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• MXML composite container initialization error

  - by mkorpela

  I'm getting an odd error from my composite canvas component: An ActionScript error has occurred: Error: null at mx.core::Container/initialize()[C:\autobuild\galaga\frameworks\projects\framework\src\mx\core\Container.as:2560] at -REMOVED THIS FOR STACK OVERFLOW-.view::EditableCanvas/initialize()[.../view/EditableCanvas .... It seems to be related to the fact that my composite component has a child and I'm trying to add one in the place I'm using the component. So how can I do this correctly? Component code looks like this (EditableCanvas.mxml): <?xml version="1.0" encoding="utf-8"?> <mx:Canvas xmlns:mx="http://www.adobe.com/2006/mxml"> <mx:Script> <mx:Image id="editTextImage" source="@Embed('/../assets/icons/small/edit.png')"/> </mx:Canvas> The code that is using the code looks like this: <view:EditableCanvas width="290" height="120" backgroundColor="#FFFFFF" horizontalScrollPolicy="off" borderStyle="solid" cornerRadius="3"> <mx:Text id="textContentBox" width="270" fontFamily="nautics" fontSize="12" text="{_text}"/> </view:EditableCanvas>

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• multipart file-upload post request from java

  - by Martin

  I'm trying to make a program that uploads a image to a webserver that accepts multipart file-uploads. More specificly i want to make a http POST request to http://iqs.me that sends a file in the variable "pic". I've made a lot of tries but i don't know if i've even been close. The hardest part seems to be to get a HttpURLConnection to make a request of the type POST. The response i get looks like it makes a GET. (And i want to do this without any third party libs) UPDATE: non-working code goes here (no errors but doesn't seem to do a POST): HttpURLConnection conn = null; BufferedReader br = null; DataOutputStream dos = null; DataInputStream inStream = null; InputStream is = null; OutputStream os = null; boolean ret = false; String StrMessage = ""; String exsistingFileName = "myScreenShot.png"; String lineEnd = "\r\n"; String twoHyphens = "--"; String boundary = "*****"; int bytesRead, bytesAvailable, bufferSize; byte[] buffer; int maxBufferSize = 1*1024*1024; String responseFromServer = ""; String urlString = "http://iqs.local.com/index.php"; try{ FileInputStream fileInputStream = new FileInputStream( new File(exsistingFileName) ); URL url = new URL(urlString); conn = (HttpURLConnection) url.openConnection(); conn.setDoInput(true); conn.setDoOutput(true); conn.setRequestMethod("POST"); conn.setUseCaches(false); conn.setRequestProperty("Connection", "Keep-Alive"); conn.setRequestProperty("Content-Type", "multipart/form-data;boundary="+boundary); dos = new DataOutputStream( conn.getOutputStream() ); dos.writeBytes(twoHyphens + boundary + lineEnd); dos.writeBytes("Content-Disposition: form-data; name=\"pic\";" + " filename=\"" + exsistingFileName +"\"" + lineEnd); dos.writeBytes(lineEnd); bytesAvailable = fileInputStream.available(); bufferSize = Math.min(bytesAvailable, maxBufferSize); buffer = new byte[bufferSize]; bytesRead = fileInputStream.read(buffer, 0, bufferSize); while (bytesRead > 0){ dos.write(buffer, 0, bufferSize); bytesAvailable = fileInputStream.available(); bufferSize = Math.min(bytesAvailable, maxBufferSize); bytesRead = fileInputStream.read(buffer, 0, bufferSize); } dos.writeBytes(lineEnd); dos.writeBytes(twoHyphens + boundary + twoHyphens + lineEnd); fileInputStream.close(); dos.flush(); dos.close(); }catch (MalformedURLException ex){ System.out.println("Error:"+ex); }catch (IOException ioe){ System.out.println("Error:"+ioe); } try{ inStream = new DataInputStream ( conn.getInputStream() ); String str; while (( str = inStream.readLine()) != null){ System.out.println(str); } inStream.close(); }catch (IOException ioex){ System.out.println("Error: "+ioex); }

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• How do I use XML Builder to create xml in rails?

  - by Angela

  I am trying the following, but the output XML is badly formed: def submit xml = Builder::XmlMarkup.new xml.instruct! xml.mail xml.documents xml.document xml.template xml.source "gallery" xml.name "Postcard: Image fill front" @xml_display = xml end I need it to look more like this: <?xml version="1.0" encoding="UTF-8"?> <mail> <documents> <document> <template> <source>gallery</source> <name>Postcard: Image fill front</name> </template> <sections> <section> <name>Text</name> <text>Hello, World!</text> </section> <section> <name>Image</name> <attachment>...attachment id...</attachment> </section> </sections> </document> </documents> <addressees> <addressee> <name>John Doe</name> <address>123 Main St</address> <city>Anytown</city> <state>AZ</state> <postal-code>10000</postal-code> </addressee> </addressees> </mail>

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• XmlSerialize a custom collection with an Attribute

