Python Pyqt 如何获取小部件的尺寸
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Pyqt how to get a widget's dimensions
提问by ibi0tux
I am currently developing an application in which i cannot use modal windows (due to some application constraints). However, in some cases i would like to simulate a popup window. To do so i dynamically create a widget that has the centralwidget as parent and i use the move() method to place it where i want. I would like to know if there is a way to get a widget's dimensions at a given time (considering the mainWindow can be resized at any time) so that i will be able to center the placeholder popup (a simple widget) at the middle of the centralwidget.
我目前正在开发一个无法使用模态窗口的应用程序（由于某些应用程序限制）。但是，在某些情况下，我想模拟一个弹出窗口。为此，我动态创建了一个以 centralwidget 作为父级的小部件，并使用 move() 方法将其放置在我想要的位置。我想知道是否有办法在给定时间获取小部件的尺寸（考虑到 mainWindow 可以随时调整大小），以便我能够将占位符弹出窗口（一个简单的小部件）居中中央小部件。
Thank you
谢谢
采纳答案by Avaris
You can use frameGeometryor geometrydepending on your needs.
您可以使用frameGeometry或geometry取决于您的需要。
回答by subin george
For gettting screen size
用于获取屏幕尺寸
import sys
from PyQt4 import QtGui, QtCore
app = QtGui.QApplication(sys.argv)
mainWindow = QtGui.QWidget()
screenShape = QtGui.QDesktopWidget().screenGeometry()
mainWindow.resize(self.screenShape.width(), self.screenShape.height())
mainWindow.show()
回答by subin george
For getting Qt Widget size:
获取 Qt Widget 大小：
import sys
from PyQt4 import QtGui, QtCore
app = QtGui.QApplication(sys.argv)
mainWindow = QtGui.QWidget()
width = mainWindow.frameGeometry().width()
height = mainWindow.frameGeometry().height()
