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  • Linq to XSD processing

    - by goel
    For Typed Xml to Linq processing ,do i need to download Linq to XSD alpha installer or is it available in VS 2008 ? (I am using VS 2008 Professional edition).

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  • Cast problem with LINQ

    - by yigit
    I'm tring to get my product's types to a list with Linq. var types = (from t in NHibernateSession.Linq<Product>() select t.Type).Distinct().ToList<Type>(); return types; But its giving an Unable to cast object of type error '...Domain.Product' to type '...Domain.Type'. Please tell where am I going wrong.

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  • XMl to XHTML using XSL with namespace

    - by user256007
    How to handle XML Elements with namespace in an XSL. Here goes my XML Document <?xml version="1.0" encoding="UTF-8"?> <?xml-stylesheet type="text/xml" href="wpanel.xsl" ?> <html xmlns="http://www.w3.org/1999/xhtml" xmlns:bong="http://bong/glob/wpanel" xml:lang="en" lang="en"> <head> <title> Transitional DTD XHTML Example </title> <link rel="stylesheet" type="text/css" href="wpanel.css" /> </head> <body> <bong:QButton text="Submit"></bong:QButton> <bong:QButton text="Reset"></bong:QButton> </body> </html> and My XSL goes here <?xml version="1.0" ?> <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:bong="http://bong/glob/wpanel" xmlns:fn="http://www.w3.org/2005/xpath-functions" > <xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes" /> <xsl:template match="*|@*"> <xsl:copy> <xsl:apply-templates select="@*"/> <xsl:apply-templates/> </xsl:copy> </xsl:template> <xsl:template match="processing-instruction()|comment()"> <xsl:copy>.</xsl:copy> </xsl:template> I want to copy the elements without bong namespace as it is But transform those using the bong namespace. I think there will be an xsl element or attribute for that.I've serched the net. but haven't found yet. Thank You.

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  • Parsing XML via jQuery, nested loops

    - by Coughlin
    I am using jQuery to parse XML on my page using $.ajax(). My code block is below and I can get this working to display say each result on the XML file, but I am having trouble because each section can have MORE THAN ONE and im trying to print ALL grades that belong to ONE STUDENT. Here is an example of the XML. <student num="505"> <name gender="male">Al Einstein</name> <course cid="1">60</course> <course cid="2">60</course> <course cid="3">40</course> <course cid="4">55</course> <comments>Lucky if he makes it to lab, hopeless.</comments> </student> Where you see the I am trying to get the results to print the grades for EACH student in each course. Any ideas on what I would do? Thanks, Ryan $.ajax({ type: "GET", url: "final_exam.xml", dataType: "xml", success: function(xml) { var student_list = $('#student-list'); $(xml).find('student').each(function(){ $(xml).find('course').each(function(){ gradeArray = $(this).text(); console.log(gradeArray); }); var name = $(this).find("name").text(); var grade = $(this).find("course").text(); var cid = $(this).find("course").attr("cid"); //console.log(cid); student_list.append("<tr><td>"+name+"</td><td>"+cid+"</td><td>"+grade+"</td></tr>"); }); } });

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  • XML::LibXML::XPathContext - Question

    - by sid_com
    This script works with and without XPathContext. Why should I use it with XPathContext? #!/usr/bin/env perl use warnings; use strict; use XML::LibXML; use 5.012; my $parser = XML::LibXML->new; my $doc = $parser->parse_string(<<EOT); <?xml version="1.0"?> <xml> Text im Dokument <element id="myID" name="myname" style="old" /> <object objid="001" objname="Object1" /> <element id="002" name="myname" /> </xml> EOT #/ # without XPathContext my $nodes = $doc->findnodes( '/xml/element[@id=002]' ); # with XPathContext #my $root = $doc->documentElement; #my $xc = XML::LibXML::XPathContext->new( $root ); #my $nodes = $xc->findnodes( '/xml/element[@id=002]' ); for my $node ( $nodes->get_nodelist ) { say "Node: ", $node->nodeName; print "Attribute: "; print $_->getName, '=', $_->getValue, ' ' for $node->attributes; say ""; }

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  • Update element values using xml.dom.minidom