  - by roomaroo

  I've got a simple class that inherits from Collection and adds a couple of properties. I need to serialize this class to XML, but the XMLSerializer ignores my additional properties. I assume this is because of the special treatment that XMLSerializer gives ICollection and IEnumerable objects. What's the best way around this? Here's some sample code: using System.Collections.ObjectModel; using System.IO; using System.Xml.Serialization; namespace SerialiseCollection { class Program { static void Main(string[] args) { var c = new MyCollection(); c.Add("Hello"); c.Add("Goodbye"); var serializer = new XmlSerializer(typeof(MyCollection)); using (var writer = new StreamWriter("test.xml")) serializer.Serialize(writer, c); } } [XmlRoot("MyCollection")] public class MyCollection : Collection<string> { [XmlAttribute()] public string MyAttribute { get; set; } public MyCollection() { this.MyAttribute = "SerializeThis"; } } } This outputs the following XML (note MyAttribute is missing in the MyCollection element): <?xml version="1.0" encoding="utf-8"?> <MyCollection xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema"> <string>Hello</string> <string>Goodbye</string> </MyCollection> What I want is <MyCollection MyAttribute="SerializeThis" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema"> <string>Hello</string> <string>Goodbye</string> </MyCollection> Any ideas? The simpler the better. Thanks.

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• Dynamic 'twitter style' urls with ASP.NET

  - by Desiny

  I am looking to produce an MVC site which has complete control of the url structure using routing. The specific requirements are: www.mysite.com/ = homepage (home controller) www.mysite.com/common/about = content page (common controller) www.mysite.com/common/contact = content page (common controller) www.mysite.com/john = twitter style user page (dynamic controller) www.mysite.com/sarah = twitter style user page (dynamic controller) www.mysite.com/me = premium style user page (premium controller) www.mysite.com/oldpage.html = 301 redirect to new page www.mysite.com/oldpage.asp?id=3333 = 301 redirect to new page My routes look as follows: routes.IgnoreRoute("{resource}.axd/{*pathInfo}"); routes.MapRoute( "Common", "common/{action}/{id}", new { controller = "common", action = "Index", id = "" } ); routes.MapRoute( "Home", "", new { controller = "Home", action = "Index", id = "" } ); routes.MapRoute( "Dynamic", "{id}", new { controller = "dynamic", action = "Index", id = "" } ); In order to handle the 301 rredirct, I have a database defining the old pages and their new page urls and a stored procdure to handle the lookup. The code (handler) looks like this: public class AspxCatchHandler : IHttpHandler, IRequiresSessionState { #region IHttpHandler Members public bool IsReusable { get { return true; } } public void ProcessRequest(HttpContext context) { if (context.Request.Url.AbsolutePath.Contains("aspx") && !context.Request.Url.AbsolutePath.ToLower().Contains("default.aspx")) { string strurl = context.Request.Url.PathAndQuery.ToString(); string chrAction = ""; string chrDest = ""; try { DataTable dtRedirect = SqlFactory.Execute( ConfigurationManager.ConnectionStrings["emptum"].ConnectionString, "spGetRedirectAction", new SqlParameter[] { new SqlParameter("@chrURL", strurl) }, true); chrAction = dtRedirect.Rows[0]["chrAction"].ToString(); chrDest = dtRedirect.Rows[0]["chrDest"].ToString(); chrDest = context.Request.Url.Host.ToString() + "/" + chrDest; chrDest = "http://" + chrDest; if (string.IsNullOrEmpty(strurl)) context.Response.Redirect("~/"); } catch { chrDest = "/";// context.Request.Url.Host.ToString(); } context.Response.Clear(); context.Response.Status = "301 Moved Permanently"; context.Response.AddHeader("Location", chrDest); context.Response.End(); } else { string originalPath = context.Request.Path; HttpContext.Current.RewritePath("/", false); IHttpHandler httpHandler = new MvcHttpHandler(); httpHandler.ProcessRequest(HttpContext.Current); HttpContext.Current.RewritePath(originalPath, false); } } #endregion } It is very simple to look up a user and in fact the above code does this. My problem is in the dynamic / premium part. I am trying to do the following: 1) in the dynamic controller, lookup the username. 2) if the username is in the user list (database), show the Index ActionResult of the Dynamic controller. 3) if the username is not found, look up the username in the premium list 4) if the username is fund in the premium list (database) then show the Index ActionResult of the Preium controller. 5) If all else fails jump to the 404 page (which will ask the user to sign up) Is this possible? Looking up the user twice is a bad idea for performance? How do I do this without redirecting?

