Search Results

Search found 35577 results on 1424 pages for 'java geek'.

Page 45/1424 | < Previous Page | 41 42 43 44 45 46 47 48 49 50 51 52  | Next Page >

  • Java API Method Run Times

    - by Mike
    Is there a good resource to get run times for standard API functions? It's somewhat confusing when trying to optimize your program. I know Java isn't made to be particularly speedy but I can't seem to find much info on this at all. Example Problem: If I am looking for a certain token in a file is it faster to scan each line using string.contains(...) or to bring in say 100 or so lines putting them to a local string them performing contains on that chunk.

    Read the article

  • Jailkit - allowing use of Java/Python

    - by James hooker
    I'm looking to allow Jailed users to use JAVA (JRE) to execute their own Java files, similarly with Python to execute py scripts. I've tried adding the JAVA binary (/usr/bin/java) to a jail using the following jk_cp /path/to/jail /usr/bin/java This seems to copy some of the libraries across, aswell as the Java binary itself, however the jailed user is still unable to execute Java. It complains first of all of a missing lib called libsli.so. I copied that across and it then complains about libjava.so which I proceeded to copy across to, to no avail. Does anyone have experience with enabling the execution of Java within a jailed environment? This is under Ubuntu 10.04 64-bit, with Jailkit 2.11.

    Read the article

  • Java run time environment mac

    - by Gatura
    I am trying to install an application on a MAC 10.6.4, however i get an installation error that no java runtime environment. However on the java website, the mac section indicates that MAC provides its own java and should use the software update software to get the latest java application. I have done this and gotten the latest java however I still get the same error no java run time is there something am missing?

    Read the article

  • Monitoring on java daemon on centos

    - by user111196
    I have a java application which I run using yasjw tool as a daemon. I need to monitor it in case it goes down I need some kind of alert or even restart it. Is there any tool can help me do this on centos environment? The results of ps -ef | grep java root 3109 1 0 Apr06 ? 00:04:35 /usr/java/jdk1.6.0_18/bin/java -Dwrapper.pidfile=/var/run/wrapper.commServer.pid -Dwrapper.service=true -Dwrapper.visible=false -jar /usr/local/yajsw-beta-10.2/wrapper.jar -c /usr/local/yajsw-beta-10.2/conf/wrapper.conf root 3132 3109 0 Apr06 ? 00:25:26 /usr/java/jdk1.6.0_18/bin/java -classpath /usr/local/yajsw-beta-10.2/./wrapperApp.jar:/usr/local -Xrs -Dwrapper.service=true -Dwrapper.console.visible=false -Dwrapper.visible=false -Dwrapper.pidfile=/var/run/wrapper.commServer.pid -Dwrapper.config=/usr/local/yajsw-beta-10.2/conf/wrapper.conf -Dwrapper.port=15003 -Dwrapper.key=4276015160565963367 -Dwrapper.teeName=4276015160565963367$1333699547154 -Dwrapper.tmpPath=/tmp org.rzo.yajsw.app.WrapperJVMMain root 23986 23945 0 16:53 pts/0 00:00:00 grep java pidof java 3132 3109

    Read the article

  • Java Applets fail to launch in Windows 7 64bit

    - by Steve
    Hi can you help? I'm having great trouble launching Java applets using Windows 7 64 bit. I have the 32 bit version of Java installed (as recommended, for use with 32 bit browsers). When I click on an Java applet to open it the Java logo shows with a circular progress bar, but the process goes no further, just the progress bar going round and round. There are no error messages. The computer is a few months old and Java has never worked properly. I've tried with both IE9 and Chrome, followed the suggested fixes, uninstall then reinstall, check certain browser settings (all were in order), check Java control panel settings (again all was in order) delete the internet cache etc, tried 32bit & 64bit Java side by side. All to no avail. I'm baffled Is there anything else to try, or do I have to accept a computer without Java (Very inconvenient!) Thanks

    Read the article

  • Upgrade to java 1.7 in eclipse on mac

    - by user2159614
    I'm sort of a beginner with eclipse but I want to update the libraries or build path or whatever to java 1.7 from java 1.6 and I can't figure it out. I'm a computer science student at the university of washington and various TA's and students have tried to figure out this problem but it's stumped them all. I've installed java 1.7 from Oracle a few times already and the java section of system preferences says I have 1.7 but java -version in terminal says: java version "1.6.0_41" Java(TM) SE Runtime Environment (build 1.6.0_41-b02-445-11M4107) Java HotSpot(TM) 64-Bit Server VM (build 20.14-b01-445, mixed mode) What's going on here? My mac is totally up to date on everything else

