每当我尝试使用ipython和qt作为gui支持时,它就会给我这个:
enter code here $ ipython --gui=qt
Python 2.7.3 (default, Apr 20 2012, 22:39:59)
Type "copyright", "credits" or "license" for more information.
IPython 0.12.1 -- An enhanced Interactive Python.
? -> Introduction and overview of IPython's features.
%quickref -> Quick reference.
help -> Python's own help system.
object? -> Details about 'object', use 'object??' for extra details.
In [1]: Got bus address: "unix:abstract=/tmp/dbus- q 1DAvsew5j,guid=a3ed4bb7c5723eeff9aaed690000006e"
Connected to accessibility bus at: "unix:abstract=/tmp/dbus- q1DAvsew5j,guid=a3ed4bb7c5723eeff9aaed690000006e"
Registered DEC: true
Registered event listener change listener: true
Registered event listener change listener: true
Registered event listener change listener: true
Registered event listener change listener: true
Registered event listener change listener: true
Registered event listener change listener: true
Registered event listener change listener: true
我的问题是:
这是什么以及为什么它会在键盘中断之前停止?
如何在ipython中嵌入qt-console?
答案 0 :(得分:2)
这个问题刚刚开始发生在我身上。我正在运行Ubuntu 12.04。删除QT辅助功能包(qt-at-spi)为我解决了这个问题。
这是从终端运行的命令。
sudo apt-get remove --purge qt-at-spi
答案 1 :(得分:0)
理论上你可以通过设置环境变量QT_ACCESSIBILITY
来完成相同的事情(停止QT AT API消息)而不用清除包即添加
export QT_ACCESSIBILITY=0
到你的shell和/或系统启动(比如〜/ .bashrc)
您可以在此处查看qt辅助功能包的自述文件
的/ usr /共享/ DOC / QT-AT-SPI /自述