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• Malware - Technical anlaysis

  - by nullptr

  Note: Please do not mod down or close. Im not a stupid PC user asking to fix my pc problem. I am intrigued and am having a deep technical look at whats going on. I have come across a Windows XP machine that is sending unwanted p2p traffic. I have done a 'netstat -b' command and explorer.exe is sending out the traffic. When I kill this process the traffic stops and obviously Windows Explorer dies. Here is the header of the stream from the Wireshark dump (x.x.x.x) is the machines IP. GNUTELLA CONNECT/0.6 Listen-IP: x.x.x.x:8059 Remote-IP: 76.164.224.103 User-Agent: LimeWire/5.3.6 X-Requeries: false X-Ultrapeer: True X-Degree: 32 X-Query-Routing: 0.1 X-Ultrapeer-Query-Routing: 0.1 X-Max-TTL: 3 X-Dynamic-Querying: 0.1 X-Locale-Pref: en GGEP: 0.5 Bye-Packet: 0.1 GNUTELLA/0.6 200 OK Pong-Caching: 0.1 X-Ultrapeer-Needed: false Accept-Encoding: deflate X-Requeries: false X-Locale-Pref: en X-Guess: 0.1 X-Max-TTL: 3 Vendor-Message: 0.2 X-Ultrapeer-Query-Routing: 0.1 X-Query-Routing: 0.1 Listen-IP: 76.164.224.103:15649 X-Ext-Probes: 0.1 Remote-IP: x.x.x.x GGEP: 0.5 X-Dynamic-Querying: 0.1 X-Degree: 32 User-Agent: LimeWire/4.18.7 X-Ultrapeer: True X-Try-Ultrapeers: 121.54.32.36:3279,173.19.233.80:3714,65.182.97.15:5807,115.147.231.81:9751,72.134.30.181:15810,71.59.97.180:24295,74.76.84.250:25497,96.234.62.221:32344,69.44.246.38:42254,98.199.75.23:51230 GNUTELLA/0.6 200 OK So it seems that the malware has hooked into explorer.exe and hidden its self quite well as a Norton Scan doesn't pick anything up. I have looked in Windows firewall and it shouldn't be letting this traffic through. I have had a look into the messages explorer.exe is sending in Spy++ and the only related ones I can see are socket connections etc... My question is what can I do to look into this deeper? What does malware achieve by sending p2p traffic? I know to fix the problem the easiest way is to reinstall Windows but I want to get to the bottom of it first, just out of interest.

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• Issue getting camera emulation to work with Tom G's HttpCamera

  - by user591524

  I am trying to use the android emulator to preview video from webcam. I have used the tom gibara sample code, minus the webbroadcaster (i am instead using VLC streaming via http). So, I have modified the SDK's "CameraPreview" app to use the HttpCamera, but the stream never appears. Debugging through doesn't give me any clues either. I wonder if anything obvious is clear to others? The preview app launches and remains black. Notes: 1) I have updated the original CameraPreview class as described here: http://www.inter-fuser.com/2009/09/live-camera-preview-in-android-emulator.html, but referencing httpCamera instead of socketcamera. 2) I updated Tom's original example to reference "Camera" type instead of deprecated "CameraDevice" type. 3) Below is my CameraPreview.java. 4) THANK YOU package com.example.android.apis.graphics; import android.app.Activity; import android.content.Context; import android.hardware.Camera; import android.os.Bundle; import android.view.SurfaceHolder; import android.view.SurfaceView; import android.view.Window; import java.io.IOException; import android.graphics.Canvas; // ---------------------------------------------------------------------- public class CameraPreview extends Activity { private Preview mPreview; @Override protected void onCreate(Bundle savedInstanceState) { super.onCreate(savedInstanceState); // Hide the window title. requestWindowFeature(Window.FEATURE_NO_TITLE); // Create our Preview view and set it as the content of our activity. mPreview = new Preview(this); setContentView(mPreview); } } // ---------------------------------------------------------------------- class Preview extends SurfaceView implements SurfaceHolder.Callback { SurfaceHolder mHolder; //Camera mCamera; HttpCamera mCamera;//changed Preview(Context context) { super(context); // Install a SurfaceHolder.Callback so we get notified when the // underlying surface is created and destroyed. mHolder = getHolder(); mHolder.addCallback(this); //mHolder.setType(SurfaceHolder.SURFACE_TYPE_PUSH_BUFFERS); mHolder.setType(SurfaceHolder.SURFACE_TYPE_NORMAL);//changed } public void surfaceCreated(SurfaceHolder holder) { // The Surface has been created, acquire the camera and tell it where // to draw. //mCamera = Camera.open(); this.StartCameraPreview(holder); } public void surfaceDestroyed(SurfaceHolder holder) { // Surface will be destroyed when we return, so stop the preview. // Because the CameraDevice object is not a shared resource, it's very // important to release it when the activity is paused. //mCamera.stopPreview();//changed mCamera = null; } public void surfaceChanged(SurfaceHolder holder, int format, int w, int h) { // Now that the size is known, set up the camera parameters and begin // the preview. //Camera.Parameters parameters = mCamera.getParameters(); //parameters.setPreviewSize(w, h); //mCamera.setParameters(parameters); //mCamera.startPreview(); this.StartCameraPreview(holder); } private void StartCameraPreview(SurfaceHolder sh) { mCamera = new HttpCamera("10.213.74.247:443", 640, 480, true);//changed try { //mCamera.setPreviewDisplay(holder); Canvas c = sh.lockCanvas(null); mCamera.capture(c); sh.unlockCanvasAndPost(c); } catch (Exception exception) { //mCamera.release(); mCamera = null; // TODO: add more exception handling logic here } } }

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• xmlEncoder not writing in netBeans

