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  • Dataset -> XML Document - Load DataSet into an XML Document - C#.Net

    - by NLV
    Hello I'm trying to read a dataset as xml and load it into an XML Document. XmlDocument contractHistoryXMLSchemaDoc = new XmlDocument(); using (MemoryStream ms = new MemoryStream()) { //XmlWriterSettings xmlWSettings = new XmlWriterSettings(); //xmlWSettings.ConformanceLevel = ConformanceLevel.Auto; using (XmlWriter xmlW = XmlWriter.Create(ms)) { xmlW.WriteStartDocument(); dsContract.WriteXmlSchema(xmlW); xmlW.WriteEndDocument(); xmlW.Close(); using (XmlReader xmlR = XmlReader.Create(ms)) { contractHistoryXMLSchemaDoc.Load(xmlR); } } } But I'm getting the error - "Root Element Missing". Any ideas? Update When i do xmlR.ReadInnerXML() it is empty. Does anyone know why? NLV

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  • Question regarding Manifest in Java jar file

    - by Tara Singh
    Hi All, Is is mandatory to have classpath inside a Manifest file inside the java jar file? can we do work without having the classpath inside it? The reason why I am asking this is because I have a jar file for a server application. When I tried to connect many connections with Server, Server went down and the error was "Too many open files", when searched about it, I found one Sun Bug http://bugs.sun.com/bugdatabase/view_bug.do?bug_id=6446657 . Then I checked and i was having a classpath entry in the Jar file. So this question arises. Thanks, Tara Singh

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  • How to include text files with Executable Jar

    - by Jake
    Hi guys rookie Java question. I have a Java project and I want to include a text file with the executable jar. Right now the text file is in the default package. InputFlatFile currentFile = new InputFlatFile("src/theFile.txt"); I grab the file with that line as you can see using src. However this doesn't work with the executable jar. Can someone please let me know how to keep this file with the executable jar so someone using the program can just click a single icon and run the program. Thanks!

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  • WCF Method is returning xml fragment but no xml UTF-8 header

    - by horls
    My method does not return the header, just the root element xml. internal Message CreateReturnMessage(string output, string contentType) { // create dictionaryReader for the Message byte[] resultBytes = Encoding.UTF8.GetBytes(output); XmlDictionaryReader xdr = XmlDictionaryReader.CreateTextReader(resultBytes, 0, resultBytes.Length, Encoding.UTF8, XmlDictionaryReaderQuotas.Max, null); if (WebOperationContext.Current != null) WebOperationContext.Current.OutgoingResponse.ContentType = contentType; // create Message return Message.CreateMessage(MessageVersion.None, "", xdr); } However, the output I get is: <Test> <Message>Hello World!</Message> </Test> I would like the output to render as: <?xml version="1.0" encoding="utf-8" standalone="yes"?> <Test> <Message>Hello World!</Message> </Test>

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  • how to merge xml string to main xml document object

    - by CliffC
    how can i merge the following xml string <employee> <name>cliff</name> </employee> to my existing xml document object XmlDocument xmlDoc = new XmlDocument(); XmlElement xmlCompany = xmlDoc.CreateElement("Company"); the final output should look like <Company> <employee> <name>cliff</name> </employee> </Company> thanks

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  • Convert XML attributes to a Dictionary in Linq to XML

    - by NateD
    I've got a program that needs to convert two attributes of a particular tag to the key and value of an Dictionary<int,string>. The XML looks like this: (fragment) <startingPoint coordinates="1,1" player="1" /> and so far my LINQ looks something like this: XNamespace ns = "http://the_namespace"; var startingpoints = from sp in xml.Elements(ns+"startingPoint") from el in sp.Attributes() select el.Value; Which gets me a nice IEnumerable full of things like "1,1" and "1", but there should be a way to adapt something like this answer to do attributes instead of elements. Little help please? Thank you!

