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• Php wrapper class for XML

  - by gms8994

  I'm working on a new class to wrap XML handling. I want my class to use simplexml if it's installed, and the built in XML functions if it's not. Can anyone give me some suggestions on a skeleton class to do this? It seems "wrong" to litter each method with a bunch of if statements, and that also seems like it would make it nearly impossible to correctly test. Any upfront suggestions would be great! EDIT: I'm talking about these built-in xml functions.

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• Creating multiple markers in Google Maps using XML

  - by Jessica Stanley

  I'm almost sure this question has been asked before, but for the love of me I just can't find the answer anywhere. Basically what I want to do is create multiple markers on a custom Google Map I'm building. I already have an XML file with the coordinates (lat/lng) and title of each item. I'd like to take the data from the XML file and use it to create markers on the map. I've found how to do this using KML files and MySQL/PHP, but I need to know how to do it in Javascript. One more thing: I have a .xml file of my own, so it won't be like I'll be getting the data from a webpage because I believe (from research I've done today) that the code for that may be different. If anyone knows if this has been posted somewhere else before, could you please direct me there? I've literally been searching all day, this is my last resort. Thanks a ton!!!

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• Parsing XML with nodes containing underscores.

  - by alvincrespo

  How do I parse an XML document that contains nodes where underscores exist? <some_xml> <child_node> <child width_info="" height_info="" /> </childnode> </some_xml> I tried this: for each (var item:XML in Environment._XMLData.some_xml.child_node.child){ trace(child.@width_info); } But that does'nt seem to work. I can't change the XML either because its from a third party. Any help would be great. Thanks in advance!

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• What processor is javax.xml.transform Using?

  - by Jeremy Witmer

  I've implented a simple webapp that transforms XML based on an XSTL stylesheet. It works fine on all the Windows servers I've deployed it on (to Tomcat), but on all Linux systems, I get a compile error on the XSLT. As best I can tell, it's because Java 1.6 isn't using the same processor behind javax.xml.transform. On the one Linux system, it's org.apache.xalan.xslt, version 2.4. What I can't figure out is how to generically figure out what any given system is using behind javax.xml.transform. Or, if anyone has any hints on what else I might do to figure out the problem, that'd be good, too.

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• Xml add element node using PHP

  - by user560411

  Hello I would like to add an element node in my entire XML file using PHP. I can not get it right and I add it below root. Please share me a clean code so that I understand how it works.. Thank you! My xml is <?xml version="1.0" encoding="utf-8"?> <GameStore> <Game type="adventure"> <TITLE>Assassin's Creed: Brotherhood</TITLE> <PUBLISHER>Ubisoft</PUBLISHER> </Game> <Game type="adventure"> <TITLE>Batman: Arkham Asylum</TITLE> <PUBLISHER>Eidos</PUBLISHER> </Game> </GameStore>

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• Create custom component based on LinearLayout, declaring layout in XML

  - by pheelicks

  I've been trying to create a Compound Control in Android 1.5 (as described here) but havn't been able to find any good examples on how to do this using an XML file to specify a layout. I'm fine with creating an Activity and then loading an xml file using the following in the constructor: setContentView(R.layout.main); However, I want to do this in subclass of LinearLayout - so I can use this compound component in other XML layouts. Something along the lines of: public class CustomView extends LinearLayout { public CustomView(Context context) { super(context); setupView(); } public CustomView(Context context, AttributeSet attrs) { super(context, attrs); setupView(); } public void setupView() { setContentView(R.layout.custom); // Not possible } } What is the correct way of going about doing this?

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• find xml element by attribute

  - by Moudy

  Using JQuery or Javascript how would I return 'Mary Boone' from the xml below starting out with the show 'id' attribute of '2'? I'm thinking something along the lines of - var result = xml.getElementByAttribute("2").gallery.text(); the XML: <shows> <show id="1"> <artist>Andreas Gursky</artist> <gallery>Matthew Marks</gallery> <medium>photography</medium> </show> <show id="2"> <artist>Eric Fischl</artist> <gallery>Mary Boone</gallery> <medium>painting</medium> </show> </shows>

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• Generating report from an xml file

  - by Vignesh

  I want to generate a report from an xml, preferably html. The html here should be dynamic to allow limiting the view based on some user entered values, preferably selecting from a drop down of categories, which inturn is populated from the xml. I also want to have links in the report to more info which is stored in another xml file. I started off with javascript with xslt for display and I'm still long way in acheiving my desires. Are there any other ways to do it?? Any automated Open sources tools, for this, rather than reinventing.