    Read the article

  • java.lang.OutOfMemoryError: bitmap size exceeds VM budget

    - by Angel
    Hi, I am trying to change the layout of my application from portrait to landscape and vice-versa. But if i do it frequently or more than once then at times my application crashes.. Below is the error log. Please suggest what can be done? < 01-06 09:52:27.787: ERROR/dalvikvm-heap(17473): 1550532-byte external allocation too large for this process. 01-06 09:52:27.787: ERROR/dalvikvm(17473): Out of memory: Heap Size=6471KB, Allocated=4075KB, Bitmap Size=9564KB 01-06 09:52:27.787: ERROR/(17473): VM won't let us allocate 1550532 bytes 01-06 09:52:27.798: DEBUG/skia(17473): --- decoder-decode returned false 01-06 09:52:27.798: DEBUG/AndroidRuntime(17473): Shutting down VM 01-06 09:52:27.798: WARN/dalvikvm(17473): threadid=3: thread exiting with uncaught exception (group=0x4001e390) 01-06 09:52:27.807: ERROR/AndroidRuntime(17473): Uncaught handler: thread main exiting due to uncaught exception 01-06 09:52:27.857: ERROR/AndroidRuntime(17473): java.lang.RuntimeException: Unable to start activity ComponentInfo{}: android.view.InflateException: Binary XML file line #2: Error inflating class 01-06 09:52:27.857: ERROR/AndroidRuntime(17473): at android.app.ActivityThread.performLaunchActivity(ActivityThread.java:2596) 01-06 09:52:27.857: ERROR/AndroidRuntime(17473): at android.app.ActivityThread.handleLaunchActivity(ActivityThread.java:2621) 01-06 09:52:27.857: ERROR/AndroidRuntime(17473): at android.app.ActivityThread.handleRelaunchActivity(ActivityThread.java:3812) 01-06 09:52:27.857: ERROR/AndroidRuntime(17473): at android.app.ActivityThread.access$2300(ActivityThread.java:126) 01-06 09:52:27.857: ERROR/AndroidRuntime(17473): at android.app.ActivityThread$H.handleMessage(ActivityThread.java:1936) 01-06 09:52:27.857: ERROR/AndroidRuntime(17473): at android.os.Handler.dispatchMessage(Handler.java:99) 01-06 09:52:27.857: ERROR/AndroidRuntime(17473): at android.os.Looper.loop(Looper.java:123) 01-06 09:52:27.857: ERROR/AndroidRuntime(17473): at android.app.ActivityThread.main(ActivityThread.java:4595) 01-06 09:52:27.857: ERROR/AndroidRuntime(17473): at java.lang.reflect.Method.invokeNative(Native Method) 01-06 09:52:27.857: ERROR/AndroidRuntime(17473): at java.lang.reflect.Method.invoke(Method.java:521) 01-06 09:52:27.857: ERROR/AndroidRuntime(17473): at com.android.internal.os.ZygoteInit$MethodAndArgsCaller.run(ZygoteInit.java:860) 01-06 09:52:27.857: ERROR/AndroidRuntime(17473): at com.android.internal.os.ZygoteInit.main(ZygoteInit.java:618) 01-06 09:52:27.857: ERROR/AndroidRuntime(17473): at dalvik.system.NativeStart.main(Native Method) 01-06 09:52:27.857: ERROR/AndroidRuntime(17473): Caused by: android.view.InflateException: Binary XML file line #2: Error inflating class 01-06 09:52:27.857: ERROR/AndroidRuntime(17473): at android.view.LayoutInflater.createView(LayoutInflater.java:513) 01-06 09:52:27.857: ERROR/AndroidRuntime(17473): at com.android.internal.policy.impl.PhoneLayoutInflater.onCreateView(PhoneLayoutInflater.java:56) 01-06 09:52:27.857: ERROR/AndroidRuntime(17473): at android.view.LayoutInflater.createViewFromTag(LayoutInflater.java:563) 01-06 09:52:27.857: ERROR/AndroidRuntime(17473): at android.view.LayoutInflater.inflate(LayoutInflater.java:385) 01-06 09:52:27.857: ERROR/AndroidRuntime(17473): at android.view.LayoutInflater.inflate(LayoutInflater.java:320) 01-06 09:52:27.857: ERROR/AndroidRuntime(17473): at android.view.LayoutInflater.inflate(LayoutInflater.java:276) 01-06 09:52:27.857: ERROR/AndroidRuntime(17473): at com.android.internal.policy.impl.PhoneWindow.setContentView(PhoneWindow.java:207) 01-06 09:52:27.857: ERROR/AndroidRuntime(17473): at android.app.Activity.setContentView(Activity.java:1629) 01-06 09:52:27.857: ERROR/AndroidRuntime(17473): at onCreate(Game.java:98) 01-06 09:52:27.857: ERROR/AndroidRuntime(17473): at android.app.Instrumentation.callActivityOnCreate(Instrumentation.java:1047) 01-06 09:52:27.857: ERROR/AndroidRuntime(17473): at android.app.ActivityThread.performLaunchActivity(ActivityThread.java:2544) 01-06 09:52:27.857: ERROR/AndroidRuntime(17473): ... 12 more 01-06 09:52:27.857: ERROR/AndroidRuntime(17473): Caused by: java.lang.reflect.InvocationTargetException 01-06 09:52:27.857: ERROR/AndroidRuntime(17473): at android.widget.LinearLayout.(LinearLayout.java:92) 01-06 09:52:27.857: ERROR/AndroidRuntime(17473): at java.lang.reflect.Constructor.constructNative(Native Method) 01-06 09:52:27.857: ERROR/AndroidRuntime(17473): at java.lang.reflect.Constructor.newInstance(Constructor.java:446) 01-06 09:52:27.857: ERROR/AndroidRuntime(17473): at android.view.LayoutInflater.createView(LayoutInflater.java:500) 01-06 09:52:27.857: ERROR/AndroidRuntime(17473): ... 22 more 01-06 09:52:27.857: ERROR/AndroidRuntime(17473): Caused by: java.lang.OutOfMemoryError: bitmap size exceeds VM budget 01-06 09:52:27.857: ERROR/AndroidRuntime(17473): at android.graphics.BitmapFactory.nativeDecodeAsset(Native Method) 01-06 09:52:27.857: ERROR/AndroidRuntime(17473): at android.graphics.BitmapFactory.decodeStream(BitmapFactory.java:464) 01-06 09:52:27.857: ERROR/AndroidRuntime(17473): at android.graphics.BitmapFactory.decodeResourceStream(BitmapFactory.java:340) 01-06 09:52:27.857: ERROR/AndroidRuntime(17473): at android.graphics.drawable.Drawable.createFromResourceStream(Drawable.java:697) 01-06 09:52:27.857: ERROR/AndroidRuntime(17473): at android.content.res.Resources.loadDrawable(Resources.java:1705) 01-06 09:52:27.857: ERROR/AndroidRuntime(17473): at android.content.res.TypedArray.getDrawable(TypedArray.java:548) 01-06 09:52:27.857: ERROR/AndroidRuntime(17473): at android.view.View.(View.java:1850) 01-06 09:52:27.857: ERROR/AndroidRuntime(17473): at android.view.View.(View.java:1799) 01-06 09:52:27.857: ERROR/AndroidRuntime(17473): at android.view.ViewGroup.(ViewGroup.java:296) 01-06 09:52:27.857: ERROR/AndroidRuntime(17473): ... 26 more

    Read the article

  • [GEEK SCHOOL] Network Security 1: Securing User Accounts and Passwords in Windows

    - by Matt Klein
    This How-To Geek School class is intended for people who want to learn more about security when using Windows operating systems. You will learn many principles that will help you have a more secure computing experience and will get the chance to use all the important security tools and features that are bundled with Windows. Obviously, we will share everything you need to know about using them effectively. In this first lesson, we will talk about password security; the different ways of logging into Windows and how secure they are. In the proceeding lesson, we will explain where Windows stores all the user names and passwords you enter while working in this operating systems, how safe they are, and how to manage this data. Moving on in the series, we will talk about User Account Control, its role in improving the security of your system, and how to use Windows Defender in order to protect your system from malware. Then, we will talk about the Windows Firewall, how to use it in order to manage the apps that get access to the network and the Internet, and how to create your own filtering rules. After that, we will discuss the SmartScreen Filter – a security feature that gets more and more attention from Microsoft and is now widely used in its Windows 8.x operating systems. Moving on, we will discuss ways to keep your software and apps up-to-date, why this is important and which tools you can use to automate this process as much as possible. Last but not least, we will discuss the Action Center and its role in keeping you informed about what’s going on with your system and share several tips and tricks about how to stay safe when using your computer and the Internet. Let’s get started by discussing everyone’s favorite subject: passwords. The Types of Passwords Found in Windows In Windows 7, you have only local user accounts, which may or may not have a password. For example, you can easily set a blank password for any user account, even if that one is an administrator. The only exception to this rule are business networks where domain policies force all user accounts to use a non-blank password. In Windows 8.x, you have both local accounts and Microsoft accounts. If you would like to learn more about them, don’t hesitate to read the lesson on User Accounts, Groups, Permissions & Their Role in Sharing, in our Windows Networking series. Microsoft accounts are obliged to use a non-blank password due to the fact that a Microsoft account gives you access to Microsoft services. Using a blank password would mean exposing yourself to lots of problems. Local accounts in Windows 8.1 however, can use a blank password. On top of traditional passwords, any user account can create and use a 4-digit PIN or a picture password. These concepts were introduced by Microsoft to speed up the sign in process for the Windows 8.x operating system. However, they do not replace the use of a traditional password and can be used only in conjunction with a traditional user account password. Another type of password that you encounter in Windows operating systems is the Homegroup password. In a typical home network, users can use the Homegroup to easily share resources. A Homegroup can be joined by a Windows device only by using the Homegroup password. If you would like to learn more about the Homegroup and how to use it for network sharing, don’t hesitate to read our Windows Networking series. What to Keep in Mind When Creating Passwords, PINs and Picture Passwords When creating passwords, a PIN, or a picture password for your user account, we would like you keep in mind the following recommendations: Do not use blank passwords, even on the desktop computers in your home. You never know who may gain unwanted access to them. Also, malware can run more easily as administrator because you do not have a password. Trading your security for convenience when logging in is never a good idea. When creating a password, make it at least eight characters long. Make sure that it includes a random mix of upper and lowercase letters, numbers, and symbols. Ideally, it should not be related in any way to your name, username, or company name. Make sure that your passwords do not include complete words from any dictionary. Dictionaries are the first thing crackers use to hack passwords. Do not use the same password for more than one account. All of your passwords should be unique and you should use a system like LastPass, KeePass, Roboform or something similar to keep track of them. When creating a PIN use four different digits to make things slightly harder to crack. When creating a picture password, pick a photo that has at least 10 “points of interests”. Points of interests are areas that serve as a landmark for your gestures. Use a random mixture of gesture types and sequence and make sure that you do not repeat the same gesture twice. Be aware that smudges on the screen could potentially reveal your gestures to others. The Security of Your Password vs. the PIN and the Picture Password Any kind of password can be cracked with enough effort and the appropriate tools. There is no such thing as a completely secure password. However, passwords created using only a few security principles are much harder to crack than others. If you respect the recommendations shared in the previous section of this lesson, you will end up having reasonably secure passwords. Out of all the log in methods in Windows 8.x, the PIN is the easiest to brute force because PINs are restricted to four digits and there are only 10,000 possible unique combinations available. The picture password is more secure than the PIN because it provides many more opportunities for creating unique combinations of gestures. Microsoft have compared the two login options from a security perspective in this post: Signing in with a picture password. In order to discourage brute force attacks against picture passwords and PINs, Windows defaults to your traditional text password after five failed attempts. The PIN and the picture password function only as alternative login methods to Windows 8.x. Therefore, if someone cracks them, he or she doesn’t have access to your user account password. However, that person can use all the apps installed on your Windows 8.x device, access your files, data, and so on. How to Create a PIN in Windows 8.x If you log in to a Windows 8.x device with a user account that has a non-blank password, then you can create a 4-digit PIN for it, to use it as a complementary login method. In order to create one, you need to go to “PC Settings”. If you don’t know how, then press Windows + C on your keyboard or flick from the right edge of the screen, on a touch-enabled device, then press “Settings”. The Settings charm is now open. Click or tap the link that says “Change PC settings”, on the bottom of the charm. In PC settings, go to Accounts and then to “Sign-in options”. Here you will find all the necessary options for changing your existing password, creating a PIN, or a picture password. To create a PIN, press the “Add” button in the PIN section. The “Create a PIN” wizard is started and you are asked to enter the password of your user account. Type it and press “OK”. Now you are asked to enter a 4-digit pin in the “Enter PIN” and “Confirm PIN” fields. The PIN has been created and you can now use it to log in to Windows. How to Create a Picture Password in Windows 8.x If you log in to a Windows 8.x device with a user account that has a non-blank password, then you can also create a picture password and use it as a complementary login method. In order to create one, you need to go to “PC settings”. In PC Settings, go to Accounts and then to “Sign-in options”. Here you will find all the necessary options for changing your existing password, creating a PIN, or a picture password. To create a picture password, press the “Add” button in the “Picture password” section. The “Create a picture password” wizard is started and you are asked to enter the password of your user account. You are shown a guide on how the picture password works. Take a few seconds to watch it and learn the gestures that can be used for your picture password. You will learn that you can create a combination of circles, straight lines, and taps. When ready, press “Choose picture”. Browse your Windows 8.x device and select the picture you want to use for your password and press “Open”. Now you can drag the picture to position it the way you want. When you like how the picture is positioned, press “Use this picture” on the left. If you are not happy with the picture, press “Choose new picture” and select a new one, as shown during the previous step. After you have confirmed that you want to use this picture, you are asked to set up your gestures for the picture password. Draw three gestures on the picture, any combination you wish. Please remember that you can use only three gestures: circles, straight lines, and taps. Once you have drawn those three gestures, you are asked to confirm. Draw the same gestures one more time. If everything goes well, you are informed that you have created your picture password and that you can use it the next time you sign in to Windows. If you don’t confirm the gestures correctly, you will be asked to try again, until you draw the same gestures twice. To close the picture password wizard, press “Finish”. Where Does Windows Store Your Passwords? Are They Safe? All the passwords that you enter in Windows and save for future use are stored in the Credential Manager. This tool is a vault with the usernames and passwords that you use to log on to your computer, to other computers on the network, to apps from the Windows Store, or to websites using Internet Explorer. By storing these credentials, Windows can automatically log you the next time you access the same app, network share, or website. Everything that is stored in the Credential Manager is encrypted for your protection.

    Read the article

  • [GEEK SCHOOL] Network Security 2: Preventing Disaster with User Account Control

    - by Ciprian Rusen
    In this second lesson in our How-To Geek School about securing the Windows devices in your network, we will talk about User Account Control (UAC). Users encounter this feature each time they need to install desktop applications in Windows, when some applications need administrator permissions in order to work and when they have to change different system settings and files. UAC was introduced in Windows Vista as part of Microsoft’s “Trustworthy Computing” initiative. Basically, UAC is meant to act as a wedge between you and installing applications or making system changes. When you attempt to do either of these actions, UAC will pop up and interrupt you. You may either have to confirm you know what you’re doing, or even enter an administrator password if you don’t have those rights. Some users find UAC annoying and choose to disable it but this very important security feature of Windows (and we strongly caution against doing that). That’s why in this lesson, we will carefully explain what UAC is and everything it does. As you will see, this feature has an important role in keeping Windows safe from all kinds of security problems. In this lesson you will learn which activities may trigger a UAC prompt asking for permissions and how UAC can be set so that it strikes the best balance between usability and security. You will also learn what kind of information you can find in each UAC prompt. Last but not least, you will learn why you should never turn off this feature of Windows. By the time we’re done today, we think you will have a newly found appreciation for UAC, and will be able to find a happy medium between turning it off completely and letting it annoy you to distraction. What is UAC and How Does it Work? UAC or User Account Control is a security feature that helps prevent unauthorized system changes to your Windows computer or device. These changes can be made by users, applications, and sadly, malware (which is the biggest reason why UAC exists in the first place). When an important system change is initiated, Windows displays a UAC prompt asking for your permission to make the change. If you don’t give your approval, the change is not made. In Windows, you will encounter UAC prompts mostly when working with desktop applications that require administrative permissions. For example, in order to install an application, the installer (generally a setup.exe file) asks Windows for administrative permissions. UAC initiates an elevation prompt like the one shown earlier asking you whether it is okay to elevate permissions or not. If you say “Yes”, the installer starts as administrator and it is able to make the necessary system changes in order to install the application correctly. When the installer is closed, its administrator privileges are gone. If you run it again, the UAC prompt is shown again because your previous approval is not remembered. If you say “No”, the installer is not allowed to run and no system changes are made. If a system change is initiated from a user account that is not an administrator, e.g. the Guest account, the UAC prompt will also ask for the administrator password in order to give the necessary permissions. Without this password, the change won’t be made. Which Activities Trigger a UAC Prompt? There are many types of activities that may trigger a UAC prompt: Running a desktop application as an administrator Making changes to settings and files in the Windows and Program Files folders Installing or removing drivers and desktop applications Installing ActiveX controls Changing settings to Windows features like the Windows Firewall, UAC, Windows Update, Windows Defender, and others Adding, modifying, or removing user accounts Configuring Parental Controls in Windows 7 or Family Safety in Windows 8.x Running the Task Scheduler Restoring backed-up system files Viewing or changing the folders and files of another user account Changing the system date and time You will encounter UAC prompts during some or all of these activities, depending on how UAC is set on your Windows device. If this security feature is turned off, any user account or desktop application can make any of these changes without a prompt asking for permissions. In this scenario, the different forms of malware existing on the Internet will also have a higher chance of infecting and taking control of your system. In Windows 8.x operating systems you will never see a UAC prompt when working with apps from the Windows Store. That’s because these apps, by design, are not allowed to modify any system settings or files. You will encounter UAC prompts only when working with desktop programs. What You Can Learn from a UAC Prompt? When you see a UAC prompt on the screen, take time to read the information displayed so that you get a better understanding of what is going on. Each prompt first tells you the name of the program that wants to make system changes to your device, then you can see the verified publisher of that program. Dodgy software tends not to display this information and instead of a real company name, you will see an entry that says “Unknown”. If you have downloaded that program from a less than trustworthy source, then it might be better to select “No” in the UAC prompt. The prompt also shares the origin of the file that’s trying to make these changes. In most cases the file origin is “Hard drive on this computer”. You can learn more by pressing “Show details”. You will see an additional entry named “Program location” where you can see the physical location on your hard drive, for the file that’s trying to perform system changes. Make your choice based on the trust you have in the program you are trying to run and its publisher. If a less-known file from a suspicious location is requesting a UAC prompt, then you should seriously consider pressing “No”. What’s Different About Each UAC Level? Windows 7 and Windows 8.x have four UAC levels: Always notify – when this level is used, you are notified before desktop applications make changes that require administrator permissions or before you or another user account changes Windows settings like the ones mentioned earlier. When the UAC prompt is shown, the desktop is dimmed and you must choose “Yes” or “No” before you can do anything else. This is the most secure and also the most annoying way to set UAC because it triggers the most UAC prompts. Notify me only when programs/apps try to make changes to my computer (default) – Windows uses this as the default for UAC. When this level is used, you are notified before desktop applications make changes that require administrator permissions. If you are making system changes, UAC doesn’t show any prompts and it automatically gives you the necessary permissions for making the changes you desire. When a UAC prompt is shown, the desktop is dimmed and you must choose “Yes” or “No” before you can do anything else. This level is slightly less secure than the previous one because malicious programs can be created for simulating the keystrokes or mouse moves of a user and change system settings for you. If you have a good security solution in place, this scenario should never occur. Notify me only when programs/apps try to make changes to my computer (do not dim my desktop) – this level is different from the previous in in the fact that, when the UAC prompt is shown, the desktop is not dimmed. This decreases the security of your system because different kinds of desktop applications (including malware) might be able to interfere with the UAC prompt and approve changes that you might not want to be performed. Never notify – this level is the equivalent of turning off UAC. When using it, you have no protection against unauthorized system changes. Any desktop application and any user account can make system changes without your permission. How to Configure UAC If you would like to change the UAC level used by Windows, open the Control Panel, then go to “System and Security” and select “Action Center”. On the column on the left you will see an entry that says “Change User Account Control settings”. The “User Account Control Settings” window is now opened. Change the position of the UAC slider to the level you want applied then press “OK”. Depending on how UAC was initially set, you may receive a UAC prompt requiring you to confirm this change. Why You Should Never Turn Off UAC If you want to keep the security of your system at decent levels, you should never turn off UAC. When you disable it, everything and everyone can make system changes without your consent. This makes it easier for all kinds of malware to infect and take control of your system. It doesn’t matter whether you have a security suite or antivirus installed or third-party antivirus, basic common-sense measures like having UAC turned on make a big difference in keeping your devices safe from harm. We have noticed that some users disable UAC prior to setting up their Windows devices and installing third-party software on them. They keep it disabled while installing all the software they will use and enable it when done installing everything, so that they don’t have to deal with so many UAC prompts. Unfortunately this causes problems with some desktop applications. They may fail to work after you enable UAC. This happens because, when UAC is disabled, the virtualization techniques UAC uses for your applications are inactive. This means that certain user settings and files are installed in a different place and when you turn on UAC, applications stop working because they should be placed elsewhere. Therefore, whatever you do, do not turn off UAC completely! Coming up next … In the next lesson you will learn about Windows Defender, what this tool can do in Windows 7 and Windows 8.x, what’s different about it in these operating systems and how it can be used to increase the security of your system.

    Read the article

  • SSLException: HelloRequest followed by an unexpected handshake message

    - by mseebach
    I'm trying to connect to a webservice over SSL using Apache Commons HttpClient 3.1, using this: String url = "https://archprod.service.eogs.dk/cvronline/esb/LegalUnitGetSSLServicePort"; HttpClient client = new HttpClient(); PostMethod post = new PostMethod(url); StringRequestEntity entity = new StringRequestEntity(requestXml, "application/soap+xml", "utf-8"); post.setRequestEntity(entity); client.executeMethod(post); String response = post.getResponseBodyAsString(); And I get this exception: javax.net.ssl.SSLException: HelloRequest followed by an unexpected handshake message at com.sun.net.ssl.internal.ssl.Alerts.getSSLException(Alerts.java:190) at com.sun.net.ssl.internal.ssl.SSLSocketImpl.fatal(SSLSocketImpl.java:1623) at com.sun.net.ssl.internal.ssl.Handshaker.fatalSE(Handshaker.java:198) at com.sun.net.ssl.internal.ssl.Handshaker.fatalSE(Handshaker.java:188) at com.sun.net.ssl.internal.ssl.ClientHandshaker.serverHelloRequest(ClientHandshaker.java:286) at com.sun.net.ssl.internal.ssl.ClientHandshaker.processMessage(ClientHandshaker.java:114) at com.sun.net.ssl.internal.ssl.Handshaker.processLoop(Handshaker.java:525) at com.sun.net.ssl.internal.ssl.Handshaker.process_record(Handshaker.java:465) at com.sun.net.ssl.internal.ssl.SSLSocketImpl.readRecord(SSLSocketImpl.java:884) at com.sun.net.ssl.internal.ssl.SSLSocketImpl.readDataRecord(SSLSocketImpl.java:746) at com.sun.net.ssl.internal.ssl.AppInputStream.read(AppInputStream.java:75) at java.io.BufferedInputStream.fill(BufferedInputStream.java:218) at java.io.BufferedInputStream.read(BufferedInputStream.java:237) at org.apache.commons.httpclient.HttpParser.readRawLine(HttpParser.java:78) at org.apache.commons.httpclient.HttpParser.readLine(HttpParser.java:106) at org.apache.commons.httpclient.HttpConnection.readLine(HttpConnection.java:1116) at org.apache.commons.httpclient.HttpMethodBase.readStatusLine(HttpMethodBase.java:1973) at org.apache.commons.httpclient.HttpMethodBase.readResponse(HttpMethodBase.java:1735) at org.apache.commons.httpclient.HttpMethodBase.execute(HttpMethodBase.java:1098) at org.apache.commons.httpclient.HttpMethodDirector.executeWithRetry(HttpMethodDirector.java:398) at org.apache.commons.httpclient.HttpMethodDirector.executeMethod(HttpMethodDirector.java:171) at org.apache.commons.httpclient.HttpClient.executeMethod(HttpClient.java:397) at org.apache.commons.httpclient.HttpClient.executeMethod(HttpClient.java:323) A request to the same URL on the same machine, using curl, works fine - and if I change the URL to e.g. https://www.verisign.com, it works fine in Java, too. So it appears to be the specific combination of Java and that host, not a general problem. Ubuntu 10.04 beta, Sun JDK 1.6.0_19 (same problem in Ubuntu's bundled OpenJDK 6b18~pre4). Any ideas what's going wrong? Thanks!

    Read the article

  • Parse.com REST API in Java (NOT Android)

    - by Orange Peel
    I am trying to use the Parse.com REST API in Java. I have gone through the 4 solutions given here https://parse.com/docs/api_libraries and have selected Parse4J. After importing the source into Netbeans, along with importing the following libraries: org.slf4j:slf4j-api:jar:1.6.1 org.apache.httpcomponents:httpclient:jar:4.3.2 org.apache.httpcomponents:httpcore:jar:4.3.1 org.json:json:jar:20131018 commons-codec:commons-codec:jar:1.9 junit:junit:jar:4.11 ch.qos.logback:logback-classic:jar:0.9.28 ch.qos.logback:logback-core:jar:0.9.28 I ran the example code from https://github.com/thiagolocatelli/parse4j Parse.initialize(APP_ID, APP_REST_API_ID); // I replaced these with mine ParseObject gameScore = new ParseObject("GameScore"); gameScore.put("score", 1337); gameScore.put("playerName", "Sean Plott"); gameScore.put("cheatMode", false); gameScore.save(); And I got that it was missing org.apache.commons.logging, so I downloaded that and imported it. Then I ran the code again and got Exception in thread "main" java.lang.NoSuchMethodError: org.slf4j.spi.LocationAwareLogger.log(Lorg/slf4j/Marker;Ljava/lang/String;ILjava/lang/String;Ljava/lang/Throwable;)V at org.apache.commons.logging.impl.SLF4JLocationAwareLog.debug(SLF4JLocationAwareLog.java:120) at org.apache.http.client.protocol.RequestAddCookies.process(RequestAddCookies.java:122) at org.apache.http.protocol.ImmutableHttpProcessor.process(ImmutableHttpProcessor.java:131) at org.apache.http.impl.execchain.ProtocolExec.execute(ProtocolExec.java:193) at org.apache.http.impl.execchain.RetryExec.execute(RetryExec.java:85) at org.apache.http.impl.execchain.RedirectExec.execute(RedirectExec.java:108) at org.apache.http.impl.client.InternalHttpClient.doExecute(InternalHttpClient.java:186) at org.apache.http.impl.client.CloseableHttpClient.execute(CloseableHttpClient.java:82) at org.apache.http.impl.client.CloseableHttpClient.execute(CloseableHttpClient.java:106) at org.apache.http.impl.client.CloseableHttpClient.execute(CloseableHttpClient.java:57) at org.parse4j.command.ParseCommand.perform(ParseCommand.java:44) at org.parse4j.ParseObject.save(ParseObject.java:450) I could probably fix this with another import, but I suppose then something else would pop up. I tried the other libraries with similar results, missing a bunch of libraries. Has anyone actually used REST API successfully in Java? If so, I would be grateful if you shared which library/s you used and anything else required to get it going successfully. I am using Netbeans. Thanks.

    Read the article

  • How to improve Java perfomance on Informix for Windows

    - by Michal Niklas
    I have problem with performance of Java UDR functions on Informix on Windows. On this server I already have some functions in C and SPL. I chose one function to write it in those 3 languages and I measured performance of this function on test table. Function calculates some kind of checksum so it not use any db libraries etc. only string and math operations. I observed performance on 30k records with SQL like: select function(txt) from _tmp_perf_test and I changed function to 'function_c, function_spl or function_java. My performance tests showed that C function is the fastest, SPL function is about 5 times slower, where Java is 100 (one hundred!) times slower than C. I checked it few times and 1:100 ratio didn't improve. I changed Java function to simply return length of the string but even this do not help so it looks, that there is general problem with Java function invocation, because there was no difference in time between Java function that calculate checksum and Java function that returns length of the string. I increased JVM_MAX_HEAP_SIZE to 128 and it not helped too. I use IBM Informix Dynamic Server Version 11.50.TC6DE. The same test on Linux server: IBM Informix Dynamic Server Version 11.50.FC6 show more "normal" results, i.e. Java is slower from C and SPL but only 2 to 5 times. What can I do to improve Java performance on Informix server on Windows? More info about Java on servers: c:\Informix\extend\krakatoa\jre\bin>java -version java version "1.5.0" Java(TM) 2 Runtime Environment, Standard Edition (build pwi32dev-20081129a (SR9-0 )) IBM J9 VM (build 2.3, J2RE 1.5.0 IBM J9 2.3 Windows Server 2003 x86-32 j9vmwi3223-20081129 (JIT enabled) J9VM - 20081126_26240_lHdSMr JIT - 20081112_1511ifx1_r8 GC - 200811_07) JCL - 20081129 [root@informix11 bin]# ./java -version java version "1.5.0" Java(TM) 2 Runtime Environment, Standard Edition (build pxa64devifx-20071025 (SR6b)) IBM J9 VM (build 2.3, J2RE 1.5.0 IBM J9 2.3 Linux amd64-64 j9vmxa6423-20071005 (JIT enabled) J9VM - 20071004_14218_LHdSMr JIT - 20070820_1846ifx1_r8 GC - 200708_10) JCL - 20071025

    Read the article

  • Lucene Error While Reading binary block : java.io.EOFException

    - by tushar Khairnar
    Hi, I am getting java.io.EOFException while reading a binary block from lucene index. I am storing java object as byte-array in lucene index field and reading it when hit occurs. Here is stack trace : Caused by: java.io.EOFException at java.io.ObjectInputStream$PeekInputStream.readFully(ObjectInputStream.java:2281) at java.io.ObjectInputStream$BlockDataInputStream.readShort(ObjectInputStream.java:2750) at java.io.ObjectInputStream.readStreamHeader(ObjectInputStream.java:780) at java.io.ObjectInputStream.(ObjectInputStream.java:280) at org.terracotta.modules.searchable.util.SerializationUtil$OIS.(SerializationUtil.java:20) I have some background threads which write into index. But i buffer them and then write them at once like 1000. Occasionally I also issue optimize() on index. When I write, I am re-opening IndexReader. Does this is happening because of IndexReader re-opening call? Thanks. Regards Tushar

    Read the article

  • Java - SAX parser on a XHTML document

    - by Peter
    Hey, I'm trying to write a SAX parser for an XHTML document that I download from the web. At first I was having a problem with the doctype declaration (I found out from here that it was because W3C have intentionally blocked access to the DTD), but I fixed that with: XMLReader reader = parser.getXMLReader(); reader.setFeature("http://apache.org/xml/features/disallow-doctype-decl",true); However, now I'm experiencing a second problem. The SAX parser throws an exception when it reaches some Javascript embedded in the XHTML document: <script type="text/javascript" language="JavaScript"> function checkForm() { answer = true; if (siw && siw.selectingSomething) answer = false; return answer; }// </script> Specifically the parser throws an error once it reaches the &&'s, as it's expecting an entity reference. The exact exception is: `org.xml.sax.SAXParseException: The entity name must immediately follow the '&' in the entity reference. at com.sun.org.apache.xerces.internal.util.ErrorHandlerWrapper.createSAXParseException(ErrorHandlerWrapper.java:198) at com.sun.org.apache.xerces.internal.util.ErrorHandlerWrapper.fatalError(ErrorHandlerWrapper.java:177) at com.sun.org.apache.xerces.internal.impl.XMLErrorReporter.reportError(XMLErrorReporter.java:391) at com.sun.org.apache.xerces.internal.impl.XMLScanner.reportFatalError(XMLScanner.java:1390) at com.sun.org.apache.xerces.internal.impl.XMLDocumentFragmentScannerImpl.scanEntityReference(XMLDocumentFragmentScannerImpl.java:1814) at com.sun.org.apache.xerces.internal.impl.XMLDocumentFragmentScannerImpl$FragmentContentDriver.next(XMLDocumentFragmentScannerImpl.java:3000) at com.sun.org.apache.xerces.internal.impl.XMLDocumentScannerImpl.next(XMLDocumentScannerImpl.java:624) at com.sun.org.apache.xerces.internal.impl.XMLDocumentFragmentScannerImpl.scanDocument(XMLDocumentFragmentScannerImpl.java:486) at com.sun.org.apache.xerces.internal.parsers.XML11Configuration.parse(XML11Configuration.java:810) at com.sun.org.apache.xerces.internal.parsers.XML11Configuration.parse(XML11Configuration.java:740) at com.sun.org.apache.xerces.internal.parsers.XMLParser.parse(XMLParser.java:110) at com.sun.org.apache.xerces.internal.parsers.AbstractSAXParser.parse(AbstractSAXParser.java:1208) at com.sun.org.apache.xerces.internal.jaxp.SAXParserImpl$JAXPSAXParser.parse(SAXParserImpl.java:525) at MLIAParser.readPage(MLIAParser.java:55) at MLIAParser.main(MLIAParser.java:75)` I suspect (but I don't know) that if I hadn't disabled the DTD then I wouldn't get this error. So, how can I avoid the DTD error and avoid the entity reference error? Cheers, Pete

    Read the article

  • Java Access Token PKCS11 Not found Provider

    - by oracleruiz
    Hello I'm trying to access the keystore from my smartcard in Java. And I'm using the following code.. I'm using the Pkcs11 implementation of OpenSc http://www.opensc-project.org/opensc File windows.cnf = name=dnie library=C:\WINDOWS\system32\opensc-pkcs11.dll Java Code = String configName = "windows.cnf" String PIN = "####"; Provider p = new sun.security.pkcs11.SunPKCS11(configName); Security.addProvider(p); KeyStore keyStore = KeyStore.getInstance("PKCS11", "SunPKCS11-dnie"); =)(= char[] pin = PIN.toCharArray(); keyStore.load(null, pin); When the execution goes by the line with =)(= throws me the following exeption java.security.KeyStoreException: PKCS11 not found at java.security.KeyStore.getInstance(KeyStore.java:635) at ObtenerDatos.LeerDatos(ObtenerDatos.java:52) at ObtenerDatos.obtenerNombre(ObtenerDatos.java:19) at main.main(main.java:27) Caused by: java.security.NoSuchAlgorithmException: no such algorithm: PKCS11 for provider SunPKCS11-dnie at sun.security.jca.GetInstance.getService(GetInstance.java:70) at sun.security.jca.GetInstance.getInstance(GetInstance.java:190) at java.security.Security.getImpl(Security.java:662) at java.security.KeyStore.getInstance(KeyStore.java:632) I think the problem is "SunPKCS11-dnie", but I don't know to put there. I had tried with a lot of combinations... Anyone can help me...

    Read the article

  • How to improve Java performance on Informix for Windows

    - by Michal Niklas
    I have problem with performance of Java UDR functions on Informix on Windows. On this server I already have some functions in C and SPL. I chose one function to write it in those 3 languages and I measured performance of this function on test table. Function calculates some kind of checksum so it does not use any db libraries etc. only string and math operations. I observed performance on 30k records with SQL like: select function(txt) from _tmp_perf_test and I changed function to 'function_c, function_spl or function_java. My performance tests showed that C function is the fastest, SPL function is about 5 times slower, where Java is 100 (one hundred!) times slower than C. I checked it few times and 1:100 ratio didn't improve. I changed Java function to simply return length of the string but even this do not help so it looks, that there is general problem with Java function invocation, because there was no difference in time between Java function that calculate checksum and Java function that returns length of the string. I increased JVM_MAX_HEAP_SIZE to 128 and it not helped too. I use IBM Informix Dynamic Server Version 11.50.TC6DE. The same test on Linux server: IBM Informix Dynamic Server Version 11.50.FC6 show more "normal" results, i.e. Java is slower from C and SPL but only 2 to 5 times. What can I do to improve Java performance on Informix server on Windows? More info about Java on servers: c:\Informix\extend\krakatoa\jre\bin>java -version java version "1.5.0" Java(TM) 2 Runtime Environment, Standard Edition (build pwi32dev-20081129a (SR9-0 )) IBM J9 VM (build 2.3, J2RE 1.5.0 IBM J9 2.3 Windows Server 2003 x86-32 j9vmwi3223-20081129 (JIT enabled) J9VM - 20081126_26240_lHdSMr JIT - 20081112_1511ifx1_r8 GC - 200811_07) JCL - 20081129 [root@informix11 bin]# ./java -version java version "1.5.0" Java(TM) 2 Runtime Environment, Standard Edition (build pxa64devifx-20071025 (SR6b)) IBM J9 VM (build 2.3, J2RE 1.5.0 IBM J9 2.3 Linux amd64-64 j9vmxa6423-20071005 (JIT enabled) J9VM - 20071004_14218_LHdSMr JIT - 20070820_1846ifx1_r8 GC - 200708_10) JCL - 20071025

