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I need to use Java 1.6.22 - need some pointers

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  • 1 有這個問題
  • 3 次檢視
  • 最近回覆由 guigs

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I have an application (an Oracle application) that relies on Java 1.6.10 - 23. I have 1.6.22 installed and, as I'm now aware, FIrefox automatically disables Java 1.6 lower than .31. My application will not run with .31. So I did some research and came up with this: https://support.mozilla.org/en-US/kb/how-allow-java-trusted-sites?esab=a&as=aaq and attempted to follow the instructions. The problem is at step one where it shows activate Java once and then either allow or always allow. My problem is that I don't get the option to run it once; I get the allow/always allow prompt but it still won't open Java or give me the Java box to run the program. Is there any way to get Firefox 31 to allow Java 1.6 lower than .31? If not what prior version (if any) will run 1.6.22?

Any help is appreciated.

I have an application (an Oracle application) that relies on Java 1.6.10 - 23. I have 1.6.22 installed and, as I'm now aware, FIrefox automatically disables Java 1.6 lower than .31. My application will not run with .31. So I did some research and came up with this: https://support.mozilla.org/en-US/kb/how-allow-java-trusted-sites?esab=a&as=aaq and attempted to follow the instructions. The problem is at step one where it shows activate Java once and then either allow or always allow. My problem is that I don't get the option to run it once; I get the allow/always allow prompt but it still won't open Java or give me the Java box to run the program. Is there any way to get Firefox 31 to allow Java 1.6 lower than .31? If not what prior version (if any) will run 1.6.22? Any help is appreciated.

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Unfortunately for the app this was started https://blog.mozilla.org/addons/2012/.../blocking-java/, but the security risks being exploited was an improvement for security in Firefox overall.

Lucky the update shows how to delete the blocklist https://blog.mozilla.org/addons/2012/.../update-on-java-blocklist/

Please let me know if this still does not work.