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  • VB.NET add an element to the XML document with LINQ to XML

    - by Bayonian
    Hi, I'm adding an element to existing XML doc with the following code: Dim theXMLSource As String = Server.MapPath("~/Demo/") & "LabDemo.xml" Dim nodeElement As XElement Dim attrAndValue As XElement = _ <LabService> <ServiceType> <%= txtServiceType.Text.Trim %> </ServiceType> <Level> <%= txtLevel.Text.Trim %> </Level> </LabService> nodeElement.Add(New XElement(attrAndValue)) nodeElement.Save(theXMLSource) It makes error like this: System.NullReferenceException: Object reference not set to an instance of an object. Object reference not set to an instance of an object. Error line: nodeElement.Add(New XElement(attrAndValue)) I debugged it but I couldn't get the error yet. Can you show what the problem is? Thank you

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  • Need Simple way to access XML in VB.Net - Pain with Linq-to-Xml

    - by aiart
    Dim myXDoc As XDocument = _ I want to access this in a simple way in VB.Net - Like: Dim Integer SizeXStr = CInt(MyZDoc.Cameras(1).Camera_Desc.@SizeX) ' where (1) is an index Why isn't this implemented in VB.Net? Better yet, type the values with a Schema and eliminate the conversion. Is this so hard? How do I access, in a simple way, data in XML - this would be VERY VERY useful! I have been using Query to try to get the values - when I use MsgBox() to display results, they display, but my main Windows Form is Trashed - changed colors, etc. The system has Bugs. Instead, I have to create an elaborate structure of arrays of objects and read the XML line-by-line and do the same for saving - this is the dark ages. Art

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  • Reading xml within xml as String in flex/AS3

    - by duder
    I'm getting XML input that looks like this <?xml version="1.0" encoding="utf-8"?> <data1>this is data 1</data1> <data2>this is data 2</data2> <data3> <3a>this is data 3a</3a> <3b>this is data 3b</3b> <3c> <TextFlow xmlns="http://ns.adobe.com/textLayout/2008"> <p direction="ltr" > <span>some text</span> <span>some additional text</span> </p> <p direction="ltr"> <span>some text</span> <span>some additional text</span> </p> </TextFlow> </3c> </data3> I can read <data1> with event.result.data1 which outputs a string this is data1 But when I do the same thing to event.result.data3.3c, it prints object [object] so I guess it's trying to dig deeper into the tree. But I need the actual string text (not xml tree) starting from and including <TextFlow></TextFlow> to be stored and printed as a string. Any idea what's the syntax for this? The string I'm looking for would look like this: <TextFlow xmlns="http://ns.adobe.com/textLayout/2008"> <p direction="ltr" > <span>some text</span> <span>some additional text</span> </p> <p direction="ltr"> <span>some text</span> <span>some additional text</span> </p> </TextFlow>

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  • how to read value of an xml node (single) using linq to xml

    - by Wondering
    Hi All, I have a xml structure similar to below one: <test> <test1>test1 value</test1> </test> Now I am reading the value of node using below LINQ to xml code. var test = from t in doc.Descendants("test") select t.Element("test1").Value; Console.WriteLine("print single node value"); Console.WriteLine(test); above code works fine, but here I have one single node, but to retrive value I am using foreach loop, which I dont think is good..any better way of doing the same thing without a foreach loop Thanks.

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  • Java - copy Jar Folder

    - by Ripei
    Hey Java - Developers Actually I am confronted with a Problem. I've got a ".apk-File" in one Package of my Application. apk is a kind of a jar File (apk = Android Package). I now want to copy this jar-file out of my Programm onto any other Location at the PC. Normally I would do this by using: FileInputStream is = new FileInputStream(this.getClass().getResource("/resources/myApp.apk").getFile()); And then write it on the disk with using a FileOutputStream. ... but since an .apk is a kind of a .jar it doesn't work. It just copies the .apk file. but without the containing other files. any help would be appreciated

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  • Adding a JAR to an Eclipse Java library

    - by Paul Reiners
    How do you add a JAR file to an already existing Java library in Eclipse? Note that this is not a user library. That is, if you look at the Java Build Path for a Java project and click on the Libraries tab, you will see the list of libraries used by the project. If you expand a given library, you will see a list of JAR files included in that library. I want to add an additional JAR file to one of these libraries. I am using Version 3.4.0 of Eclipse.