    Read the article

  • Unable to deploy WAR file to Websphere using IBM Rational Application Developer

    - by Matt1776
    Hello - I am using the IBM RAD IDE and building a dynamic web project. When I build the project and attempt to add it to the server by selecting 'add or remove projects' I get the response that there are no projects to add or remove. Does this mean I will have to create a EAR file (J2EE Project) and add my web project to it in order to deploy to the local WAS? Might I be missing some essential configuration?

    Read the article

  • My java.util.Scanner won't work

    - by Kevin Steen Hansen
    Hello Stackoverflow my code is getting this error: Skriv din alder herunder og tryk enter: Exception in thread "main" java.util.NoSuchElementException at java.util.Scanner.throwFor(Scanner.java:907) at java.util.Scanner.next(Scanner.java:1530) at java.util.Scanner.nextInt(Scanner.java:2160) at java.util.Scanner.nextInt(Scanner.java:2119) at Tasteturindtastning.main(Tasteturindtastning.java:20) [Finished in 1.7s with exit code 1] Adn my code is: // Starter java som man plejer, læs i HejVerden.java public class Tasteturindtastning { public static void main(String[] arg) { /* Jeg skal nu angive en variable, men jeg kan ikke bestemme denne variable * Da jeg ønsker at indtastningen fra dette tastetur skal være variablen. * I stedet for int og double bruger jeg så java.util.Scanner, som aflæser * brugerens indtastninger. */ java.util.Scanner tastetur = new java.util.Scanner(System.in); // Printer en opgave/spørgsmål til brugeren System.out.println("Skriv din alder herunder og tryk enter:"); int alder; // Angiver et variablenavn alder = tastetur.nextInt(); // Angiver variablen med værdien fra indtastningen /* Herunder gør jeg brug af et if statement der tjekker værdien for * variablen alder, og ser om den er lig med eller højere end 18, og hvis * dette er tilfældet, så udprinter den en sætning */ if (alder >= 18) System.out.println("Du er myndig, da du er " + alder + " år gammel"); // Printes hvis han er 18 eller ældre } } Can snyone tell me what is wrong?

    Read the article

  • java.io.FileNotFoundException for valid URL

    - by Alexei
    Hello. I use library rome.dev.java.net to fetch RSS. Code is URL feedUrl = new URL("http://planet.rubyonrails.ru/xml/rss"); SyndFeedInput input = new SyndFeedInput(); SyndFeed feed = input.build(new XmlReader(feedUrl)); You can check that http://planet.rubyonrails.ru/xml/rss is valid URL and the page is shown in browser. But I get exception from my application java.io.FileNotFoundException: http://planet.rubyonrails.ru/xml/rss at sun.net.www.protocol.http.HttpURLConnection.getInputStream(HttpURLConnection.java:1311) at com.sun.syndication.io.XmlReader.<init>(XmlReader.java:237) at com.sun.syndication.io.XmlReader.<init>(XmlReader.java:213) at rssdaemonapp.ValidatorThread.run(ValidatorThread.java:32) at java.util.concurrent.ThreadPoolExecutor$Worker.runTask(ThreadPoolExecutor.java:886) at java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:908) at java.lang.Thread.run(Thread.java:619) I don't use any proxy. I get this exception on my PC and on the production server and only for this URL, other URLs are working.

    Read the article

  • NoClassDefFoundError without any class name

    - by Safder
    Hello, I am trying to run a java task from ant. I am trying to run the "org.apache.tools.ant.launch.Launcher" class. I keep on getting the "NoClassDefFoundError" without any class name being specified. I am also getting a "ClassNotFoundException" along with that displaying a message "Could not find the main class: . Program will exit". Here's a snippet of the error [java] Exception in thread "main" java.lang.NoClassDefFoundError: [java] Caused by: java.lang.ClassNotFoundException: [java] at java.net.URLClassLoader$1.run(URLClassLoader.java:200) [java] at java.security.AccessController.doPrivileged(Native Method) [java] at java.net.URLClassLoader.findClass(URLClassLoader.java:188) [java] at java.lang.ClassLoader.loadClass(ClassLoader.java:307) [java] at sun.misc.Launcher$AppClassLoader.loadClass(Launcher.java:301) [java] at java.lang.ClassLoader.loadClass(ClassLoader.java:252) [java] at java.lang.ClassLoader.loadClassInternal(ClassLoader.java:320) [java] Could not find the main class: . Program will exit. [java] Java Result: 1 Now I am trying to run an ant class from an ant jar and i specifiy the classpath where this class file resides using the "classpathref" attribute, however I still get this message. I checked the ant jar to check the Manifest and the "main" class is specified properly (it's "org.apache.tools.ant.launch.Launcher") . I have exhausted all my resources. Please help ! ! ! ps: My environment is Eclipse on Ubuntu 9.04

    Read the article

  • RMI java doesn't create skeleton class

    - by pavithra
    I wrote a remote service MyremoteImpl.java and used following command after compiled it. rmic MyRemoteImpl I learned that this method suppose to create stub class and a skeleton class but I can only see the stub class, why is that? The other problem I faced after run rmiregistry I started the service but it gives following error, I doubt I get this error as I'm missing skeleton class? java.net.MalformedURLException: invalid URL String: Remote Hello at java.rmi.Naming.parseURL(Unknown Source) at java.rmi.Naming.rebind(Unknown Source) at RMIservice.MyRemoteImpl.main(MyRemoteImpl.java:22) Caused by: java.net.URISyntaxException: Illegal character in path at index 6: Remote Hello at java.net.URI$Parser.fail(Unknown Source) at java.net.URI$Parser.checkChars(Unknown Source) at java.net.URI$Parser.parseHierarchical(Unknown Source) at java.net.URI$Parser.parse(Unknown Source) at java.net.URI.<init>(Unknown Source) at java.rmi.Naming.intParseURL(Unknown Source) ... 3 more Please help me to solve this, Thanx in advance!!!

