如何在Windows Vista上不使用快捷方式的情况下启动IDLE(Python编辑器)?
我正在尝试教科莫多在击中目标时触发 IDLE 正确的击键.我可以在Windows资源管理器位置栏中的开始"菜单中使用快捷方式的确切路径来启动IDLE,因此我希望Komodo也能够使用它.但是,将此路径提供给Komodo会使它说返回1.这似乎是失败的,因为IDLE无法启动.
I'm trying to teach Komodo to fire up IDLE when I hit the right keystrokes. I can use the exact path of the shortcut in start menu in the Windows Explorer location bar to launch IDLE so I was hoping Komodo would be able to use it as well. But, giving this path to Komodo causes it to say that 1 is returned. This appears to be a failure as IDLE doesn't start up.
我认为我会避免使用快捷方式,而只使用确切的路径.我转到开始菜单,找到IDLE的快捷方式,右键单击以查看属性.目标显示为灰色,但显示为"Python 2.5.2". 开始于"设置为"C:\ Python25 \". 打开文件位置"按钮也显示为灰色.
I thought I'd avoid the shortcut and just use the exact path. I go to the start menu, find the shortcut for IDLE, right click to look at the properties. The target is grayed out, but says "Python 2.5.2". The "Start in" is set to, "C:\Python25\". The "Open File Location" button is also grayed out.
我如何找出此快捷方式的真正指向?我尝试在C:\ Python25中都启动python.exe和pythonw.exe,但都没有启动IDLE.
How do I find out where this shortcut is really pointing? I have tried starting python.exe and pythonw.exe both in C:\Python25, but neither starts up IDLE.
在Lib\idlelib\idle.py
中的Python安装目录中有一个名为idle.py
的文件.
There's a file called idle.py
in your Python installation directory in Lib\idlelib\idle.py
.
如果您使用Python运行该文件,则IDLE应该启动.
If you run that file with Python, then IDLE should start.
c:\ Python25 \ pythonw.exe c:\ Python25 \ Lib \ idlelib \ idle.py
c:\Python25\pythonw.exe c:\Python25\Lib\idlelib\idle.py