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  • adding org.springframework.test package to spring-2.5.6-SEC01.jar in netbeans

    - by John
    This is probably very simple, however I can't get this to work. I want to use AbstractDependencyInjectionSpringContextTest class which is in org.springframework.test package. This package is not included in Netbeans' spring library, so I want to add it. So what I have tried so far is: copy and paste "test" directory (downloaded from spring) into the Netbeans' spring-2.5.6-SEC01.jar file (copy it to org.springframework directory in that jar so I can use org.springframework.test to import it). If I go to project/libraries in Netbeans it is there, but when I try to import org.springframework.test.*; the autocomplition doesn't give me the option to choose test directory from org.soringframework package. create a new library which points to "test" directory and add it to the project- as there is no any jar file in "test" I'm not sure what path should I use to import it. I'm pretty sure this is something very simple but I'm still a novice and can't figure this out.

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  • Java - Loading dlls by a relative path and hide them inside a jar

    - by supertreta
    Hi guys, I am developing a Java application that should be release as a jar. This program depends on c++ external libraries called by JNI. To load them, I use the method System.load with an absolute path and this works fine. However, I really want to "hide" them inside the jar, so I have created a package to collect them. This forces me to load an relative path - the package path. By this approach, I let the user run the jar in any directory, without being worried about linking the dll's or bored with a previous installation process. This throws the expected exception: Exception in thread "main" java.lang.UnsatisfiedLinkError: Expecting an absolute path of the library How can I get this working? Thanks for your help!

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  • what happen if I delete the xx.jar file after I started to execute the xx.jar

    - by ogzylz
    I have a server program running a java binary code (xx.jar file) While it is running I erranously delete the xx.jar file. The program continues to run. But I am not sure if the results will be correct, and I am not sure if the program will fail? When I delete the xx.jar file, the program was in a method for a long time and still it is in that method call. When it calls another method call will my program fail? I am asking this question because If deleting the file has no harm I will be gaining about 3-4h on a server machine

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  • Unable to use JAR in Eclipse

    - by Myn
    Hi guys, I have just created my first JAR in Eclipse, just a simple program with a single class Database.class. It is not in a package. public class Database { public Database() { int dbInit = 1; } } I have added it as an external JAR to the build path libraries for another project in Eclipse, but for some reason I cannot get Database db = new Database(), the default constructor, to work - it's as if the contents of the JAR are not being recognised. Could anyone please offer any advice on this? Thanks very much, M

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  • NoInitialContextException: in ejb 3

    - by JavaDeveloper
    My client class code Hashtable hs = new Hashtable(); hs.put(Context.INITIAL_CONTEXT_FACTORY, "org.jnp.interfaces.NamingContextFactory"); hs.put(Context.PROVIDER_URL, "jnp://10.20.52.41:1200"); hs.put(Context.URL_PKG_PREFIXES, "org.jboss.naming:org.jnp.interfaces"); Context context = new InitialContext(hs); System.out.println("-----"); try { Example helloWorld = (Example) context .lookup("Example"); helloWorld.display("HelloWorld"); } catch (NamingException e) { System.out.println("naming exception occoured"); e.printStackTrace(); } My session bean code public class Example implements javax.ejb.SessionBean { //method here.. } getting following exception. how to resolve? Exception in thread "main" javax.naming.NoInitialContextException: Cannot instantiate class: org.jnp.interfaces.NamingContextFactory [Root exception is java.lang.ClassNotFoundException: org.jnp.interfaces.NamingContextFactory] at javax.naming.spi.NamingManager.getInitialContext(NamingManager.java:657) at javax.naming.InitialContext.getDefaultInitCtx(InitialContext.java:288) at javax.naming.InitialContext.init(InitialContext.java:223) at javax.naming.InitialContext.(InitialContext.java:197) at comtest.client.Example.main(Test.java:22) Caused by: java.lang.ClassNotFoundException: org.jnp.interfaces.NamingContextFactory at java.net.URLClassLoader$1.run(URLClassLoader.java:202) at java.security.AccessController.doPrivileged(Native Method) at java.net.URLClassLoader.findClass(URLClassLoader.java:190) at java.lang.ClassLoader.loadClass(ClassLoader.java:306) at sun.misc.Launcher$AppClassLoader.loadClass(Launcher.java:301) at java.lang.ClassLoader.loadClass(ClassLoader.java:247) at java.lang.Class.forName0(Native Method) at java.lang.Class.forName(Class.java:249) at com.sun.naming.internal.VersionHelper12.loadClass(VersionHelper12.java:46) at javax.naming.spi.NamingManager.getInitialContext(NamingManager.java:654) ... 4 more

