运行时错误而关闭的小程序 [英] run time error while closing the applet

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问题描述

我想知道如何读这些错误;请人帮助我。将显示以下错误关闭一个小程序后,

  C:\\用户\\拉达摹\\桌面\\实践​​与GT;的appletviewer NewApplet.html
在异常线程AWT-EventQueue的-1java.security.AccessControlException:ACC
ESS否认(java.lang.RuntimePermission exitVM)
        在java.security.AccessControlContext.checkPermission(AccessControlConte
xt.java:264)
        在java.security.AccessController.checkPermission(AccessController.java:
427)
        在java.lang.SecurityManager.checkPermission(SecurityManager.java:532)
        在java.lang.SecurityManager.checkExit(SecurityManager.java:744)
        在java.lang.Runtime.exit(Runtime.java:88)
        在java.lang.System.exit(System.java:869)
        在SimpleApplet.actionPerformed(SimpleApplet.java:50)
        在java.awt.Button.processActionEvent(Button.java:388)
        在java.awt.Button.processEvent(Button.java:356)
        在java.awt.Component.dispatchEventImpl(Component.java:3984)
        在java.awt.Component.dispatchEvent(Component.java:3819)
        在java.awt.EventQueue.dispatchEvent(EventQueue.java:463)
        在java.awt.EventDispatchThread.pumpOneEventForHierarchy(EventDispatchTh
read.java:242)
        在java.awt.EventDispatchThread.pumpEventsForHierarchy(EventDispatchThre
ad.java:163)
        在java.awt.EventDispatchThread.pumpEvents(EventDispatchThread.java:157)        在java.awt.EventDispatchThread.pumpEvents(EventDispatchThread.java:149)        在java.awt.EventDispatchThread.run(EventDispatchThread.java:110)


解决方案

从您的堆栈跟踪:

 在java.lang.System.exit(System.java:869)LT; ----拿去!
    在SimpleApplet.actionPerformed(SimpleApplet.java:50)

您不能调用 System.exit()从applet。

请参阅如何通过code关闭Java小程序对于如何可以关闭你的小应用程序的正确方法的建议。

i want to know how to read these errors; please anyone help me out. The following error is displayed after closing an applet

C:\Users\Radha G\Desktop\practice>appletviewer NewApplet.html
Exception in thread "AWT-EventQueue-1" java.security.AccessControlException: acc
ess denied (java.lang.RuntimePermission exitVM)
        at java.security.AccessControlContext.checkPermission(AccessControlConte
xt.java:264)
        at java.security.AccessController.checkPermission(AccessController.java:
427)
        at java.lang.SecurityManager.checkPermission(SecurityManager.java:532)
        at java.lang.SecurityManager.checkExit(SecurityManager.java:744)
        at java.lang.Runtime.exit(Runtime.java:88)
        at java.lang.System.exit(System.java:869)
        at SimpleApplet.actionPerformed(SimpleApplet.java:50)
        at java.awt.Button.processActionEvent(Button.java:388)
        at java.awt.Button.processEvent(Button.java:356)
        at java.awt.Component.dispatchEventImpl(Component.java:3984)
        at java.awt.Component.dispatchEvent(Component.java:3819)
        at java.awt.EventQueue.dispatchEvent(EventQueue.java:463)
        at java.awt.EventDispatchThread.pumpOneEventForHierarchy(EventDispatchTh
read.java:242)
        at java.awt.EventDispatchThread.pumpEventsForHierarchy(EventDispatchThre
ad.java:163)
        at java.awt.EventDispatchThread.pumpEvents(EventDispatchThread.java:157)

        at java.awt.EventDispatchThread.pumpEvents(EventDispatchThread.java:149)

        at java.awt.EventDispatchThread.run(EventDispatchThread.java:110)

解决方案

From your stack trace:

    at java.lang.System.exit(System.java:869)  <---- nooo!
    at SimpleApplet.actionPerformed(SimpleApplet.java:50)

You cannot call System.exit() from an applet.

See How to close java Applet through code for a suggestion of how you can close your applet the correct way.

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