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  • running an android app on the device instead of on the emulator.

    - by gil
    hi, I've installed the usb driver, i'm running win7. I can see that the driver is installed in the window-android SDK and AVD manager-installed packages but when i'm writing "adb devices" in the cmd it doesnt show like the phone is connected (it is - it has the orange led on..) I'm using the HTC G1. I also did the "Turn on "USB Debugging" on your device" step... anyone got an idea??

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  • How do I determine whether it's a mobile device with PHP?

    - by Thomaschaaf
    I am writing a website with PHP. Since it will need to be accessed by anyone on the network to access the internet I have to create a mobile version. How do I best check if it's a mobile device? I don't want to have a switch statement with 50 devices at the end since I don't only want to support the iPhone. Is there a PHP class I could use?

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  • Starting ActiveMQ message listener in Tomcat?

    - by Nick Swarr
    I've created an ActiveMQ MessageListener and configured it using Spring. I'm hosting the listener in Tomcat. When I start up the web application (that features only the listener), should the listener automatically start? Or do I need additional configuration? Here's what I have. First, updated the web.xml to allow spring to configure itself on startup, <?xml version="1.0" encoding="UTF-8"?> <web-app version="2.5" xmlns="http://java.sun.com/xml/ns/javaee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd"> <listener> <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class> </listener> <context-param> <param-name>contextConfigLocation</param-name> <param-value>/WEB-INF/classes/spring/applicationContext.xml</param-value> </context-param> </web-app> Then I created the applicationContext.xml to configure the ActiveMQ listener, <?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:context="http://www.springframework.org/schema/context" xmlns:amq="http://activemq.apache.org/schema/core" xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-3.0.xsd http://www.springframework.org/schema/context http://www.springframework.org/schema/context/spring-context-3.0.xsd"> <context:annotation-config /> <context:component-scan base-package="com.somepackage"/> <context:property-placeholder location="classpath:env.properties"/> <bean id="jmsFactory" class="org.apache.activemq.ActiveMQConnectionFactory"> <property name="brokerURL" value="tcp://localhost:61616" /> </bean> <bean id="documentListener" class="com.somepackage.SomeListener" /> <bean id="container" class="org.springframework.jms.listener.DefaultMessageListenerContainer"> <property name="connectionFactory" ref="cachingConnectionFactory"/> <property name="messageListener" ref="documentListener"/> <property name="destinationName" value="STOCKS.MSFT" /> </bean> <bean id="cachingConnectionFactory" class="org.springframework.jms.connection.CachingConnectionFactory"> <property name="targetConnectionFactory" ref="jmsFactory" /> <property name="sessionCacheSize" value="1" /> </bean> </beans> And that's it. Based on what I've seen around the web, I can't tell if that's all I need? Maybe I need some other configuration in Tomcat to kick off the listener?

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  • How to load a springframework ApplicationContext from Jython

