模拟PyQt中的鼠标释放

Question

我该如何使用Qt.SIGNAL来模拟鼠标释放的动作？我需要在没有用户操作的情况下模拟一次鼠标释放。谢谢！

Answer 1

你可以使用：

from PyQt4.QtTest import QTest

#(...) Where you want to release
QTest.mouseRelease(widget_to_release, Qt.LeftButton)

这样做会把鼠标释放到小部件的中心位置。

还有一些其他的方法，比如 mousePress()、mouseClick() 等等。不过，如果你在Windows上测试拖放功能，要注意，QTest.mousePress() 会被阻塞，因为 QDrag.exec_() 也会阻塞。

Answer 2

这里有一个例子，展示了如何使用clicked信号，这个信号来自于QPushButton按钮：

#!/usr/bin/env python
#-*- coding:utf-8 -*-

from PyQt4 import QtGui, QtCore

class MyWindow(QtGui.QWidget):
    def __init__(self, parent=None):
        super(MyWindow, self).__init__(parent)

        self.pushButtonSimulate = QtGui.QPushButton(self)
        self.pushButtonSimulate.setText("Simulate Mouse Release!")
        self.pushButtonSimulate.clicked.connect(self.on_pushButtonSimulate_clicked)

        self.layoutHorizontal = QtGui.QHBoxLayout(self)
        self.layoutHorizontal.addWidget(self.pushButtonSimulate)

    @QtCore.pyqtSlot()
    def on_pushButtonSimulate_clicked(self):
        mouseReleaseEvent = QtGui.QMouseEvent(
            QtCore.QEvent.MouseButtonRelease,
            self.cursor().pos(),
            QtCore.Qt.LeftButton,
            QtCore.Qt.LeftButton,
            QtCore.Qt.NoModifier,
        )

        QtCore.QCoreApplication.postEvent(self, mouseReleaseEvent)

    def mouseReleaseEvent(self, event):
        if event.button() == QtCore.Qt.LeftButton:
            print "Mouse Release"

        super(MyWindow, self).mouseReleaseEvent(event)

if __name__ == "__main__":
    import sys

    app = QtGui.QApplication(sys.argv)
    app.setApplicationName('MyWindow')

    main = MyWindow()
    main.show()

    sys.exit(app.exec_())

模拟PyQt中的鼠标释放

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