    - by amnesia-55
    Hello, I have an XML structure which looks similar to: <Store> <foo> <book> <isbn>123456</isbn> </book> <title>XYZ</title> <checkout>no</checkout> </foo> <bar> <book> <isbn>7890</isbn> </book> <title>XYZ2</title> <checkout>yes</checkout> </bar> </Store> Using xml.dom.minidom only (restrictions) i would like to 1)traverse through the XML file 2)Search/Get for particular element, depending on its parent Example: checkout element for author1, isbn for author2 3)Change/Set that element's value 4)Write the new XML structure to a file Can anyone help here? Thank you! UPDATE: This is what i have done till now import xml.dom.minidom checkout = "yes" def getLoneChild(node, tagname): assert ((node is not None) and (tagname is not None)) elem = node.getElementsByTagName(tagname) if ((elem is None) or (len(elem) != 1)): return None return elem def getLoneLeaf(node, tagname): assert ((node is not None) and (tagname is not None)) elem = node.getElementsByTagName(tagname) if ((elem is None) or (len(elem) != 1)): return None leaf = elem[0].firstChild if (leaf is None): return None return leaf.data def setcheckout(node, tagname): assert ((node is not None) and (tagname is not None)) child = getLoneChild(node, 'foo') Check = getLoneLeaf(child[0],'checkout') Check = tagname return Check doc = xml.dom.minidom.parse('test.xml') root = doc.getElementsByTagName('Store')[0] output = setcheckout(root, checkout) tmp_config = '/tmp/tmp_config.xml' fw = open(tmp_config, 'w') fw.write(doc.toxml()) fw.close()

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  • Problem with XML encoding of database contents with Latin characters

    - by user89691
    I have an ASP Access database that contains strings in various European languages. The database was populated prior by agents in the respective countries. It contains entries with accented etc characters as you would expect. If I open the database with MS Access these characters show up fine. For example the the German equivalent of "Open" shows as "Öffnen" (hopefully you can see an "O" with 2 dots above it!). I have ASP code that reads the database and returns records in XML. The text is passed to XMLEncode to construct the XML, but that only seems to deal with the 5 specials like "<", "&", etc. If I dump the XML the accented characters are unchanged. <English>Open</English> <German>Öffnen</German> If I look at the raw packets with Wireshark I see that the "Ö" byte is hex D6, which appears to be it's decimal Unicode and ISO 8859-1 value. The problem starts when I try to parse the XML in client-side JS. I get: "An invalid character was found in text content" from IE. FF and Chrome happily accept the XML without hiccup but the browser shows the "Ö" character as a diamond with a question mark inside. http://www.validome.org/xml/validate/ reports "encoding error." http://www.w3schools.com/dom/dom_validate.asp thinks it is fine. The XML is UTF-8 encoded. What do I need to do to have IE accept my XML without complaint? What do I need to do to have browsers display the stuff correctly?

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  • Writing Russian in XML

    - by zavié
    Hi, I am writing a Xml Tag Renamer class with Java which reads in a XML, renames the tags and write them back into another XML file using DocumentBuilderFactory and TransformerFactory (text nodes are preserved). It worked fine before with German and English texts, until today, when I tried to rename a XML file with russian text. Instead of the source texts I got ????? in the newly created XML file. I've tried setting Encoding Any idea how to correct this? Thanks! PS. Strings were correct before entering TransformerFactory, as I checked in the debugger. I've tried setting OutputKeys.ENCODING to UTF-8 and ISO-8859-5. None of them helped. The Transformer part: // Output the XML // Set up a transformer TransformerFactory transFactory = TransformerFactory.newInstance(); Transformer transformer = transFactory.newTransformer(); transformer.setOutputProperty(OutputKeys.OMIT_XML_DECLARATION, "no"); // Fix to a bug about indent in transformer transformer.setOutputProperty ("{http://xml.apache.org/xslt}indent-amount", "4"); transformer.setOutputProperty(OutputKeys.INDENT, "yes"); // TODO encoding parameter transformer.setOutputProperty(OutputKeys.ENCODING, "UTF-8"); // Create string from xml tree StringWriter sw = new StringWriter(); StreamResult result = new StreamResult(sw); DOMSource source = new DOMSource(doc); transformer.transform(source, result); String xmlString = sw.toString(); xmlString.replaceAll("\n", System.getProperty("line.separator")); // Write to file BufferedWriter output = new BufferedWriter(new FileWriter(outputPath)); output.write(xmlString); output.close();