  - by Greg

  Hi, I am trying to use the xmlEncoder to write to xml file in net-beans but it doesnt work. Here is the call to the writing function: dbManipulator.writeStudents(deps); where deps = new Hashtable<String, Department>(); dbManipulator = new DataBaseManipulator(); Department is an class-object I made, and here is writeStudents method which is located in the DataBaseManipulator class: public void writeStudents(Hashtable<Integer, Student> students) { XMLEncoder encoder = null; try { encoder = new XMLEncoder(new FileOutputStream(".\\test\\Students.xml")); } catch(Exception e){} encoder.writeObject(students); encoder.close(); }//end of function writeStudents() Any ideas why it isnt working? I tried changing the hashtable to vector but still the xml file looks like that after the writing: <?xml version="1.0" encoding="UTF-8"?> <java version="1.6.0_18" class="java.beans.XMLDecoder"> <object class="java.util.Hashtable"/> </java> Thanks in advance, Greg

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• parse xml with elementtree, custom sorting

  - by microspace

  I want to parse xml file in utf-8 and sort it by some field. Soring is made by custom alphabet (s1 from sourcecode). History of question is here: sorting of list containing utf-8 charachters. I've found how to sort xml here. Sorting work correctly, the problem is with elementtree, I must admit that it doesn't work on python3 Here is source code: #!/usr/bin/env python # -*- coding: utf-8 -*- #import xml.etree.ElementTree as ET # Python 2.5 import elementtree.ElementTree as ET s1='aáàAâÂbBcCçÇdDeéEfFgGgGhHiIîÎíiiIjJkKlLmMnNóoOöÖpPqQrRsSsStTuUûúÛüÜvVwWxXyYzZ' s2='11111122334455666aabbccddeeeeeeffgghhiijjkklllllmmnnooppqqrrsssssttuuvvwwxxyy' trans = str.maketrans(s1, s2) def unikey(seq): return seq[0].translate(trans) tree = ET.parse("tosort.xml") container = tree.find("entries") data = [] for elem in container: keyd = elem.findtext("k") data.append((keyd, elem)) print (data) data.sort(key=unikey) print (data) container[:] = [item[-1] for item in data] tree.write("sorted.xml", encoding="utf-8") Here are instructions to import elementtree module. When I import module this way :import xml.etree.ElementTree as ET, I get a message: Traceback (most recent call last): File "pcs.py", line 19, in <module> container[:] = [item[-1] for item in data] File "/usr/lib/python3.1/xml/etree/ElementTree.py", line 210, in __setitem__ assert iselement(element) AssertionError When I use this method to import: import elementtree.ElementTree as ET, I get this message: Traceback (most recent call last): File "pcs.py", line 4, in <module> import elementtree.ElementTree as ET File "/usr/local/lib/python3.1/dist-packages/elementtree/ElementTree.py", line 794, in <module> _escape = re.compile(eval(r'u"[&<>\"\u0080-\uffff]+"')) File "<string>", line 1 u"[&<>\"\u0080-\uffff]+" ^ SyntaxError: invalid syntax I use Python 3.1.3 (r313:86834, Nov 28 2010, 11:28:10). In python2.6 elementtree work without a problem. Content of tosort.xml: <xdxf> <entries> <ar><k>zaaaa</k>definition1</ar> <ar><k>saaaa</k>definition2</ar> ... ... </entries> </xdxf>

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• hint and textview with right gravity and a singleline

  - by codeScriber

  I've opened a bug but i was wondering if anyone encountered this issue and knows a workaround. If you define a text view with a hint inside it, give it right gravity (android:gravity="right") then if you define android:singleLine=true or android:maxLines="1" or android:scrollHorizonatally="true" you don't see the hint. removing the right gravity returns the hint to the left side, removing all the tree params i mentioned above puts the hint on the right side. i want my hint on the right, but i need a single horizontal line... here's the sample layout that doesn't show the hint: <?xml version="1.0" encoding="utf-8"?> <LinearLayout xmlns:android="http://schemas.android.com/apk/res/android" android:layout_width="fill_parent" android:layout_height="wrap_content" android:padding="5dp"> <EditText android:layout_width="fill_parent" android:layout_gravity="center_vertical|right" android:layout_height="wrap_content" android:layout_margin="6dp" android:textSize="16sp" android:paddingRight="5dp" android:id="@+id/c" android:gravity="right" android:hint="hello!!!" android:scrollHorizontally="true" android:maxLines="1" android:singleLine="true"/> </LinearLayout> i checked on 1.6 and 2.1 emulators and it reproduces 100%, i'm prettysure it's a bug, i don't see the connection between single line and the hint.... what's more the hint got it's own layout in the TextView (mLayout and mHintLayout both exists, in onDraw if the text length is 0 mHintLayout if mHint is not null is used).

