Fungovanie tejto stránky je z dôvodu údržby dočasne obmedzené. Ak článok nevyrieši váš problém a chcete položiť otázku, napíšte našej komunite podpory na Twitter @FirefoxSupport alebo Reddit /r/firefox.

Vyhľadajte odpoveď

Vyhnite sa podvodom s podporou. Nikdy vás nebudeme žiadať, aby ste zavolali alebo poslali SMS na telefónne číslo alebo zdieľali osobné informácie. Nahláste prosím podozrivú aktivitu použitím voľby “Nahlásiť zneužitie”.

Ďalšie informácie

The program will not close properly unless I turn off the computer.

  • 1 odpoveď
  • 13 má tento problém
  • 1 zobrazenie
  • Posledná odpoveď od mccreigh

more options

The program will not close properly. If I close it and then try to restart it I get a message saying that Firefox is already running.

The program will not close properly. If I close it and then try to restart it I get a message saying that Firefox is already running.

Všetky odpovede (1)

more options

Even under normal circumstances it takes a few seconds after exiting Firefox before you can launch Firefox again successfully.

But you're right, sometimes a zombie Firefox process lives on, consuming CPU cycles and memory long after all the user interface for it has disappeared. The zombie Firefox process blocks the launch of new ones. It's annoying, but there is a lighter-weight work-around than rebooting. If you key Del while holding down Ctrl and Alt (the old three-finger DOS reboot), either the Windows Task manager will pop up immediately or you can select it from a list of options. In the Windows Task Manager's Processes tab, you should be able to find "firefox.exe" in the Image Name column. (Left-clicking the Image Name column will sort the names alphabetically, and then you can scroll up or down looking for "f".) Left-click on that "firefox.exe" and then left-click the "End Process" box. A warning box should then pop up, asking if you're really sure you want to terminate the process. Answer Yes. Then wait a few seconds for disk activity to die down. Then re-launch Firefox as you would normally.