    - by staticman
    I have a class that loads a springframework application context like so: package com.offlinesupport; import org.springframework.context.ApplicationContext; import org.springframework.context.support.ClassPathXmlApplicationContext; public class OfflineScriptSupport { private static ApplicationContext appCtx; public static final void initialize() { appCtx = new ClassPathXmlApplicationContext( new String[] { "mycontext.spring.xml" } ); } public static final ApplicationContext getApplicationContext() { return appCtx; } public static final void main( String[] args ) { System.out.println( "Starting..." ); initialize(); System.out.println( "loaded" ); } } The class OfflineScriptSupport, and the file mycontext.spring.xml are each deployed into separate jars (along with other classes and resources in their respective modules). Lets say the jar files are OfflineScriptSupport.jar and *MyContext.jar". mycontext.spring.xml is put at the root of the jar. In a Jython script (*myscript.jy"), I try to call the initialize method to create the application context: from com.offlinesupport import OfflineScriptSupport OfflineScriptSupport.initialize(); I execute the Jython script with the following command (from Linux): jython -Dpython.path=spring.jar:OfflineScriptSupport.jar:MyContext.jar myscript.jy The Springframework application context cannot find the mycontext.spring.xml file. It displays the following error: java.io.FileNotFoundException: class path resource [mycontext.spring.xml] cannot be opened because it does not exist at org.springframework.core.io.ClassPathResource.getInputStream(ClassPathResource.java:137) at org.springframework.beans.factory.xml.XmlBeanDefinitionReader.loadBeanDefinitions(XmlBeanDefinitionReader.java:167) at org.springframework.beans.factory.xml.XmlBeanDefinitionReader.loadBeanDefinitions(XmlBeanDefinitionReader.java:148) at org.springframework.beans.factory.support.AbstractBeanDefinitionReader.loadBeanDefinitions(AbstractBeanDefinitionReader.java:126) at org.springframework.beans.factory.support.AbstractBeanDefinitionReader.loadBeanDefinitions(AbstractBeanDefinitionReader.java:142) at org.springframework.context.support.AbstractXmlApplicationContext.loadBeanDefinitions(AbstractXmlApplicationContext.java:113) at org.springframework.context.support.AbstractXmlApplicationContext.loadBeanDefinitions(AbstractXmlApplicationContext.java:81) at org.springframework.context.support.AbstractRefreshableApplicationContext.refreshBeanFactory(AbstractRefreshableApplicationContext.java:89) at org.springframework.context.support.AbstractApplicationContext.refresh(AbstractApplicationContext.java:269) at org.springframework.context.support.ClassPathXmlApplicationContext.<init>(ClassPathXmlApplicationContext.java:87) at org.springframework.context.support.ClassPathXmlApplicationContext.<init>(ClassPathXmlApplicationContext.java:72) at com.offlinesupport.OfflineScriptSupport.initialize(OfflineScriptSupport.java:27) at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method) at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:39) at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:25) at java.lang.reflect.Method.invoke(Method.java:597) If I run the jar directly from Java (using the main entry point in OfflineScriptSupport) it works and there is no error thrown. Is there something special about the way Jython handles classpaths making the Springframework's ClassPathXmlApplicationContext not work (i.e. not be able to find resource files in the classpath)?

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  • Android: Any way to auto-pair to another device over Bluetooth without prompting for a pin?

    - by D.
    I am looking for a way to connect to Android devices via Bluetooth without user intervention(assuming at least on device is set to "Discoverable"). Since 2.0, it seems the devices prompt for a random pin to be entered when connecting to each other for the first time. I've tried some Bluetooth projects, but none seem to work as the underlying Bluetooth Adapter code always kicks in. Is there any way around this? Thanks.

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  • javax.swing.JDialog is appearing twice on windows device.

    - by Bhaiyaji Patel
    I have created a JDialog to be opened when I click on the edit button of my JFrame, it is being opened properly and does'nt have any issue, but when I took this code on the windows ce 5.0 device this dialog is being opened twice. hat is i am clicking only once on the edit button but the dialog is appearing twice, I want there should be only one dialog appear on edit button click.

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  • Change HttpContext.Request.InputStream

    - by user320478
    I am getting lot of errors for HttpRequestValidationException in my event log. Is it possible to HTMLEncode all the inputs from override of ProcessRequest on web page. I have tried this but it gives context.Request.InputStream.CanWrite == false always. Is there any way to HTMLEncode all the feilds when request is made? public override void ProcessRequest(HttpContext context) { if (context.Request.InputStream.CanRead) { IEnumerator en = HttpContext.Current.Request.Form.GetEnumerator(); while (en.MoveNext()) { //Response.Write(Server.HtmlEncode(en.Current + " = " + //HttpContext.Current.Request.Form[(string)en.Current])); } long nLen = context.Request.InputStream.Length; if (nLen > 0) { string strInputStream = string.Empty; context.Request.InputStream.Position = 0; byte[] bytes = new byte[nLen]; context.Request.InputStream.Read(bytes, 0, Convert.ToInt32(nLen)); strInputStream = Encoding.Default.GetString(bytes); if (!string.IsNullOrEmpty(strInputStream)) { List<string> stream = strInputStream.Split('&').ToList<string>(); Dictionary<int, string> data = new Dictionary<int, string>(); if (stream != null && stream.Count > 0) { int index = 0; foreach (string str in stream) { if (str.Length > 3 && str.Substring(0, 3) == "txt") { string textBoxData = str; string temp = Server.HtmlEncode(str); //stream[index] = temp; data.Add(index, temp); index++; } } if (data.Count > 0) { List<string> streamNew = stream; foreach (KeyValuePair<int, string> kvp in data) { streamNew[kvp.Key] = kvp.Value; } string newStream = string.Join("", streamNew.ToArray()); byte[] bytesNew = Encoding.Default.GetBytes(newStream); if (context.Request.InputStream.CanWrite) { context.Request.InputStream.Flush(); context.Request.InputStream.Position = 0; context.Request.InputStream.Write(bytesNew, 0, bytesNew.Length); //Request.InputStream.Close(); //Request.InputStream.Dispose(); } } } } } } base.ProcessRequest(context); }