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• Validating XML tag by tag

  - by Greatful

  Hi All I'm having some issues validating some XML against a Schema, using .net and C#. I am using XmlReaderSettings with the ValidationEventHandler. However, this seems to stop catching errors after it has encountered the first error at a particular level within the XML file, instead of checking the next tag at the same level, so basically it does not check each and every tag within the XML file instead skipping a level when it has found an error. I was hoping to get some advice from somebody who has successfully accomplished this type of validation. Thanks very much

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• C# XML Serialisation Only Serialise Single Element in List

  - by guazz

  Given some sample XML such as: <XML> <EMPLOYEES> <EMPLOYEE isBestEmployee="false">John"<"/EMPLOYEE> <EMPLOYEE isBestEmployee="true">Joe"<"/EMPLOYEE> <EMPLOYEE isBestEmployee="false">Bill"<"/EMPLOYEE> </EMPLOYEES> </XML> How do I serialise just the employee with isBestEmployee="true" to a single Employee object?

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• parser the xml and build a ui for user

  - by hguser

  Hi: Now I have to handle some xml in my java swing application. I have to build a swing ui according to the special schema ,then user can fill some values.After user completed,I will collect the information,validate the value and then build a xml file. For building xml file I can use the xmlbeans,however how to parse the schema and build a swing ui? Since the schema is rather complex. A schema can be found here: example schema I have to parser this schema,for the LiteralInputType ,a JTextArea should be built. However there are other types "complexType" and etc.. These types may not occur at the sametime. Some times only the LiteralInputType is needed,somethimes the ComplexType is needed,also maybe all of them are needed. So, how to implement it? Anyone can help me?

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• Generating 8000 text files from xml files

  - by Ray

  Hi all, i need to generate the same number of text files as the xml files i have. Within the text files, i need the title and maybe some other tags of it. I can generate text files with the elements i wanted but not all xml files can be generated. Only some of them are generated. Something might be wrong with my parser so help out please thanks. This is my code. Please have a look and give me suggestions. Thanks in advance. import java.io.File; import javax.xml.parsers.DocumentBuilder; import javax.xml.parsers.DocumentBuilderFactory; import org.w3c.dom.*; import java.io.*; public class AccessingXmlFile1 { public static void main(String argv[]) { try { //File file = new File("C:\\MyFile.xml"); // create a file that is really a directory File aDirectory = new File("C:/Documents and Settings/I2R/Desktop/test"); // get a listing of all files in the directory String[] filesInDir = aDirectory.list(); System.out.println(""+filesInDir.length); // sort the list of files (optional) // Arrays.sort(filesInDir); //////////////////////////////////////////////////////////////////////////////////// //////////////////////////////////////////////////////////////////////////////////// // have everything i need, just print it now for ( int a=0; a<filesInDir.length; a++ ) { String xmlFile = filesInDir[a]; String newLine = System.getProperty("line.separator"); File file = new File(xmlFile); DocumentBuilderFactory dbf = DocumentBuilderFactory.newInstance(); DocumentBuilder db = dbf.newDocumentBuilder(); Document document = db.parse(file); document.getDocumentElement().normalize(); //System.out.println("Root element " + document.getDocumentElement().getNodeName()); NodeList node = document.getElementsByTagName("metadata"); System.out.println("Information of Xml File"); System.out.println(xmlFile.substring(0, xmlFile.length() - 4)); //////////////////////////////////////////////////////////////////////////////////// String titleStoreText = ""; String descriptionStoreText = ""; String collectionStoreText = ""; String textToWrite = ""; //////////////////////////////////////////////////////////////////////////////////// for (int i = 0; i < node.getLength(); i++) { Node firstNode = node.item(i); if (firstNode.getNodeType() == Node.ELEMENT_NODE) { Element element = (Element) firstNode; NodeList titleElementList = element.getElementsByTagName("title"); Element titleElement = (Element) titleElementList.item(0); NodeList title = titleElement.getChildNodes(); //////////////////////////////////////////////////////////////////////////////////// if(titleElement == null) titleStoreText = " There is no title for this file."+ newLine; else titleStoreText = titleStoreText+((Node) title.item(0)).getNodeValue() + newLine; //titleStoreText = titleStoreText+((Node) title.item(0)).getNodeValue()+ newLine; //////////////////////////////////////////////////////////////////////////////////// System.out.println("Title : " + titleStoreText); NodeList collectionElementList = element.getElementsByTagName("collection"); Element collectionElement = (Element) collectionElementList.item(0); NodeList collection = collectionElement.getChildNodes(); //////////////////////////////////////////////////////////////////////////////////// if(collectionElement == null) collectionStoreText = " There is no collection for this file."+ newLine; else collectionStoreText = collectionStoreText+((Node) collection.item(0)).getNodeValue() + newLine; //collectionStoreText = collectionStoreText+((Node) collection.item(0)).getNodeValue()+ newLine; //////////////////////////////////////////////////////////////////////////////////// System.out.println("Collection : " + collectionStoreText); NodeList descriptionElementList = element.getElementsByTagName("description"); Element descriptionElement = (Element) descriptionElementList.item(0); NodeList description = descriptionElement.getChildNodes(); //////////////////////////////////////////////////////////////////////////////////// if(descriptionElement == null) descriptionStoreText = " There is no description for this file."+ newLine; else descriptionStoreText = descriptionStoreText+((Node) description.item(0)).getNodeValue() + newLine; //descriptionStoreText = descriptionStoreText+((Node) description.item(0)).getNodeValue() + newLine; //////////////////////////////////////////////////////////////////////////////////// System.out.println("Description : " + descriptionStoreText); //////////////////////////////////////////////////////////////////////////////////// textToWrite = "=====Title=====" + newLine + titleStoreText + newLine + "=====Collection=====" + newLine + collectionStoreText + newLine + "=====Description=====" + newLine + descriptionStoreText;// + newLine + "=====Subject=====" + newLine + subjectStoreText; //////////////////////////////////////////////////////////////////////////////////// } } ///////////////////////////////////////////write to file part is here///////////////////////////////////////// Writer output = null; File file2 = new File(xmlFile.substring(0, xmlFile.length() - 4)+".txt"); output = new BufferedWriter(new FileWriter(file2)); output.write(textToWrite); output.close(); System.out.println("Your file has been written"); //////////////////////////////////////////////////////////////////////////////////// } } catch (Exception e) { e.printStackTrace(); } } }

