Question

我有一个应用程序，它会生成一个包含进度条的弹出窗口。与弹出窗口相关的代码如下。我想强制这个弹出窗口保持在生成此窗口的应用程序的顶部。我尝试使用wx.STAY_ON_TOP，但使用此样式会强制弹出窗口保持在所有应用程序之上，这不是我想要的。任何建议/意见将不胜感激！

class ProgressBar(wx.Frame):
    def __init__(self, parent):
        wx.Frame.__init__(self, None, title="In progress...", 
                      size=(300, 125))        
        GridBagSizer = wx.GridBagSizer()
        TextFont = wx.Font(pointSize = 9, family = wx.SWISS, style = wx.NORMAL, weight = wx.NORMAL, faceName = 'Tahoma')

        self.SetBackgroundColour('white')
        self.gauge = wx.Gauge(self, range = 100, size = (-1, 30), style =  wx.GA_HORIZONTAL, name = 'In Progress') 
        self.gauge.SetValue(0)
        GridBagSizer.Add(self.gauge, pos = (0, 0), span = (1, 1), flag = wx.EXPAND|wx.ALL, border = 15)

        self.txt = wx.StaticText(self, label = 'Retrieving data...', style = wx.ALIGN_CENTER)
        self.txt.SetFont(TextFont)
        box = wx.BoxSizer(wx.HORIZONTAL)
        box.Add(self.txt, 0, wx.CENTER)
        GridBagSizer.Add(box, pos = (1, 0), span = (1, 1), 
                     flag = wx.EXPAND|wx.LEFT|wx.RIGHT|wx.BOTTOM, 
                     border = 15)

        GridBagSizer.AddGrowableCol(0)
        #GridBagSizer.AddGrowableCol(1)

        self.SetSizer(GridBagSizer)
        self.SetMinSize((300, 125))
        self.SetMaxSize((300, 125))
        self.Layout()
        self.Center()

    def Update(self, step):        
        self.gauge.SetValue(step)
        if step == 100:
            self.Close()

    def SetLabel(self, label):
        self.txt.SetLabel(label)
        self.Refresh()
        self.Layout()

Answer 1

使用样式wx.FRAME_FLOAT_ON_PARENT

class ProgressBar(wx.Frame):
    def __init__(self, parent): # you must set the parent ...
        wx.Frame.__init__(self, parent, title="In progress...", 
                      size=(300, 125),style=wx.FRAME_FLOAT_ON_PARENT) 
        ...

如果你想让它拥有你可以做的默认框架内容

style=wx.DEFAULT_FRAME_STYLE|wx.FRAME_FLOAT_ON_PARENT

wxpython：如何使弹出窗口保持在当前应用程序之上

1 个答案: