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  • Java -Android. Parser problem

    - by Kano
    I am making a very simple app with an RSS reader. The reader works great, but it's only giving me the title, and i want the description too. I'am very new to android, and I have tried a lot of things, but I can't get it to work. I've found a lot of parsers but they are to complicated for me to understand, so I was hoping to find a simple solution, since it's only title and description i want. Can anyone help me? import java.io.IOException; import java.net.MalformedURLException; import java.net.URL; import javax.xml.parsers.ParserConfigurationException; import javax.xml.parsers.SAXParser; import javax.xml.parsers.SAXParserFactory; import org.xml.sax.Attributes; import org.xml.sax.InputSource; import org.xml.sax.SAXException; import org.xml.sax.XMLReader; import org.xml.sax.helpers.DefaultHandler; import android.app.Activity; import android.os.Bundle; import android.widget.TextView; public class NyhedActivity extends Activity { String streamTitle = ""; @Override protected void onCreate(Bundle savedInstanceState) { // TODO Auto-generated method stub super.onCreate(savedInstanceState); setContentView(R.layout.nyheder); TextView result = (TextView)findViewById(R.id.result); try { URL rssUrl = new URL("http://tv2sport.dk/rss/*/*/*/248/*/*"); SAXParserFactory mySAXParserFactory = SAXParserFactory.newInstance(); SAXParser mySAXParser = mySAXParserFactory.newSAXParser(); XMLReader myXMLReader = mySAXParser.getXMLReader(); RSSHandler myRSSHandler = new RSSHandler(); myXMLReader.setContentHandler(myRSSHandler); InputSource myInputSource = new InputSource(rssUrl.openStream()); myXMLReader.parse(myInputSource); result.setText(streamTitle); } catch (MalformedURLException e) { // TODO Auto-generated catch block e.printStackTrace(); result.setText("Cannot connect RSS!"); } catch (ParserConfigurationException e) { // TODO Auto-generated catch block e.printStackTrace(); result.setText("Cannot connect RSS!"); } catch (SAXException e) { // TODO Auto-generated catch block e.printStackTrace(); result.setText("Cannot connect RSS!"); } catch (IOException e) { // TODO Auto-generated catch block e.printStackTrace(); result.setText("Cannot connect RSS!"); } } private class RSSHandler extends DefaultHandler { final int stateUnknown = 0; final int stateTitle = 1; int state = stateUnknown; int numberOfTitle = 0; String strTitle = ""; String strElement = ""; @Override public void startDocument() throws SAXException { // TODO Auto-generated method stub strTitle = "Nyheder fra "; } @Override public void endDocument() throws SAXException { // TODO Auto-generated method stub strTitle += ""; streamTitle = "" + strTitle; } @Override public void startElement(String uri, String localName, String qName, Attributes attributes) throws SAXException { // TODO Auto-generated method stub if (localName.equalsIgnoreCase("title")) { state = stateTitle; strElement = ""; numberOfTitle++; } else { state = stateUnknown; } } @Override public void endElement(String uri, String localName, String qName) throws SAXException { // TODO Auto-generated method stub if (localName.equalsIgnoreCase("title")) { strTitle += strElement + "\n"+"\n"; } state = stateUnknown; } @Override public void characters(char[] ch, int start, int length) throws SAXException { // TODO Auto-generated method stub String strCharacters = new String(ch, start, length); if (state == stateTitle) { strElement += strCharacters; } } } }

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  • Insert a doctype into an XML document (Java/ SAX)

    - by Thom Nichols
    Imagine you have an XML document and imagine you have the DTD but the document itself doesn't actually specify a DOCTYPE ... How would you insert the DOCTYPE declaration, preferably by specifying it on the parser (similar to how you can set the schema for a document that will be parsed) or by inserting the necessary SAX events via an XMLFilter or the like? I've found many references to EntityResolver, but that is what's invoked once a DOCTYPE is found during parsing and it's used to point to a local DTD file. EntityResolver2 appears to have what I'm looking for but I haven't found any examples of usage. This is the closest I've come thus far: (code is Groovy, but close enough that you should be able to understand it...) import org.xml.sax.* import org.xml.sax.ext.* import org.xml.sax.helpers.* class XmlFilter extends XMLFilterImpl { public XmlFilter( XMLReader reader ) { super(reader) } @Override public void startDocument() { super.startDocument() super.resolveEntity( null, 'file:///./entity.dtd') println "filter startDocument" } } class MyHandler extends DefaultHandler2 { public InputSource resolveEntity(String name, String publicId, String baseURI, String systemId) { println "entity: $name, $publicId, $baseURI, $systemId" return new InputSource(new StringReader('<!ENTITY asdf "&#161;">')) } } def handler = new MyHandler() def parser = XMLReaderFactory.createXMLReader() parser.setFeature 'http://xml.org/sax/features/use-entity-resolver2', true def filter = new XmlFilter( parser ) filter.setContentHandler( handler ) filter.setEntityResolver( handler ) filter.parse( new InputSource(new StringReader('''<?xml version="1.0" ?> <test>one &asdf; two! &nbsp; &iexcl;&pound;&cent;</test>''')) ); I see resolveEntity called but still hit org.xml.sax.SAXParseException: The entity "asdf" was referenced, but not declared. at com.sun.org.apache.xerces.internal.parsers.AbstractSAXParser.parse(AbstractSAXParser.java:1231) at org.xml.sax.helpers.XMLFilterImpl.parse(XMLFilterImpl.java:333) I guess this is because there's no way to add SAX events that the parser knows about, I can only add events via a filter that's upstream from the parser which are passed along to the ContentHandler. So the document has to be valid going into the XMLReader. Any way around this? I know I can modify the raw stream to add a doctype or possibly do a transform to set a DTD... Any other options?

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  • "Java & XML" job title

    - by Aspiring developer
    When people say they are a "Java & XML" developer, what do they mean? Are they saying they create XML data to be used in applications, or they use XML to configure a framework like Spring or Hibernate? What other ways are there to use XML in a Java application?

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  • Middle East XML Currency Conversion

    - by Tim
    Using the following script to do currency conversion which relies on an XML feed. http://www.white-hat-web-design.co.uk/articles/php-currency-conversion.php It grabs the data from the following feed... var $xml_file = "www.ecb.int/stats/eurofxref/eurofxref-daily.xml"; However this XML feed has limited currencies and I require currencies for the Middle East. Does anyone know where I can find an XML file with Middle East currencies or have any better suggestions?

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  • Create an Xml file from an object

    - by remi bourgarel
    I work as a web developer with a web designer and we usually do like this : - I create the system , I generate some Xml files - the designer display the xml files with xslt Nothing new. My problem is that I use Xml Serialization to create my xml files, but I never use Deserialization. So I'd like to know if there is a way to avoid fix like these : empty setter for my property empty parameter-less constructor implement IXmlSerializable and throw "notimplementedexception" on deserialization do a copy of the class with public fields

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  • Parsing a UTF-16 encoded xml file in ruby

    - by Matthew Toohey
    Hello I've been trying to parse a UTF-16 encoded xml file in Ruby (1.8.7), and I can't seem to find how to do it by searching (google and stack overflow) Here's the xml file url: http://www.abc.net.au/triplej/feeds/playout/triplejsydneyplayout.xml?_5366 Getting the xml string from Net::HTTP and passing it to REXML, then calling logger.info xmlDoc.inspect produces: <UNDEFINED> ... </> Any ideas? Cheers

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  • Create an Xml file from an object (c#)

    - by remi bourgarel
    Hi All, I work as a web developer with a web designer and we usually do like this : - I create the system , I generate some Xml files - the designer display the xml files with xslt Nothing new. My problem is that I use Xml Serialization to create my xml files, but I never use Deserialization. So I'd like to know if there is a way to avoid fix like these : empty setter for my property empty parameter-less constructor implement IXmlSerializable and throw "notimplementedexception" on deserialization do a copy of the class with public fields thanks.

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  • Log4Net & RollingFileAppender to generate Xml files

    - by SaguiItay
    I've managed to configure Log4Net with a RollingFileAppender in order to generate Xml files. However, the generated files are not valid XML files until a "roll" is performed - the XML doesn't have a closing XML tag. Basically, this prevents to files from being read until that are "closed"/"rolled". Anyone else encountered this issue? I my previous (custom) solution I had to write the closing tag after writing each entry, and overwrite it with the next entry... :(

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  • XML security in world wide web

    - by nikky
    Hi, Im a newbie in XML and i have some questions Can XML be used in stead of normal database (store data in a tuple and column) in website? XML is built to share information easier (from my understanding) such as can share cross platform and in different language used so Is it secure to store secure data in XML? thank you so much

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  • Modify an XML file in Python

    - by michele
    Hi, I have two file. I have to modify the file one in a particular node and add in a list of child. The list is in the file2. Can I do it, and how? from xml.dom.minidom import Document from xml.dom import minidom file1=modificare.xml file2=sorgente.xml xmldoc=minidom.parse(file1) for Node in xmldoc.getElementsByTagName("Sampler"): # put in the file2 content Thanks a lot.

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  • Blocking problem, C#, .net, Deserializing XML to object problem

    - by fernando
    Hi I have a blocking problem I have XML file under some url http://myserver/mywebApp/myXML.xml In the below code which I run in Console Application, bookcollection has null Books field :( <books> <book id="5352"> <date>1986-05-05</date> <title> Alice in chains </title> </book> <book id="4334"> <date>1986-05-05</date> <title> 1000 ways to heaven </title> </book> <book id="1111"> <date>1986-05-05</date> <title> Kitchen and me </title> </book> </books> XmlDocument doc = new XmlDocument(); doc.Load("http://myserver/mywebapp/myXML.xml"); BookCollection books = new BookCollection(); XmlNodeReader reader2 = new XmlNodeReader(doc.DocumentElement); XmlSerializer ser2 = new XmlSerializer(books.GetType()); object obj = ser2.Deserialize(reader2); BookCollection books2= (BookCollection)obj; using System; using System.Collections.Generic; using System.Linq; using System.Text; namespace ConsoleApplication1 { [Serializable()] public class Book { [System.Xml.Serialization.XmlAttribute("id")] public string id { get; set; } [System.Xml.Serialization.XmlElement("date")] public string date { get; set; } [System.Xml.Serialization.XmlElement("title")] public string title { get; set; } } } using System; using System.Collections.Generic; using System.Linq; using System.Text; using System.Xml.Serialization; namespace ConsoleApplication1 { [Serializable()] [System.Xml.Serialization.XmlRootAttribute("books", Namespace = "", IsNullable = false)] public class BookCollection { [XmlArray("books")] [XmlArrayItem("book", typeof(Book))] public Book[] Books { get; set; } } }

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  • Validate xml against xsd using c++

    - by manu
    Hi , i am very new to xml and c++. i want to validate xml against xsd using c++ api.can any one gimme any tutorial link? or sample program.i don want to do using msxml as its works in windows and visual studio. i tried to use xerces and libxml but i have failed.now i am trying using tinyxml. please help me to validate xml against xml schema using c++(not vc++,should work in cross platform) regards, manu

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  • XSL transformation and special XML entities escaping

    - by Tomas R
    I have an XML file which is transformed with XSL. Some elements have to be changed, some have to be left as is - specifically, text with entities &quot;, &amp;, &apos;, &lt;, &gt; should be left as is, and in my case &quot; and &apos; are changed to " and ' accordingly. Test XML: <?xml version="1.0" encoding="UTF-8" ?> <root> <element> &quot; &amp; &apos; &lt; &gt; </element> </root> transformation file: <?xml version="1.0" encoding="UTF-8"?> <xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0"> <xsl:output method="xml" encoding="UTF-8" omit-xml-declaration="no" indent="no" /> <xsl:template match="element"> <xsl:copy> <xsl:value-of disable-output-escaping="no" select="." /> </xsl:copy> </xsl:template> </xsl:stylesheet> result: <?xml version="1.0" encoding="UTF-8"?> <element> " &amp; ' &lt; &gt; </element> desired result: <?xml version="1.0" encoding="UTF-8"?> <element> &quot; &amp; &apos; &lt; &gt; </element> I have 2 questions: why does some of those entities are transformed and other not? how can I get a desired result?