    Read the article

  • java applet won't work

    - by scoobi_doobi
    hey guys, this is homework stuff, but the question is not much about coding. the task is to write a java applet to work on an m-grid server. i have the server running on apache. it has a few sample applets in .jar and .class form. the .class versions work; the .jar versions work on appletviewer, but they break if I submit them as a job to the server with this: load: class examples/pixelcount/PixelCount.class not found. java.lang.ClassNotFoundException: examples.pixelcount.PixelCount.class at sun.plugin2.applet.Applet2ClassLoader.findClass(Unknown Source) at java.lang.ClassLoader.loadClass(Unknown Source) at java.lang.ClassLoader.loadClass(Unknown Source) at sun.plugin2.applet.Plugin2ClassLoader.loadCode(Unknown Source) at sun.plugin2.applet.Plugin2Manager.createApplet(Unknown Source) at sun.plugin2.applet.Plugin2Manager$AppletExecutionRunnable.run(Unknown Source) at java.lang.Thread.run(Unknown Source) Caused by: java.net.ConnectException: Connection refused: connect at java.net.PlainSocketImpl.socketConnect(Native Method) at java.net.PlainSocketImpl.doConnect(Unknown Source) at java.net.PlainSocketImpl.connectToAddress(Unknown Source) at java.net.PlainSocketImpl.connect(Unknown Source) at java.net.SocksSocketImpl.connect(Unknown Source) at java.net.Socket.connect(Unknown Source) at sun.net.NetworkClient.doConnect(Unknown Source) at sun.net.www.http.HttpClient.openServer(Unknown Source) at sun.net.www.http.HttpClient.openServer(Unknown Source) at sun.net.www.http.HttpClient.<init>(Unknown Source) at sun.net.www.http.HttpClient.New(Unknown Source) at sun.net.www.http.HttpClient.New(Unknown Source) at sun.net.www.protocol.http.HttpURLConnection.getNewHttpClient(Unknown Source) at sun.net.www.protocol.http.HttpURLConnection.plainConnect(Unknown Source) at sun.net.www.protocol.http.HttpURLConnection.connect(Unknown Source) at sun.net.www.protocol.http.HttpURLConnection.getInputStream(Unknown Source) at java.net.HttpURLConnection.getResponseCode(Unknown Source) at sun.plugin2.applet.Applet2ClassLoader.getBytes(Unknown Source) at sun.plugin2.applet.Applet2ClassLoader.access$000(Unknown Source) at sun.plugin2.applet.Applet2ClassLoader$1.run(Unknown Source) at java.security.AccessController.doPrivileged(Native Method) ... 7 more Exception: java.lang.ClassNotFoundException: examples.pixelcount.PixelCount.class I'm not really sure where exactly is the problem in here, given that they work on appletviewer. any help would be appreciated.. EDIT: don't know if I wrote it clearly. by ".class version" i refer to html file with this content: <applet height="300" width="450" code="examples/pixelcount/PixelCount.class"></applet> and ".jar" with this content: <applet height="300" width="450" archive="PixelCount.jar" code="examples.pixelcount.PixelCount.class"></applet>

    Read the article

  • "java.lang.OutOfMemoryError: Java heap space" in image and array storage

    - by totalconscience
    I am currently working on an image processing demonstration in java (Applet). I am running into the problem where my arrays are too large and I am getting the "java.lang.OutOfMemoryError: Java heap space" error. The algorithm I run creates an NxD float array where: N is the number of pixel in the image and D is the coordinates of each pixel plus the colorspace components of each pixel (usually 1 for grayscale or 3 for RGB). For each iteration of the algorithm it creates one of these NxD float arrays and stores it for later use in a vector, so that the user of the applet may look at the individual steps. My client wants the program to be able to load a 500x500 RGB image and run as the upper bound. There are about 12 to 20 iterations per run so that means I need to be able to store a 12x500x500x5 float in some fashion. Is there a way to process all of this data and, if possible, how? Example of the issue: I am loading a 512 by 512 Grayscale image and even before the first iteration completes I run out of heap space. The line it points me to is: Y.add(new float[N][D]) where Y is a Vector and N and D are described as above. This is the second instance of the code using that line.

    Read the article

  • ApplicationDispatcher exception

    - by JFB
    Whenever I try to redirect to a certain page using this dispatch method that is called from my doGet method, I get the following exception. I have no idea why! account controller servlet protected void dispatch(HttpServletRequest request, HttpServletResponse response, String page) throws javax.servlet.ServletException, java.io.IOException { RequestDispatcher dispatcher = getServletContext() .getRequestDispatcher(page); try { dispatcher.forward(request, response); } catch (java.lang.NullPointerException e) { System.out.println("NullPointerException: attribute expected in view"); } } Error msg java.lang.NullPointerException org.apache.jasper.JasperException: java.lang.NullPointerException org.apache.jasper.servlet.JspServletWrapper.handleJspException(JspServletWrapper.java:502) org.apache.jasper.servlet.JspServletWrapper.service(JspServletWrapper.java:430) org.apache.jasper.servlet.JspServlet.serviceJspFile(JspServlet.java:313) org.apache.jasper.servlet.JspServlet.service(JspServlet.java:260) javax.servlet.http.HttpServlet.service(HttpServlet.java:717) controller.AccountController.dispatch(AccountController.java:91) controller.AccountController.doExecute(AccountController.java:72) controller.AccountController.doGet(AccountController.java:34) javax.servlet.http.HttpServlet.service(HttpServlet.java:617) javax.servlet.http.HttpServlet.service(HttpServlet.java:717) java.lang.NullPointerException org.apache.jsp.content.edit_jsp._jspService(edit_jsp.java:109) org.apache.jasper.runtime.HttpJspBase.service(HttpJspBase.java:70) javax.servlet.http.HttpServlet.service(HttpServlet.java:717) org.apache.jasper.servlet.JspServletWrapper.service(JspServletWrapper.java:388) org.apache.jasper.servlet.JspServlet.serviceJspFile(JspServlet.java:313) org.apache.jasper.servlet.JspServlet.service(JspServlet.java:260) javax.servlet.http.HttpServlet.service(HttpServlet.java:717) controller.AccountController.dispatch(AccountController.java:91) controller.AccountController.doExecute(AccountController.java:72) controller.AccountController.doGet(AccountController.java:34) javax.servlet.http.HttpServlet.service(HttpServlet.java:617) javax.servlet.http.HttpServlet.service(HttpServlet.java:717)

    Read the article

  • R looking for the wrong java version

    - by Veit
    Hi, I installed/uninstalled java jre/jdk now many times and finally installed the older version 1.6.0_17 which is now located at "C:\Program Files\Java\jre6\bin". Now after all if I call 'java -version' within R i can see that R is looking for Java at the old path which is now wrong. The question is: Why is R looking for Java at the wrong path even so the windows path is set correctly? There are no double entrys within the windows path as far as I can see and I restarted R as well as Windows more then once since then. Any Ideas where R takes the wrong path from? On windows shell: $set [..] OS=Windows_NT Path=C:\Program Files\Java\jre6\bin; [..] $ java -version java version "1.6.0_17" Java(TM) SE Runtime Environment (build 1.6.0_17-b04) Java HotSpot(TM) 64-Bit Server VM (build 14.3-b01, mixed mode) within R: $system("java -version") Error: could not open `C:\Program Files (x86)\Java\jre6\lib\i386\jvm.cfg'

    Read the article

< Previous Page | 41 42 43 44 45 46 47 48 49 50 51 52  | Next Page >