaise1
    Hello! Not a programmer, I begin with xml, html forms and xslt on Mac. I plan to use a form to post short texts in a xhtml page and invite end users to add some annotations to the said text. The users would select a specific part of the text posted and each annotation would stand for one specific chain of characters. My goal is to consolidate the tags obtained from various user's annotations to one xml "knowledge base" containing the original text with all the revision indicators. Then I plan to use xslt sheets to product various reports based on the tags obtained. my two questions are : 1- am I dreaming ? Is it really possible to do that with xml, xforms, xslt without using java, php, ajax or other seasoned programmer's tools ? 2- What should be my focus for further explorations aiming in that direction ? Which schema, events, sequences should I study ? Je vous remercie à l'avance, Please excuse my English. Blaise

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