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  • Flash caroussel xml parse html link

    - by Marvin
    Hello I am trying to modify a carousel script I have in flash. Its normal function is making some icons rotate and when clicked they zoom in, fade all others and display a little text. On that text I would like to have a link like a "read more". If I use CDATA it wont display a thing, if I use alt char like &#60;a href=&#34;www.google.com&#34;&#62; Read more + &#60;/a&#62; It just displays the text as: <a href="www.google.com"> Read more + </a>. The flash dynamic text box wont render it as html. I dont enough as2 to figure out how to add this. My code: var xml:XML = new XML(); xml.ignoreWhite = true; //definições do xml xml.onLoad = function() { var nodes = this.firstChild.childNodes; numOfItems = nodes.length; for(var i=0;i<numOfItems;i++) { var t = home.attachMovie("item","item"+i,i+1); t.angle = i * ((Math.PI*2)/numOfItems); t.onEnterFrame = mover; t.toolText = nodes[i].attributes.tooltip; t.content = nodes[i].attributes.content; t.icon.inner.loadMovie(nodes[i].attributes.image); t.r.inner.loadMovie(nodes[i].attributes.image); t.icon.onRollOver = over; t.icon.onRollOut = out; t.icon.onRelease = released; } } And the xml: <?xml version="1.0" encoding="UTF-8"?> <icons> <icon image="images/product.swf" tooltip="Product" content="Hello this is some random text &#60;a href=&#34;www.google.com&#34;&#62; Read More + &#60;/a&#62; "/> </icons> Any suggestions? Thanks.

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  • Blocking problem Deserializing XML to object problem

    - by fernando
    I have a blocking problem I have XML file under some url http://myserver/mywebApp/myXML.xml In the below code which I run in Console Application, bookcollection has null Books field :( <books> <book id="5352"> <date>1986-05-05</date> <title> Alice in chains </title> </book> <book id="4334"> <date>1986-05-05</date> <title> 1000 ways to heaven </title> </book> <book id="1111"> <date>1986-05-05</date> <title> Kitchen and me </title> </book> </books> XmlDocument doc = new XmlDocument(); doc.Load("http://myserver/mywebapp/myXML.xml"); BookCollection books = new BookCollection(); XmlNodeReader reader2 = new XmlNodeReader(doc.DocumentElement); XmlSerializer ser2 = new XmlSerializer(books.GetType()); object obj = ser2.Deserialize(reader2); BookCollection books2= (BookCollection)obj; using System; using System.Collections.Generic; using System.Linq; using System.Text; namespace ConsoleApplication1 { [Serializable()] public class Book { [System.Xml.Serialization.XmlAttribute("id")] public string id { get; set; } [System.Xml.Serialization.XmlElement("date")] public string date { get; set; } [System.Xml.Serialization.XmlElement("title")] public string title { get; set; } } } using System; using System.Collections.Generic; using System.Linq; using System.Text; using System.Xml.Serialization; namespace ConsoleApplication1 { [Serializable()] [System.Xml.Serialization.XmlRootAttribute("books", Namespace = "", IsNullable = false)] public class BookCollection { [XmlArray("books")] [XmlArrayItem("book", typeof(Book))] public Book[] Books { get; set; } } }

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  • Writing to a xml file in java