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• WebService Stubs WSDL Axis2

  - by tt0686

  Good afternoon in my timezone. I have a wsdl file with the following snippet of code: <schema targetNamespace="http://util.cgd.pt" xmlns="http://www.w3.org/2001/XMLSchema"> <import namespace="http://schemas.xmlsoap.org/soap/encoding/" /> <complexType name="Attribute"> <sequence> <element name="fieldType" nillable="true" type="xsd:string" /> <element name="dataType" nillable="true" type="xsd:string" /> <element name="name" nillable="true" type="xsd:string" /> <element name="value" nillable="true" type="xsd:string" /> </sequence> </complexType> <complexType name="ArrayOffAttribute"> <complexContent> <restriction base="soapenc:Array"> <attribute ref="soapenc:arrayType" wsdl:arrayType="tns3:Attribute[]" /> </restriction> </complexContent> </complexType> When i use jax-rpc or Axis1 to generate the stubs the type Attribute is generated, but when i use Axis2 the type Attribute is not generated and a new type is created ArrayOffAttribute, this new type extends the axis2.databinding.types.soapencoding.Array and permits to add elements through the array.addObject(object), my question is, i am migrating one Java EE application from webservices using jax-rpc to start using Axis2, and the methods are using the Attribute type to fullfill attributes fields, now in Axis2 and do not have attribute type , what should i use in the ArrayOffAttribute.addObject(?) ? Could be something wrong with Axis2 ? i am stop here :( Thanks in advance Best regards

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• Reading text files line by line, with exact offset/position reporting

  - by Benjamin Podszun

  Hi. My simple requirement: Reading a huge ( a million) line test file (For this example assume it's a CSV of some sorts) and keeping a reference to the beginning of that line for faster lookup in the future (read a line, starting at X). I tried the naive and easy way first, using a StreamWriter and accessing the underlying BaseStream.Position. Unfortunately that doesn't work as I intended: Given a file containing the following Foo Bar Baz Bla Fasel and this very simple code using (var sr = new StreamReader(@"C:\Temp\LineTest.txt")) { string line; long pos = sr.BaseStream.Position; while ((line = sr.ReadLine()) != null) { Console.Write("{0:d3} ", pos); Console.WriteLine(line); pos = sr.BaseStream.Position; } } the output is: 000 Foo 025 Bar 025 Baz 025 Bla 025 Fasel I can imagine that the stream is trying to be helpful/efficient and probably reads in (big) chunks whenever new data is necessary. For me this is bad.. The question, finally: Any way to get the (byte, char) offset while reading a file line by line without using a basic Stream and messing with \r \n \r\n and string encoding etc. manually? Not a big deal, really, I just don't like to build things that might exist already..

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• java.lang.ClassNotFoundException: org.springframework.ui.ModelMap

  - by aelshereay

  I create a simple webapp using tomcat 6, spring 2.5.6 and maven. The problem is when I boot up tomcat, I am getting the following errors: SEVERE: StandardWrapper.Throwable java.lang.NoClassDefFoundError: org/springframework/ui/ModelMap ... Caused by: java.lang.ClassNotFoundException: org.springframework.ui.ModelMap The ModelMap class does exist in spring-2.5.6.jar and spring-context-2.5.6.jar, I also have some other spring jars. All of them are being deployed to tomcat correctly, when I check the application WEB-INF (deployed to tomcat) I found all those jars there! I have only one @Controller that has a @RequestMapping("/home.htm") showForm(ModelMap model) method. My applicationContext is quite simple: <?xml version="1.0" encoding="utf-8"?> <beans xmlns="http://www.springframework.org/schema/beans" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:aop="http://www.springframework.org/schema/aop" xmlns:context="http://www.springframework.org/schema/context" xmlns:dwr="http://www.directwebremoting.org/schema/spring-dwr" xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-2.5.xsd http://www.springframework.org/schema/aop http://www.springframework.org/schema/aop/spring-aop-2.5.xsd http://www.springframework.org/schema/context http://www.springframework.org/schema/context/spring-context-2.5.xsd http://www.directwebremoting.org/schema/spring-dwr http://www.directwebremoting.org/schema/spring-dwr-3.0.xsd" default-autowire="byName"> <context:component-scan base-package="org.myapp"/> <bean class="org.springframework.web.servlet.mvc.annotation.DefaultAnnotationHandlerMapping"/> <bean class="org.springframework.web.servlet.mvc.annotation.AnnotationMethodHandlerAdapter"/> <bean id="viewResolver" class="org.springframework.web.servlet.view.InternalResourceViewResolver"> <property name="viewClass" value="org.springframework.web.servlet.view.JstlView"></property> <property name="prefix" value="/WEB-INF/view/"></property> <property name="suffix" value=".jsp"></property> </bean> </beans>

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• UTF8 charset, diacritical elements, conversion problems - and Zend Framework form escaping

  - by imanc

  Hey, I am writing a webapp in ZF and am having serious issues with UTF8. It's using multi lingual content through Zend Form and it seems that ZF heavily escapes all of these characters and basically just won't show a field if there's diacritical elements 'é' and if I use the HTML entity equivalent e.g. é it gets escaped so that the user will see 'é'. Zend Form allows for having non escaped data, but trying to use this is confusing, and it seems it'd need to be used all over the place. So, I have been told that if the page and the text is in UTF8, no conversion to htmlentities is required. Is this true? And if the last question is true, then how do I convert the source text to UTF8? I am comfortable setting up apache so that it sends a default UTF8 charset heading, and also adding the charset meta tag to the html, but doing this I am still getting messed up encoding. I have also tried opening the translation csv file in TextWrangler on OSX as UTF8, but it has done nothing. Thanks! L

