如何使用python单击窗口的“下一步"按钮 [英] How to click a 'next' button of a window using python

问题描述

我使用下面的代码连接一个打开的窗口(类类型是 SunAwtFrame)，确实我能够连接到它.但无法单击它的下一步按钮，我怀疑下一步"按钮可能位于该窗口的框架内.即使我使用swapy"工具，我也无法浏览控件，实际上并没有显示控件.那么，如果是这种情况，如何切换到框架并单击下一步"按钮.

I used the below code to connect a opened window(Class type is SunAwtFrame), indeed I able to connect to it. but unable to click next button of it, my doubt is that 'Next' button may reside inside the frame of that window. Even when I use 'swapy' tool, I am unable to navigate through the controls, indeed not showing the controls actually. So, how to switch to frame if it is that case and click the 'Next' button.

app2 = application.Application()

app2.connect(title_re = u'abc')

dialog = app2.abc

print dialog

next =dialog.Next

print next

next.Click()

当我运行上面的代码时，我得到了这样的错误，请帮我解决这个问题

When I ran the above code I got the error like, Please help me to over come over this

#Error#:-

  <pywinauto.application.WindowSpecification object at 0x025F26F0>
  <pywinauto.application.WindowSpecification object at 0x025FA3B0>
  next.Click()
  File "C:\Python27\lib\site-packages\pywinauto\application.py", line 229, in __getattr__
    ctrls = _resolve_control(self.criteria)
  File "C:\Python27\lib\site-packages\pywinauto\application.py", line 788, in _resolve_contro
l
    raise e.original_exception
pywinauto.findwindows.WindowNotFoundError

推荐答案

您似乎一直在尝试通过 pywinauto 自动化非标准控件.我建议你按坐标点击，当然如果除了按下按钮没有其他任务.

It seems like you have been trying automate non standard controls by pywinauto. I am recommending you make a click by coordinates, of course if there are no other tasks except press the button.

dialog.Click(coords=(x, y))

dialog.Click(coords=(x, y))

或

dialog.ClickInput(coords=(x, y))

dialog.ClickInput(coords=(x, y))

这篇关于如何使用python单击窗口的“下一步"按钮的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！