我从一个小应用程序中拿出勇气来证明这种行为。在osx下,这是我所期望的工作方式:当你按下按钮时,它会打印“被要求开始工作”,然后出现一个消息框,事情会暂停,直到按下OK按钮,然后它开始打印“yielding”,GUI保持适度的活动。在
import wx
import time
class MainWindow(wx.Frame):
def __init__(self, parent, id, title):
wx.Frame.__init__(self, parent, id, title, size=(800, 700))
self.tabbed = wx.Notebook(self, -1, style=(wx.NB_TOP))
self.running = RunningPane(self.tabbed)
self.submissions = SubmissionPane(self.tabbed, self.running)
self.tabbed.AddPage(self.submissions, "Submit Job")
self.tabbed.AddPage(self.running, "Running Jobs")
self.Show()
class SubmissionPane(wx.Panel):
def __init__(self, parent, run_pane):
wx.Panel.__init__(self, parent, -1)
self.run_pane = run_pane
self.buttonGo = wx.Button(self, -1, "Submit", pos=(290,170))
self.buttonGo.Bind(wx.EVT_BUTTON, self.OnSubmit)
self.Show()
def OnSubmit(self, event):
self.run_pane.StartWork()
print "requested work start"
wx.MessageBox('Job Submitted')
print "displayed message box"
class RunningPane(wx.Panel):
def __init__(self, parent):
wx.Panel.__init__(self, parent, -1)
self.running_log = wx.TextCtrl(self, -1, pos=(35, 210), size=(720,400))
self.Show()
def StartWork(self):
print "Asked to Start Work..."
wx.CallAfter(self.DoTheWork)
print "registered the CallAfter"
def DoTheWork(self):
print "Actually starting work"
self.running_log.WriteText("doing..."+"\n")
for i in range(20):
print "yielding"
wx.Yield()
time.sleep(1)
print "I pretended to do the work :) "
app = wx.App()
MainWindow(None, -1, 'Application')
app.MainLoop()
但是,在Windows下,直到打印了20个yielding(按下按钮后立即开始),并且GUI在此期间没有响应,对话框才会出现。在
我是否误解了对wx.Yield()
的期望?在
您必须记住,wxPython在每个操作系统上都包装了本机小部件。如果本机widget在Windows上的行为与在Linux或Mac上不同,那么对于该操作系统来说,这可能是正常的。我猜这就是这里发生的事情。一个简单的解决方法是切换到wx.message对话框你可以用情态表现出来。所以换掉你创建wx.MessageBox有了这两个:
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