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• Deploy GWT Application to Google App Engine using NetBeans

  - by Yan Cheng CHEOK

  Hello, I try to deploy a GWT application, to Google App Engine using NetBeans. I had successful run GWT sample http://code.google.com/webtoolkit/doc/latest/tutorial/create.html using Personal GlassFish v3 Prelude Domain, by 1) Copy generated source code from StockWatcher to C:\Projects\StockWatcherNetbeans\src\java\com\google\ 2) Modify C:\Projects\StockWatcherNetbeans\nbproject\gwt.properties gwt.module=com.google.gwt.stockwatcher.StockWatcher 3) Select Personal GlassFish v3 Prelude Domain, and run. All works fine! Now, I try to select Google App Engine server, and run. However, I get the error "There is no appengine web project opened!" I check... There is file called C:\Projects\StockWatcherNetbeans\war\WEB-INF\appengine-web.xml with content <?xml version="1.0" encoding="UTF-8"?> <appengine-web-app xmlns="http://appengine.google.com/ns/1.0" xmlns:xsi='http://www.w3.org/2001/XMLSchema-instance' xsi:schemaLocation='http://kenai.com/projects/nbappengine/downloads/download/schema/appengine-web.xsd appengine-web.xsd'> <application>StockWatcherNetbeans</application> <version>1</version> </appengine-web-app> I am using NetBeans 6.7.1 GWT4NB (GWT Plugin for NetBeans) 2.6.12 Google App Engine plugin for NetBeans from http://kenai.com/downloads/nbappengine/1.0_NetBeans671/updates.xml Anything I had missed out? Even when I right click to the project, the Deploy to Google App Engine options is disabled. And yes, please do not ask me why not use Eclipse.

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• JavaScript snippet to read and output XML file on page load?

  - by Banderdash

  Hey guys, hoping I might get some help. Have XML file here of a list of books each with unique id and numeral value for whether they are checked out or not. I need a JavaScript snippet that requests the XML file after the page loads and displays the content of the XML file. XML file looks like this: <?xml version="1.0" encoding="UTF-8" ?> <response> <library name="My Library"> <book id="1" checked-out="1"> <authors> <author>John Resig</author> </authors> <title>Pro JavaScript Techniques (Pro)</title> <isbn-10>1590597273</isbn-10> </book> <book id="2" checked-out="0"> <authors> <author>Erich Gamma</author> <author>Richard Helm</author> <author>Ralph Johnson</author> <author>John M. Vlissides</author> </authors> <title>Design Patterns: Elements of Reusable Object-Oriented Software</title> <isbn-10>0201633612</isbn-10> </book> ... </library> </response> Would LOVE any and all help!

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• innerHTML doesn't work correctly with xhtml in Chrome

  - by Desperadeus

  Hi! I've got a trouble with Chrome5.0.375.70, but FF 3.6.3 and Opera 10.53 are OK. Below is the line of code: document.getElementById('content').innerHTML = data.documentElement.innerHTML; The data object from the code is a document (typeof(data) == 'object') and I've got it by ajax request to chapter01.xhtml: <?xml version="1.0" encoding="utf-8"?> <!DOCTYPE html [ <!ENTITY D "&#x2014;"> <!ENTITY o "&#x2018;"> <!ENTITY c "&#x2019;"> <!ENTITY O "&#x201C;"> <!ENTITY C "&#x201D;"> ]> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <title>Alice's Adventures in Wonderland by Lewis Carroll. Chapter I: Down the Rabbit-Hole</title> <link rel="stylesheet" type="text/css" href="style.css"/> <link rel="stylesheet" type="application/vnd.adobe-page-template+xml" href="page-template.xpgt"/> </head> <body> <div class="title_box"> <h2 class="chapnum">Chapter I</h2> <h2 class="chaptitle">Down the Rabbit-Hole</h2> <hr/> </div> The Chrome cuts all before body and as a result link to css in header is missed; user can't see formatted text and images. How can I fix it or bypass?

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• How come my form input sometimes moves when I refresh the page?