    - by user243680
    import java.io.*; import javax.xml.parsers.*; import javax.xml.transform.*; import javax.xml.transform.dom.*; import javax.xml.transform.stream.*; import org.w3c.dom.*; public class CreatXMLFile { public static void main(String[] args) throws Exception { BufferedReader bf = new BufferedReader(new InputStreamReader(System.in)); // System.out.print("Enter number to add elements in your XML file: "); // String str = bf.readLine(); int no=2; // System.out.print("Enter root: "); String root = "SMS"; DocumentBuilderFactory documentBuilderFactory =DocumentBuilderFactory.newInstance(); DocumentBuilder documentBuilder =documentBuilderFactory.newDocumentBuilder(); Document document = documentBuilder.newDocument(); Element rootElement = document.createElement(root); document.appendChild(rootElement); // for (int i = 1; i <= no; i++) // System.out.print("Enter the element: "); // String element = bf.readLine(); String element ="Number"; System.out.print("Enter the Number: "); String data = bf.readLine(); Element em = document.createElement(element); em.appendChild(document.createTextNode(data)); rootElement.appendChild(em); String element1 ="message"; System.out.print("Enter the SMS: "); String data1 = bf.readLine(); Element em1 = document.createElement(element1); em1.appendChild(document.createTextNode(data1)); rootElement.appendChild(em1); TransformerFactory transformerFactory = TransformerFactory.newInstance(); Transformer transformer = transformerFactory.newTransformer(); DOMSource source = new DOMSource(document); StreamResult result = new StreamResult(System.out); transformer.transform(source, result); } } i am working on the above code and it gives the following output run: Enter the Number: 768678 Enter the SMS: ytu <?xml version="1.0" encoding="UTF-8" standalone="no"?><SMS><Number>768678</Number><message>ytu</message></SMS>BUILD SUCCESSFUL (total time: 8 seconds) Now i want to write the output generated(<?xml version="1.0" encoding="UTF-8" standalone="no"?><SMS><Number>768678</Number><message>ytu</message></SMS>) to a XML file on the hard disk.How do i do it?

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  • XML parse file from HTTP

    - by Travis
    I have an XML file located at a location such as http://example.com/test.xml I'm trying to parse the XML file to use it in my program with xPath like this but it is not working. Document doc = builder.parse(new File(url)); How can I get the XML file?

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  • Optimal xml storage engine

    - by nixau
    I'm considering optimal open source solution for storing xml documents with further querying on them effectively. Amount of data will be small. As far as I understand native xml databases might form a proper solution for my case. They obviously store xml documents in highly efficient way. It would be great to learn your experience. Any suggestions on proper solution? Have you got any experience employing xml storage engines in your apps?

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  • Redirection of xml file for iphone app

    - by gotye
    Hey guys, I have a xml file on my server like the one below : www.myWebSite.com/myXmlFile.xml which is used by my iphone application. In case the address of my xml file changes to www.myOtherWebsite.com/myXmlFile.xml How can I make my app to work anyway ? What kind of PHP server-side code do I need to write ? Is NSURLConnection supporting reirections ? Thanks for any incomings ;) Gotye.

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  • Qt C++ XML, validating against a DTD?

    - by Airjoe
    Is there a way to validate an XML file against a DTD with Qt's XML handling? I've tried googling around but can't seem to get a straight answer. If Qt doesn't include support for validating an XML file, what might be the process of implementing validation myself? Any good reference to start with in regards to validating XML against a spec? Thanks for the help!

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  • Is there a way to enforce/preserve order of XML elements in an XML Schema?

    - by MarcoS
    Let's consider the following XML Schema: <?xml version="1.0" encoding="UTF-8"?> <schema targetNamespace="http://www.example.org/library" elementFormDefault="qualified" xmlns="http://www.w3.org/2001/XMLSchema" xmlns:lib="http://www.example.org/library"> <element name="library" type="lib:libraryType"></element> <complexType name="libraryType"> <sequence> <element name="books" type="lib:booksType"></element> </sequence> </complexType> <complexType name="booksType"> <sequence> <element name="book" type="lib:bookType" maxOccurs="unbounded" minOccurs="1"></element> </sequence> </complexType> <complexType name="bookType"> <attribute name="title" type="string"></attribute> </complexType> </schema> and a corresponding XML example: <?xml version="1.0" encoding="UTF-8"?> <lib:library xmlns:lib="http://www.example.org/library" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://www.example.org/library src/library.xsd "> <lib:books> <lib:book title="t1"/> <lib:book title="t2"/> <lib:book title="t3"/> </lib:books> </lib:library> Is there a way to guarantee that the order of <lib:book .../> elements is preserved? I want to be sure that any parser reading the XML will return books in the specified oder, that is first the book with title="t1", then the book with title="t2", and finally the book with title="t3". As far as I know XML parsers are not required to preserve order. I wonder whether one can enforce this through XML Schema? One quick solution for me would be adding an index attribute to the <lib:book .../> element, and delegate order preservation to the application reading the XML. Comments? Suggestions?

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