[PyQt] Call a window from other parent window

Michael Towers gradgrind at online.de
Wed Oct 17 08:44:51 BST 2007


A couple of (also untested) tweaks?

Mark Summerfield wrote:
> On 2007-10-16, Sergio Jovani Guzmán wrote:
>   
>> Hi all!
>>
>> I am learning PyQt and I have a question.
>>
>> I have two classes, one for each widget:
>>
>> ==================
>> class Main(QMainWindow):
>> 	def __init__(self):
>> 		QMainWindow.__init__(self)
>> 		loadUi("frmmain.ui", self)
>>
>> class Conectar(QDialog):
>>
>> 	def __init__(self):
>> 		QDialog.__init__(self)
>> 		loadUi("frmconectar.ui", self)
>> =====================
>>
>> My question is, how can I call and show a widget from other with parent?
>>
>> For example, open Conectar() window from Main() window?
>>
>> Many thanks!
>>     
>
> You need a method in your Main() class, something like this (untested):
>
>     def runConectar(self):
> 	conectar = Conectar(self)
> 	if conectar.exec():
>   

if conectar.exec_():

> 	    pass # User clicked OK
>
> And in the Main() __init__ method, assuming you have an action called
> runConectarAction:
>
>     self.connect(runConectarAction, SIGNAL("triggered()"),
> 		 self.runConectar)
>
>   

self.connect(runConectarAction, SIGNAL("triggered(bool)"),
                      self.runConectar)





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