我有一个pyqt小部件,启动时会在所有窗口上显示。我需要它保持关闭状态,直到用户决定为止。无论在哪里进行鼠标单击,每次执行鼠标单击事件时,都可能捕获到鼠标释放事件:在QtWidget窗口内部还是外部?
这里是我使用的样本:
class Release_check(QWidget):
def __init__(self):
super().__init__()
self.initUI()
def initUI(self):
self.canvas = iface.mapCanvas()
self.setWindowFlags(self.windowFlags() | QtCore.Qt.WindowStaysOnTopHint)
self.grid = QGridLayout()
self.grid.setSpacing(10)
self.setGeometry(500, 500, 400, 100)
self.text_out = QTextEdit()
self.setLayout(self.grid)
self.grid.addWidget(self.text_out, 0, 1, 1, 2)
self.show()
def mouseReleaseEvent(self, e):
screen_coordinate = f"x:{e.x()}, y:{e.y()}"
self.text_out.setText(screen_coordinate)
super(Release_check, self).mouseReleaseEvent(e)
app = Release_check()
Qt仅检测窗口小部件内部的单击,如果要检测窗口小部件外部,则必须使用另一个使用OS资源来监视OS事件的库,例如pyinput:
import sys
from pynput import mouse
from PyQt5 import QtCore, QtGui, QtWidgets
class ButtonReleaseManager(QtCore.QObject):
released = QtCore.pyqtSignal(int, int)
def __init__(self, parent=None):
super().__init__(parent)
self._listener = mouse.Listener(on_click=self._handle_click)
self._listener.start()
def _handle_click(self, x, y, button, pressed):
if not pressed:
self.released.emit(x, y)
class Release_check(QtWidgets.QWidget):
def __init__(self):
super().__init__()
self.initUI()
self.manager = ButtonReleaseManager()
self.manager.released.connect(self.show_position)
def initUI(self):
self.setWindowFlags(self.windowFlags() | QtCore.Qt.WindowStaysOnTopHint)
self.setGeometry(500, 500, 400, 100)
self.text_out = QtWidgets.QTextEdit()
grid = QtWidgets.QGridLayout(self)
grid.setSpacing(10)
grid.addWidget(self.text_out, 0, 1, 1, 2)
@QtCore.pyqtSlot(int, int)
def show_position(self, x, y):
screen_coordinate = f"x:{x}, y:{x}"
self.text_out.setText(screen_coordinate)
if __name__ == "__main__":
app = QtWidgets.QApplication(sys.argv)
w = Release_check()
w.show()
sys.exit(app.exec_())