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  • Flash AS3 load file xml

    - by Elias
    Hello, I'm just trying to load an xml file witch can be anywere in the hdd, this is what I have done to browse it, but later when I'm trying to load the file it would only look in the same path of the swf file here is the code package { import flash.display.Sprite; import flash.events.; import flash.net.; public class cargadorXML extends Sprite { public var cuadro:Sprite = new Sprite(); public var file:FileReference; public var req:URLRequest; public var xml:XML; public var xmlLoader:URLLoader = new URLLoader(); public function cargadorXML() { cuadro.graphics.beginFill(0xFF0000); cuadro.graphics.drawRoundRect(0,0,100,100,10); cuadro.graphics.endFill(); cuadro.addEventListener(MouseEvent.CLICK,browser); addChild(cuadro); } public function browser(e:Event) { file = new FileReference(); file.addEventListener(Event.SELECT,bien); file.browse(); } public function bien(e:Event) { xmlLoader.addEventListener(Event.COMPLETE, loadXML); req=new URLRequest(file.name); xmlLoader.load(req); } public function loadXML(e:Event) { xml=new XML(e.target.data); //xml.name=file.name; trace(xml); } } } when I open a xml file that isnt it the same directory as the swf, it gives me an unfound file error. is there anything I can do? cause for example for mp3 there is an especial class for loading the file, see http://www.flexiblefactory.co.uk/flexible/?p=46 thanks

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  • Flex through Flash Builder 4; Connecting to a dynamic XML feed: "The response is not a valid XML or

    - by jtromans
    I am learning how to use Flex with Adobe Flash Builder 4 standalone. I am working my through the Adobe Flash Build 4 Bible by David Gassner. This has led me to create my own micro problems to try and solve. I am trying to connect to a dynamix XML feed created by the following aspx page: generate_xml.aspx When I create the data connection through the Data/Service panel, I can pick between XML and HTTP. I figured because the generate_xml.aspx has to generate the XML file first, I should use the HTTP service as opposed to the XML. The HTTP service offers GET, which seems to be the kinda thing I want. However, I am really struggling to do this. I keep getting: "The response is not a valid XML or a JSON string" The actual STATIC generated XML file that is created by this page works perfectly when I save it and manually connect with the XML service. Therefore I know my XML code is properly formatted and contains no other HTML of JavaScript. I figure my problem occurs because the page itself is .aspx, but I cannot work out how to successfully ask Flex to request the output of this page, rather than the page itself.

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  • faster way to change xml to array(grails to flex)

    - by Anthony Umpad
    I have a large xml passed from grails to flex. When flex receives the xml, it converts the xml into an associative array object. Given the large xml file, it takes too long to complete the loop, is there any way in flex to make conversion faster? Below is my sample code. <xml> <car> <model>Vios</model> <type>Sedan</type> <color>Blue</color> </car> <car> <model>Camry</model> <type>Luxury</type> <color>Black</color> </car> </xml> *converted to the flex associative array below.* [Vios].type = Sedan .color = Blue [Camry].type = Luxury .color = Black *Below is a code I used in flex to convert the xml to the associative array object* var tempXML=xml.children() var tempArray:Array= new Array() for(var i:int=0;i<tempXML.length();i++) { tempArray[tempXML[i].@model]= new Object(); tempArray[tempXML[i].@model].color = tempXML[i][email protected](); tempArray[tempXML[i].@model].type = tempXML[i][email protected](); }

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  • PHP4: Send XML over HTTPS/POST via cURL?

    - by starmonkey
    I wrote a class/function to send xml over https via PHP4/cURL, just wondering if this is the correct approach, or if there's a better one. Note that PHP5 is not an option at present. /** * Send XML via http(s) post * * curl --header "Content-Type: text/xml" --data "<?xml version="1.0"?>...." http://www.foo.com/ * */ function sendXmlOverPost($url, $xml) { $ch = curl_init(); curl_setopt($ch, CURLOPT_URL, $url); // For xml, change the content-type. curl_setopt ($ch, CURLOPT_HTTPHEADER, Array("Content-Type: text/xml")); curl_setopt($ch, CURLOPT_POST, 1); curl_setopt($ch, CURLOPT_POSTFIELDS, $xml); curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1); // ask for results to be returned if(CurlHelper::checkHttpsURL($url)) { curl_setopt($ch, CURLOPT_SSL_VERIFYPEER, 0); curl_setopt($ch, CURLOPT_SSL_VERIFYHOST, 0); } // Send to remote and return data to caller. $result = curl_exec($ch); curl_close($ch); return $result; } cheers!

