问题描述
我想在Windows下将几个文件扩展名链接到我的java应用程序中。当用户双击带有链接扩展名的文件时，我想打开我的应用程序，我需要知道启动应用程序的文件路径。
I would like to "link" few file extensions to my java application under windows. When user double clicks file with "linked" extension, I would like to open my app and I need to know path to file that launched app.
推荐答案
如果部署应用程序。使用，可以在启动文件中声明对文件类型的兴趣。请参阅，...
If you deploy the app. using Java Web Start, an interest in file-types can be declared in the launch file. See the demo. of the file services, which..
当用户双击 .zzz 文件时，它应该是在应用程序中打开。实际上，提示这个词并不是整个故事。如果启动沙盒版本，系统将提示您关联文件类型。受信任的版本不会提示。
When the user double clicks a .zzz file, it should open in the app. Actually, the word 'prompts' there is not the whole story. If you launch the sand-boxed version you will be prompted as to associating the file-type. The trusted version does not prompt.
要向流程添加更多用户控制，请查看在1.6.0_18（我还没有那个演示。）在与用户核实后，您可以在启动时运行它。
To add more user-control to the process, look to the IntegrationService that was introduced in 1.6.0_18 (I don't have a demo. of that one yet). You might run it at start-up, after checking with the user.
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