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  • BroadcastReciever doesn't show SMS not Send or Not delivered

    - by user1657111
    While using broadcastreciver for checking sms status, it shows the toast when sms is sent but shows nothing when sms is not sent or delivered (im testing it by putting an abrupt number). the code im using is the one ive seen the most on every site of checking sms delivery status. But my code is only showing the status when sms is sent successfully. Can any one get a hint of what am i doing wrong ? I hav this method in doInBackground() and so obviously im using AsyncTask. Thanks guys public void send_SMS(String list, String msg, AtaskClass task) { String SENT = "SMS_SENT"; String DELIVERED = "SMS_DELIVERED"; SmsManager sms = SmsManager.getDefault(); PendingIntent sentPI = PendingIntent.getBroadcast(this, 0, new Intent(SENT), 0); PendingIntent deliveredPI = PendingIntent.getBroadcast(this, 0, new Intent(DELIVERED), 0); //---when the SMS has been sent--- registerReceiver(new BroadcastReceiver(){ @Override public void onReceive(Context context, Intent arg1) { switch (getResultCode()) { case Activity.RESULT_OK: Toast.makeText(context, "SMS sent", Toast.LENGTH_SHORT).show(); break; case SmsManager.RESULT_ERROR_GENERIC_FAILURE: Toast.makeText(context, "Generic failure", Toast.LENGTH_SHORT).show(); break; case SmsManager.RESULT_ERROR_NO_SERVICE: Toast.makeText(context, "No service", Toast.LENGTH_SHORT).show(); break; case SmsManager.RESULT_ERROR_NULL_PDU: Toast.makeText(context, "Null PDU", Toast.LENGTH_SHORT).show(); break; case SmsManager.RESULT_ERROR_RADIO_OFF: Toast.makeText(context, "Radio off", Toast.LENGTH_SHORT).show(); break; } } }, new IntentFilter(SENT)); //---when the SMS has been delivered--- registerReceiver(new BroadcastReceiver(){ @Override public void onReceive(Context context, Intent arg1) { switch (getResultCode()) { case Activity.RESULT_OK: Toast.makeText(context, "SMS delivered", Toast.LENGTH_SHORT).show(); break; case Activity.RESULT_CANCELED: Toast.makeText(context, "SMS not delivered", Toast.LENGTH_SHORT).show(); break; } } }, new IntentFilter(DELIVERED)); StringTokenizer st = new StringTokenizer(list,","); int count= st.countTokens(); int i =1; count = 1; while(st.hasMoreElements()) { // PendingIntent pi = PendingIntent.getActivity(this,0,new Intent(this, SMS.class),0); String tempMobileNumber = (String)st.nextElement(); //SmsManager sms = SmsManager.getDefault(); sms.sendTextMessage(tempMobileNumber, null, msg , sentPI, deliveredPI); Double cCom = ((double)i/count) * 100; int j = cCom.intValue(); task.doProgress(j); i++; count ++; } // class ends }

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  • Determining if an Activity exists on the current device?

    - by stormin986
    Is there a way to check and see if an Activity exists on your device? If I have a youtube video link I want to specify it open in the YouTube PlayerActivity. However, I don't want to crash if for some reason they don't have it. Is there a way to check and see if the activity exists? I don't think I can catch the runtime exception since startActivity() doesn't throw it.

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