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• Best tools / formats for documenting XML API?

  - by ivant

  We are developing XML over HTTP service and we need a way to document the XML interface. Our supported XMLs generally are subsets of some industry standard XML API, which unfortunately lacks any good documentation. It has XSDs though with some annotations, so we use this go generate the initial documentation and then remove the unsupported parts and refine the result. But this process is tedious and the results aren't at all great. Any suggestion about documentation formats and tools? I prefer to have HTML output and possibly more, e.g. PDF. Oh, and we are mixed Linux and Windows shop, so the tools should work well on both.

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• Generalised XML Serialization

  - by Tom W

  I apologise for asking a question that's probably been asked hundreds of times before, but I don't seem to be able to find an answer in the archives; probably because my question is too basic. I know that XML Serialization by default only touches public members and properties. Properties very often mask a private variable; particularly if they're readonly. Serializing these is fine; the value that the instance exposes to the world is what goes into the XML. But if Deserialization of the same data can't put the value back where it belongs, what's the point of doing it? Is there something I'm missing about how XML Serialization is normally used for classes with masking properties? Surely it can't be that the only answer is explicitly implementing Read/WriteXML - because that's more effort than it's worth!

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• RSS-Feed or XML-Stream?

  - by Maenny

  HI Folks, I am programming a website for online-reservation-management, preferably for holiday homes. I want to implement a service, which allows the user to pass all their occupation dates to other websites. My question is, what would be the most professional way to do this? Generate a XML-stream Create a valid RSS Feed I think that a XML file has the advantages of being straight to the point, without any information that noone needs. Webmasters of other websites surely prefer the structure of a simple XML file. On the other hand, when someone without programming knowledge wants to use the RSS-Feed with a feedreader or simply with the browser, it may be user-friendlier to use a RSS-FEED. What so you think, I'm unsure...? Greetz Maenny

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• Can I serialize a BitArray to XML?

  - by SimonNet

  Hello I have a business class which I need to serialize to xml. It has a BitArray property. I have decorated it with [XmlAttribute] but the serialization is failing with To be XML serializable, types which inherit from ICollection must have an implementation of Add(System.Boolean) at all levels of their inheritance hierarchy. System.Collections.BitArray does not implement Add(System.Boolean). I am not sure whether its possible to serialize to xml? If not what would be an efficient means of serializing the BitArray Thanks for looking

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• how to sort xml data in jQuery

  - by pixeltocode

  how can i sort all officers based on their ranks jQuery $.get('officers.xml', function(grade){ $(grade).find('officer').each(function(){ var $rank = $(this).attr('rank'); }); }); XML (officer.xml) <grade> <officer rank="2"></student> <officer rank="3"></student> <officer rank="1"></student> </grade> thanks.

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• Excel XML Newline in Cell

  - by spasarto

  I am using LINQ to XML to create an Excel XML file. I want to include newlines within the cells. Excel uses the &#10; literal to represent a new line. If I try to add this using an XText: XText newlineElement = new XText( "Foo&#10;Bar" ); I get: Foo&amp;#10;Bar Which shows up in Excel as: Foo&#10;Bar Is there a way to write &#10 to the XML file without doing a String.Replace( "&amp;#10;", "&#10;" ) over the resulting file text?