  - by samoz

  On a page that I'm designing I have a form with one input of type text. Normally, this form and input render properly in my browser, Chrome, but occasionally, it renders about 20 pixels to the left of where it is supposed to be. When I refresh the page, it goes back to the original, correct place. I have only tested in Chrome so far, so this isn't a cross-browser issue (it happens in the same browser). Is there anything wrong with my code below? Here's my HTML code: <!DOCTYPE htmls> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <link rel="stylesheet" href="style.css" type="text/css" /> <title>Test Site</title> </head> <body > <div id="supercontainer" class="style1"> <img class="floater" src="top.jpg" alt="Top" /> <img class="floater" src="left.jpg" alt="Left" /> <div id="content"> <p id="theText"> Welcome. Please type a username. </p> <form id="prompt"> <div><input type="text" name="promptLine" autocomplete="off" id="promptLine" onkeypress="return submitenter(event);" value="% " /></div> </form> </div> <img class="floater" src="right.jpg" alt="Right" /> <img class="floater" src="bottom.jpg" alt="Bottom" /> </div> Here's my CSS code: #supercontainer { margin: 0 auto; width: 900px; display: block; } img.floater { display: inline; float: left; } #content { background-color:black; display: inline; float: left; padding-left:5px; padding-right:5px; min-height:458px; max-height:458px; min-width: 803px; max-width: 803px; color: lime; } #theText { text-align:left; margin-bottom:0; margin-top:0; line-height: 0.3; font-family:"Courier New", Courier, monospace; } #prompt { position: fixed; top: 470px; } #promptLine { width: 100%; background-color: black; color: lime; border: none; outline:none; }

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• How to parse this xml document?

  - by dfjhdfjhdf

  Get such a XML document with the help of ajax (var data = request.responseXML;), how do I parse the contacts?: <?xml version="1.0" encoding="UTF-8"?> <Alladresses xmlns="http://somedomain.org/doc/2007-08-02/"> <Owner> <ID>gut74hfowesdfj49fjsifhryh8e8rta3uyhw4</ID> <Name>Mr.Bin</Name> </Owner> <Contacts> <Person> <Name>Greg</Name> <Phone>3254566756</Phone> </Person> <Person> <Name>Smith</Name> <Phone>342446446</Phone> </Person> <Person> <Name>Yuliya</Name> <Phone>675445566867</Phone> </Person> </Contacts> </Alladresses>

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• XSLT: need to replace document('')

  - by Daziplqa

  I've the following xslt file: <?xml version="1.0" encoding="UTF-8"?> <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <!-- USDomesticCountryList - USE UPPERCASE LETTERS ONLY --> <xsl:variable name="USDomesticCountryList"> <entry name="US"/> <entry name="UK"/> <entry name="EG"/> </xsl:variable> <!--// USDomesticCountryList --> <xsl:template name="IsUSDomesticCountry"> <xsl:param name="countryParam"/> <xsl:variable name="country" select="normalize-space($countryParam)"/> <xsl:value-of select="normalize-space(document('')//xsl:variable[@name='USDomesticCountryList']/entry[@name=$country]/@name)"/> </xsl:template> </xsl:stylesheet> I need to replace the "document('')" xpath function, what should I use instead? I've tried to remove it completely but the xsl document doesn't work for me! I need to to so because the problem is : I am using some XSLT document that uses the above file, say document a. So I have document a that includes the above file (document b). I am using doc a from java code, I am do Caching for doc a as a javax.xml.transform.Templates object to prevent multiple reads to the xsl file on every transformation request. I found that, the doc b is re-calling itself from the harddisk, I believe this is because of the document('') function above, so I wanna replace/remove it. Thanks.

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• java.lang.ExceptionInInitializerError Exception when creating Application Context in Spring

  - by cyotee

  I am practicing with Spring, and am getting a java.lang.ExceptionInInitializerError exception when I try to instantiate the context. The Exception appears below, with my code following it. I have simplified my experiment from before. The Exception Oct 17, 2012 5:54:22 PM org.springframework.context.support.AbstractApplicationContext prepareRefresh INFO: Refreshing org.springframework.context.support.ClassPathXmlApplicationContext@570c16b7: startup date [Wed Oct 17 17:54:22 CDT 2012]; root of context hierarchy Exception in thread "main" java.lang.ExceptionInInitializerError at org.springframework.context.support.AbstractRefreshableApplicationContext.createBeanFactory(AbstractRefreshableApplicationContext.java:195) at org.springframework.context.support.AbstractRefreshableApplicationContext.refreshBeanFactory(AbstractRefreshableApplicationContext.java:128) at org.springframework.context.support.AbstractApplicationContext.obtainFreshBeanFactory(AbstractApplicationContext.java:535) at org.springframework.context.support.AbstractApplicationContext.refresh(AbstractApplicationContext.java:449) at org.springframework.context.support.ClassPathXmlApplicationContext.<init>(ClassPathXmlApplicationContext.java:139) at org.springframework.context.support.ClassPathXmlApplicationContext.<init>(ClassPathXmlApplicationContext.java:83) at helloworld.HelloWorldTest.main(HelloWorldTest.java:13) Caused by: java.lang.NullPointerException at org.springframework.beans.factory.support.DefaultListableBeanFactory.<clinit>(DefaultListableBeanFactory.java:105) ... 7 more My configuration XML <?xml version="1.0" encoding="UTF-8"?> <beans xmlns="http://www.springframework.org/schema/beans" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:c="http://www.springframework.org/schema/c" xmlns:p="http://www.springframework.org/schema/p" xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-3.1.xsd"> <bean id="messageContainer" class="helloworld.MessageContainer"> <property name="message" value="Hello World"> </property> </bean> <bean id="messageOutputService" class="helloworld.MessageOutputService"> </bean> My test class. package helloworld; import org.springframework.context.ApplicationContext; import org.springframework.context.support.ClassPathXmlApplicationContext; public class HelloWorldTest { /** * @param args */ public static void main(String[] args) { ApplicationContext context = new ClassPathXmlApplicationContext("HelloWorldTest-context.xml"); MessageContainer message = context.getBean(MessageContainer.class); MessageOutputService service = context.getBean(MessageOutputService.class); service.outputMessageToConsole(message); } }