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  • Stairway to XML: Level 3 - Working with Typed XML

    You can enforce the validation of an XML data type, variable or column by associating it with an XML Schema Collection. SQL Server validates a typed XML value against the rules defined in the schema collection so that INSERT or UPDATE operations will succeed only if the value being inserted or updated is valid as per the rules defined in the Schema Collection. NEW! Deployment Manager Early Access ReleaseDeploy SQL Server changes and .NET applications fast, frequently, and without fuss, using Deployment Manager, the new tool from Red Gate. Try the Early Access Release to get a 20% discount on Version 1. Download the Early Access Release.

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  • Spring Framework 3.0.5 MVC Issue

    - by user578923
    I know that this may be absolutely dumb but for the life of me I cannot figure out why I'm getting these errors in my Spring Project, it is basically from the MVC tutorial with a few modifications. This is the error I get when running my tomcat server. `Caused by: java.lang.NoClassDefFoundError: org/springframework/web/servlet/mvc/SimpleFormController at java.lang.ClassLoader.defineClass1(Native Method) at java.lang.ClassLoader.defineClass(ClassLoader.java:634) at java.security.SecureClassLoader.defineClass(SecureClassLoader.java:142) at java.net.URLClassLoader.defineClass(URLClassLoader.java:277) at java.net.URLClassLoader.access$000(URLClassLoader.java:73) at java.net.URLClassLoader$1.run(URLClassLoader.java:212) at java.security.AccessController.doPrivileged(Native Method) at java.net.URLClassLoader.findClass(URLClassLoader.java:205) at java.lang.ClassLoader.loadClass(ClassLoader.java:321) at sun.misc.Launcher$AppClassLoader.loadClass(Launcher.java:294) at java.lang.ClassLoader.loadClass(ClassLoader.java:266) at org.apache.catalina.loader.WebappClassLoader.loadClass(WebappClassLoader.java:1581) at org.apache.catalina.loader.WebappClassLoader.loadClass(WebappClassLoader.java:1511) at org.springframework.util.ClassUtils.forName(ClassUtils.java:257) at org.springframework.beans.factory.support.AbstractBeanDefinition.resolveBeanClass(AbstractBeanDefinition.java:408) at org.springframework.beans.factory.support.AbstractBeanFactory.doResolveBeanClass(AbstractBeanFactory.java:1271) at org.springframework.beans.factory.support.AbstractBeanFactory.resolveBeanClass(AbstractBeanFactory.java:1242) ... 54 more Caused by: java.lang.ClassNotFoundException: org.springframework.web.servlet.mvc.SimpleFormController at java.net.URLClassLoader$1.run(URLClassLoader.java:217) at java.security.AccessController.doPrivileged(Native Method) at java.net.URLClassLoader.findClass(URLClassLoader.java:205) at java.lang.ClassLoader.loadClass(ClassLoader.java:321) at sun.misc.Launcher$AppClassLoader.loadClass(Launcher.java:294) at java.lang.ClassLoader.loadClass(ClassLoader.java:266) ... 71 more` I just cannot figure out the issue with my classpath...I would appreciate any help. Here are all the jars in my classpath. I know that the class is inside the web-servlet jar but it's not seeing it. Is there a conflict? aopalliance.jar aspectjweaver.jar commons-codec.jar commons-dbcp.jar commons-logging.jar commons-pool.jar jstl.jar org.springframework.aop-3.0.5.RELEASE.jar org.springframework.asm-3.0.5.RELEASE.jar org.springframework.aspects-3.0.5.RELEASE.jar org.springframework.beans-3.0.5.RELEASE.jar org.springframework.context.support-3.0.5.RELEASE.jar org.springframework.context-3.0.5.RELEASE.jar org.springframework.core-3.0.5.RELEASE.jar org.springframework.expression-3.0.5.RELEASE.jar org.springframework.instrument.tomcat-3.0.5.RELEASE.jar org.springframework.instrument-3.0.5.RELEASE.jar org.springframework.jdbc-3.0.5.RELEASE.jar org.springframework.jms-3.0.5.RELEASE.jar org.springframework.orm-3.0.5.RELEASE.jar org.springframework.oxm-3.0.5.RELEASE.jar org.springframework.test-3.0.5.RELEASE.jar org.springframework.transaction-3.0.5.RELEASE.jar org.springframework.web.portlet-3.0.5.RELEASE.jar org.springframework.web.servlet-3.0.5.RELEASE.jar org.springframework.web.struts-3.0.5.RELEASE.jar org.springframework.web-3.0.5.RELEASE.jar postgresql-9.0-801.jdbc3.jar servlet-api.jar spring-security-config-3.0.5.RELEASE.jar spring-security-core-3.0.5.RELEASE.jar spring-security-web-3.0.5.RELEASE.jar standard.jar