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• Are their any good way to genrate XML file through C# in .net

  - by steven spielberg

  i wanto make a xml file like <?xml version="1.0" encoding="Windows-1252" ?> <settings> <typeofsetting> <wordname="add" /> </typeofsettings> </settings> the wordname can be depend on what user need. how i can make a application that user can generate the XML file of thing they want. are their any good way to do this. the wordname not user defined it's come from database the application have inbuilt. are their any good practice to do this in c# win-forms.

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• soap response to xml (simple_xml)

  - by ntan

  Hi i am trying to convert a soap response to XML. Soap has an envelop and a body <soap:Envelope xmlns:soap="http://schemas.xmlsoap.org/soap/envelope/" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema"> <soap:Body> When try to convert $responseXML = simplexml_load_string($string); I get object(SimpleXMLElement)#20 (0) { } If i edit the $string as soap-Envelope and soap-Body i can get the XML Whats wrong with : and can not get XML I hope its clear. Anyone

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• receive xml file via post in php

  - by thegunner

  Hi, I'm looking for a PHP script that can accept an XML file via a POST, then send a response.... Does anyone have any code that could do this? So far the only code I have is this but not sure about the response or if indeed I am even going in the right direction as XML characters are not saved correctly. Any ideas? <?php if ( $_SERVER['REQUEST_METHOD'] === 'POST' ){ $postText = file_get_contents('php://input'); } $datetime=date('ymdHis'); $xmlfile = "myfile" . $datetime . ".xml"; $FileHandle = fopen($xmlfile, 'w') or die("can't open file"); fwrite($FileHandle, $postText); fclose($FileHandle); ?> Thanks,

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• Opening a new screen from xml file text.

  - by Clozecall

  Hey all I'm looking to edit my app so that when a text within an xml file is selected, a new window opens. I've found various ways of making a new screen open via a class, but I need it done from a xml file. My program as a tablayout, and here is some of the stuff displayed within my first tab: <LinearLayout android:orientation="vertical" android:layout_width="fill_parent" android:layout_height="fill_parent"> android:id="@+id/ScrollView01" android:layout_height="wrap_content" android:layout_width="fill_parent" <TextView android:textColor="#FFFFFF" android:text="First Header here" android:textStyle="bold" android:layout_width="fill_parent" android:layout_height="wrap_content" /> <TextView android:textColor="#FFFFFF" android:text="First body text here" android:layout_width="fill_parent" android:layout_height="wrap_content" /> So basically I need it so that when a text is selected, a new screen pops up. As said the reason this specifically has to be done in xml is because this is all apart of a tab layout, It would be great if the next within each tab could open a a new screen. Thanks.

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• Is there any Android XML documentation?

  - by Eddified

  Is there any sort of xml reference? I found this which turned out to be invaluable for me http://groups.google.com/group/android-developers/msg/d334017d72909c79 but I can't figure out how I was supposed to know how to do that, had I not found that post. I know that the api reference has xml attributes listed for many of the classes... but what about xml tags? Where is it documented that I could build a shape using , , tags? I'd really like to know where I can find such documentation.

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• Python universal feedparser xml

  - by timg

  I am trying to read an xml feed with the python feedparser, but cannot seem to navigate the elements. Here is what I am trying: import feedparser d = feedparser.parse('http://www.website.com/feed') text= d.status.test and here is the xml: <?xml version="1.0" encoding="UTF-8"?> <statuses type="array"> <status> <created>Tue Dec 21 14:16:12 +0000 2010</created> <id>123</id> <text>Hello</text> </status> </statuses>

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• Parse XML with XPath & namespaces in Java

  - by ripper234

  Can you help me adjust this code so it manages to parse the XML? If I drop the XML namespace it works: String webXmlContent = "<?xml version=\"1.0\" encoding=\"UTF-8\"?>\n" + "<foo xmlns=\"http://foo.bar/boo\"><bar>baz</bar></foo>"; DocumentBuilderFactory domFactory = DocumentBuilderFactory.newInstance(); domFactory.setNamespaceAware(true); DocumentBuilder builder = domFactory.newDocumentBuilder(); org.w3c.dom.Document doc = builder.parse(new StringInputStream(webXmlContent)); NamespaceContextImpl namespaceContext = new NamespaceContextImpl(); namespaceContext.startPrefixMapping("foo", "http://www.w3.org/2001/XMLSchema-instance"); XPath xpath = XPathFactory.newInstance().newXPath(); xpath.setNamespaceContext(namespaceContext); XPathExpression expr = xpath.compile("/foo/bar"); Object result = expr.evaluate(doc, XPathConstants.NODESET); NodeList nodes = (NodeList) result; System.out.println("Got " + nodes.getLength() + " nodes");

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