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• accessing a value of a nested hash

  - by st

  Hello! I am new to perl and I have a problem that's very simple but I cannot find the answer when consulting my perl book. When printing the result of Dumper($request); I get the following result: $VAR1 = bless( { '_protocol' => 'HTTP/1.1', '_content' => '', '_uri' => bless( do{\(my $o = 'http://myawesomeserver.org:8081/counter/')}, 'URI::http' ), '_headers' => bless( { 'user-agent' => 'Mozilla/5.0 (X11; U; Linux i686; en; rv:1.9.0.4) Gecko/20080528 Epiphany/2.22 Firefox/3.0', 'connection' => 'keep-alive', 'cache-control' => 'max-age=0', 'keep-alive' => '300', 'accept' => 'text/html,application/xhtml+xml,application/xml;q=0.9,*/*;q=0.8', 'accept-language' => 'en-us,en;q=0.5', 'accept-encoding' => 'gzip,deflate', 'host' => 'localhost:8081', 'accept-charset' => 'ISO-8859-1,utf-8;q=0.7,*;q=0.7' }, 'HTTP::Headers' ), '_method' => 'GET', '_handle' => bless( \*Symbol::GEN0, 'FileHandle' ) }, 'HTTP::Server::Simple::Dispatched::Request' ); How can I access the values of '_method' ('GET') or of 'host' ('localhost:8081'). I know that's an easy question, but perl is somewhat cryptic at the beginning. Thank you, St.

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• Data of the web services in a RSS

  - by vymz

  I only have the .wsdl and I want to put the data that the web services return in the RSS. And in a web page of SAP, I can only upload RSS, so I need put the information of the web services in the RSS For example, I put the information manually(name and total value) in the fields <title> and <descripcion>, these data are extracted from the web services. But sometimes I don´t know how much information brings the web service. Also, I know that RSS is not to store information such as web services. <?xml version="1.0" encoding="UTF-8" ?> <rss version="2.0"> <channel> <title>Test RSS</title> <link>http://solutions.com</link> <description>RSS</description> <item> <title>Luiz</title> <link>http://www.solutions.com/prueba1</link> <description>10</description> </item> <item> <title>Clodoaldo</title> <link>http://www.solutions.com/prueba2</link> <description>5</description> </item>

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• how to dispaly image in grid view reading imageUrl from xml using sax parser in android

  - by Pramod kuamr

  thanks for answer but i am able to read xml file from url but i need if in xml imageUrl is there so show in grid view ..this is my xml file and read URL <?xml version="1.0" encoding="UTF-8"?> <channels> <channel> <name>ndtv</name> <logo>http://a3.twimg.com/profile_images/670625317/aam-logo--twitter.png</logo> <description>this is a news Channel</description> <rssfeed>ndtv.com</rssfeed> </channel> <channel> <name>star news</name> <logo>http://a3.twimg.com/profile_images/740897825/AndroidCast-350_normal.png</logo> <description>this is a newsChannel</description> <rssfeed>starnews.com</rssfeed> </channel> </channels>

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• Is the order of params important in NHibernate?

  - by Blake Blackwell

  If I have an int parameter followed by a string parameter in a sproc I get the following error: Input string was not in the correct format However, if I switch those parameters in the sproc than I get the result set I expect. Are params sorted by data type, or do I have to do anything special in my config file? I've included my code for reference: Config File <?xml version="1.0" encoding="utf-8" ?> <hibernate-mapping xmlns="urn:nhibernate-mapping-2.2" assembly="NHibernateDemo" namespace="NHibernateDemo.Domain"> <class name="Blake_Test" table="Blake_Test"> <id name="TestId" column="TESTID"></id> <property name="TestName" column="TESTNAME" /> <loader query-ref="GetBlakeTest"/> </class> <sql-query name="GetBlakeTest" callable="true"> <return class="Blake_Test" /> call procedure AREA51.NHIBERNATE_TEST.GetBlakeTest(:int_TestId, :vch_TestName) </sql-query> </hibernate-mapping> Sproc Code: PROCEDURE GetBlakeTest ( ret_cursor OUT SYS_REFCURSOR, int_testid integer, vch_testname varchar2 ) AS BEGIN OPEN ret_cursor FOR SELECT TestId, TestName FROM blake_test WHERE testid = int_testid ORDER BY TestName DESC; END GetBlakeTest; END NHIBERNATE_TEST; Executing Code: IQuery query1 = session.GetNamedQuery( "GetBlakeTest" ); query1.SetParameter( "int_TestId", 1 ); query1.SetParameter( "vch_TestName", "TEST" ); IList<Blake_Test> mystuff = query1.List<Blake_Test>();

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