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  • Identify valid server in XML-RPC request using PHP

    - by Ian
    I'm working on a client-server system, where the client makes XMLRPC requests to the server. The client part of the system is handed to a third-party, meaning that he could eventually modify the code or re-route the xmlrpc requests. Now, hoping the third-party won't modify the code, I need a way to make sure that the server the client script is contacting is actually MY server (cause, the person could somehow reroute the requests to his own server where he could make up some xml responses, not what I want). Is there a way to identify a server using PHP? Some sort of SSL connection? Hope you guys understand me. Cheers.

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  • xml schema and using a choice as the document root

    - by mikey
    I have a bit of a newbie xml schema question. I believe the answer is that what I need to do is not possible with schema, but I'd like to be sure. The problem is that I have a webservice that returns a response with one type of root element on success (say <Response), and on a complete failure, returns a document with a different root element (say, <Exception). So, basically, two completely different documents: <Response......</Response OR <Exception....</Exception Is it possible to describe these two different documents with one schema document? It's like I want a choice as the first element under the schema element -- but that isn't valid syntax. I've tried a couple of variants that parse as valid xsd, but don't validate the documents. Any suggestions? Or is this simply not possible? Thanks very much in advance -- m

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  • XSD Restrictions based on target xml elements

    - by ??????
    Is it possible in xsd to create restriction based on elements of some type in target (processed) document? For example I have XML like this: <Pets> <Pet name="Murka" /> <Pet name="Browko" /> <Pet name="Tuzik" /> </Pets> <Children> <Child name="Petruk" favoritePet="Browko" /> </Children> so what I want to restrict the attribute "favoritePet" of element "Child" based on existing "Pet" elements. How can I do this?

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  • XMl Data Structure

    - by metdos
    Which one of two XML structures below do you prefer? Why? Any other suggestion is welcome :) <Parameters> <Parameter id=username>metdos</Parameter> <Parameter id=password>123</Parameter> </Parameters> or <Parameters> <username>metdos</username> <password>123</password> </Parameters>

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