python - Python - PyQT4 如何检测窗口中任意位置的鼠标点击位置？

Question

我有 1024x768 分辨率窗口，当点击或鼠标悬停时，我想找到 x、y 值。我怎样才能做到这一点？

import sys
from PyQt4 import QtGui, QtCore


class Example(QtGui.QWidget):

    def __init__(self):
        super(Example, self).__init__()

        self.initUI()

    def initUI(self):               

        qbtn = QtGui.QPushButton('Quit', self)
        #qbtn.clicked.connect(QtCore.QCoreApplication.instance().quit)
        qbtn.clicked.connect(self.test)
        qbtn.resize(qbtn.sizeHint())
        qbtn.move(50, 50)       

        self.setGeometry(0, 0, 1024, 768)
        self.setWindowTitle('Quit button')    
        self.setWindowFlags(self.windowFlags() | QtCore.Qt.FramelessWindowHint)
        self.show()

    def test(self):
      print "show the position of mouse cursor in screen resolution: x is ?? , y is ??"

def main():

    app = QtGui.QApplication(sys.argv)
    ex = Example()
    sys.exit(app.exec_())


if __name__ == '__main__':
    main()

Answer 1

import sys
from PyQt5 import QtWidgets, QtGui, QtCore

class Example(QtWidgets.QMainWindow):
    def __init__(self):
        super(Example, self).__init__()
        self.initUI()
    def mousePressEvent(self, QMouseEvent):
        print(QMouseEvent.pos())
    def mouseReleaseEvent(self, QMouseEvent):
        cursor = QtGui.QCursor()
        print(cursor.pos())
    def initUI(self):
        qbtn = QtWidgets.QPushButton('Quit', self)
        qbtn.resize(qbtn.sizeHint())
        qbtn.move(50, 50)
        self.setGeometry(0, 0, 1024, 768)
        self.setWindowTitle('Quit button')
        self.setWindowFlags(self.windowFlags() | QtCore.Qt.FramelessWindowHint)
        self.show()
def main():
    app = QtWidgets.QApplication(sys.argv)
    ex = Example()
    sys.exit(app.exec_())
if __name__ == '__main__':
    main()

输出：

PyQt4.QtCore.QPoint(242, 285)
PyQt4.QtCore.QPoint(1741, 423)
PyQt4.QtCore.QPoint(439, 372)

Answer 2

获取 (x,y) 坐标的另一种方法：

def mouseReleaseEvent(self, QMouseEvent):
    print('(', QMouseEvent.x(), ', ', QMouseEvent.y(), ')')