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  • Save a form in an XML file using Ajax and JSP

    - by novellino
    Hello, I want to create a simple form with a name and an email and save these data in an XML file. So far I found that using Ajax with jQuery is quite easy. So I used the usual code: //dataString have the values taken from the form var dataString = 'name='+ name + '&email=' + email; $.ajax({ type: "POST", url: "users.xml", data: dataString, dataType: "xml", success: function() { .... } }); If I understood well, in the url I should add the name of the XML file that will be created. When the user clicks a button I call the function with the Ajax request, and then I should call somewhere a function for generating the xml. I am using also two beans. One is for setting the elements of the user and the other is for saving the data in the XML. I am using the XStream library for the xml although I don't know if is the best solution. The problem now it that I can not connect all these together in order to save the data in the XML. Does anyone know what should I do? Thanks a lot!

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  • How to find/extract data from xml with jQuery

    - by darryl
    I'm trying to extract the StateLongName and StateShortName values from the xml below. I know there has to be a simple elegant way to do this with jQuery. <NewDataSet> <Table> <StateLongName>Alabama</StateLongName> <StateShortName>AL</StateShortName> </Table> <Table> <StateLongName>Alaska</StateLongName> <StateShortName>AK</StateShortName> </Table> ...elments removed for brevity </NewDataSet> Here's what I've tried. Load the xml from above into a Javascript variable name xml. Try #1 $(xml).find("TABLE").each(function() { var stateName = $(this).find("StateLongName").innerText; var stateCode = $(this).find("StateShortName").innerText; }); Try #1 doesn't find anything and never goes inside to load the stateName and stateCode variables. Try #2 $(xml).find("StateLongName").each(function() { var stateName = $(this).find("StateLongName").innerText; var stateCode = $(this).find("StateShortName").innerText; }); Try #2 does find matches, however the stateName and stateCode are left undefined. Try #3 $(xml).find("StateLongName").each(function() { var stateName = $($(xml).find('StateLongName').parent()[0].innerHTML)[1].data; var stateCode = $($(xml).find('StateLongName').parent()[0].innerHTML)[5].data; }); Try #3 works but there has to be a better way. Please enlighten me. Thanks for you time!

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  • Idiomatic way to build a custom structure from XML zipper in Clojure

    - by Checkers
    Say, I'm parsing an RSS feed and want to extract a subset of information from it. (def feed (-> "http://..." clojure.zip/xml-zip clojure.xml/parse)) I can get links and titles separately: (xml-> feed :channel :item :link text) (xml-> feed :channel :item :title text) However I can't figure out the way to extract them at the same time without traversing the zipper more than once, e.g. (let [feed (-> "http://..." clojure.zip/xml-zip clojure.xml/parse)] (zipmap (xml-> feed :channel :item :link text) (xml-> feed :channel :item :title text))) ...or a variation of thereof, involving mapping multiple sequences to a function that incrementally builds a map with, say, assoc. Not only I have to traverse the sequence multiple times, the sequences also have separate states, so elements must be "aligned", so to speak. That is, in a more complex case than RSS, a sub-element may be missing in particular element, making one of sequences shorter by one (there are no gaps). So the result may actually be incorrect. Is there a better way or is it, in fact, the way you do it in Clojure?

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  • ajax html vs xml/json responses - perfomance or other reasons

    - by pedalpete
    I've got a fairly ajax heavy site and some 3k html formatted pages are inserted into the DOM from ajax requests. What I have been doing is taking the html responses and just inserting the whole thing using jQuery. My other option is to output in xml (or possibly json) and then parse the document and insert it into the page. I've noticed it seems that most larger site do things the json/xml way. Google Mail returns xml rather than formatted html. Is this due to performance? or is there another reason to use xml/json vs just retrieving html? From a javascript standpoint, it would seem injecting direct html is simplest. In jQuery I just do this jQuery.ajax({ type: "POST", url: "getpage.php", data: requestData, success: function(response){ jQuery('div#putItHear').html(response); } with an xml/json response I would have to do jQuery.ajax({ type: "POST", url: "getpage.php", data: requestData, success: function(xml){ $("message",xml).each(function(id) { message = $("message",xml).get(id); $("#messagewindow").prepend(""+$("author",message).text()+ ": "+$("text",message).text()+ ""); }); } }); clearly not as efficient from a code standpoint, and I can't expect that it is better browser performance, so why do things the second way?

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  • Transform RSS-Feed into another "standard" XML-Format with PHP

    - by ChrisBenyamin
    Hey friends, quick question: I need to transform a default RSS Structure into another XML-format. The RSS File is like.... Name des RSS Feed Feed Beschreibung de http://xml-rss.de Sat, 1 Jan 2000 00:00:00 GMT Titel der Nachricht Die Nachricht an sich http://xml-rss.de/link-zur-nachricht.htm Sat, 1. Jan 2000 00:00:00 GMT 01012000-000000 Titel der Nachricht Die Nachricht an sich http://xml-rss.de/link-zur-nachricht.htm Sat, 1. Jan 2000 00:00:00 GMT 01012000-000000 Titel der Nachricht Die Nachricht an sich http://xml-rss.de/link-zur-nachricht.htm Sat, 1. Jan 2000 00:00:00 GMT 01012000-000000 ...and I want to extract only the item-elements (with childs and attributes) XML like: <?xml version="1.0" encoding="ISO-8859-1"?> <item> <title>Titel der Nachricht</title> <description>Die Nachricht an sich</description> <link>http://xml-rss.de/link-zur-nachricht.htm</link> <pubDate>Sat, 1. Jan 2000 00:00:00 GMT</pubDate> <guid>01012000-000000</guid> </item> ... It hasn't to be stored into a file. I need just the output. I tried different approaches with DOMNode, SimpleXML, XPath, ... but without success. Thanks chris

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  • C# XML export for Excel table using XmlDocument

    - by Mark
    I am trying to write the following into an XML document: <head> <xml> <x:ExcelWorkbook> <x:ExcelWorksheets> <x:ExcelWorksheet> other code here </x:ExcelWorksheet> </x:ExcelWorksheets> </x:ExcelWorkbook> </xml> </head> However, if I use the following code, it strips out the 'x:'. System.Xml.XmlDocument document = new System.Xml.XmlDocument(); System.Xml.XmlElement htmlNode = document.CreateElement("html"); htmlNode.SetAttribute("xmlns:o", "urn:schemas-microsoft-com:office:office"); htmlNode.SetAttribute("xmlns:x", "urn:schemas-microsoft-com:office:excel"); htmlNode.SetAttribute("xmlns", "http://www.w3.org/TR/REC-html40"); document.AppendChild(htmlNode); System.Xml.XmlElement headNode = document.CreateElement("head"); htmlNode.AppendChild(headNode); headNode.AppendChild( document.CreateElement("xml")).AppendChild( document.CreateElement("x:ExcelWorkbook"))).AppendChild( document.CreateElement("x:ExcelWorksheets")).AppendChild( document.CreateElement("x:ExcelWorksheet")).InnerText="other code here"; How can I stop this from happening? Thanks!

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  • How to check if a node's value in one XML is present in another XML with a specific attribute?

    - by Manish
    The question seems to be little confusing. So let me describe my situation through an example: Suppose I have an XML: A.xml <Cakes> <Cake>Egg</Cake> <Cake>Banana</Cake> </Cakes> Another XML: B.xml <Cakes> <Cake show="true">Egg</Cake> <Cake show="true">Strawberry</Cake> <Cake show="false">Banana</Cake> </Cakes> Now I want to show some text say "TRUE" if all the Cake in A.xml have show="true" in B.xml else "FALSE". In the above case, it will print FALSE. I need to develop an XSL for that. I can loop through all the Cake in A.xml and check if that cake has show="true" in B.xml but I don't know how to break in between (and set a variable) if a show="false" is found. Is it possible? Any help/comments appreciated.

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  • Having trouble parsing XML with jQuery

    - by Jack
    Hi Guys, I'm trying to parse some XML data using jQuery, and as it stands I have extracted the 'ID' attribute of the required nodes and stored them in an array, and now I want to run a loop for each array member and eventually grab more attributes from the notes specific to each ID. The problem currently is that once I get to the 'for' loop, it isn't looping, and I think I may have written the xml path data incorrectly. It runs once and I recieve the 'alert(arrayIds.length);' only once, and it only loops the correct amount of times if I remove the subsequent xml path code. Here is my function: var arrayIds = new Array(); $(document).ready(function(){ $.ajax({ type: "GET", url: "question.xml", dataType: "xml", success: function(xml) { $(xml).find("C").each(function(){ $("#attr2").append($(this).attr('ID') + "<br />"); arrayIds.push($(this).attr('ID')); }); for (i=0; i<arrayIds.length; i++) { alert(arrayIds.length); $(xml).find("C[ID='arrayIds[i]']").(function(){ // pass values alert('test'); }); } } }); }); Any ideas?

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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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