我正在尝试将鼠标按下事件链接到在QLabel中单击时显示鼠标的坐标。一些问题..。当我传递一个通用的QWidget.mousePressEvent时,坐标只会在第一次单击时显示。当我试图使鼠标事件特定于一个GraphicsScene(self.p1)时,我得到了以下错误:
Traceback (most recent call last):
File "C:\Users\Tory\Desktop\DIDSONGUIDONOTCHANGE.py", line 59, in mousePressEvent
self.p1.mousePressEvent(event)
TypeError: QGraphicsWidget.mousePressEvent(QGraphicsSceneMouseEvent): argument 1 has unexpected type 'QMouseEvent'这是我使用的代码..。我知道这是错误的,但我是个新手,不知道从哪里开始。
def mousePressEvent(self, event):
self.p1.mousePressEvent(event)
x=event.x()
y=event.y()
if event.button()==Qt.LeftButton:
self.label.setText("x=%0.01f,y=%0.01f" %(x,y))如何让鼠标点击显示图形场景self.p1的坐标?
发布于 2012-08-25 06:21:03
看起来event filter可以做你想做的事情。
下面是一个简单的演示:
from PyQt4 import QtGui, QtCore
class Window(QtGui.QWidget):
def __init__(self):
QtGui.QWidget.__init__(self)
self.scene = QtGui.QGraphicsScene(self)
self.scene.addPixmap(QtGui.QPixmap('image.jpg'))
self.scene.installEventFilter(self)
self.view = QtGui.QGraphicsView(self)
self.view.setScene(self.scene)
self.label = QtGui.QLabel(self)
layout = QtGui.QVBoxLayout(self)
layout.addWidget(self.view)
layout.addWidget(self.label)
def eventFilter(self, source, event):
if (source is self.scene and
event.type() == QtCore.QEvent.GraphicsSceneMouseRelease and
event.button() == QtCore.Qt.LeftButton):
pos = event.scenePos()
self.label.setText('x=%0.01f,y=%0.01f' % (pos.x(), pos.y()))
return QtGui.QWidget.eventFilter(self, source, event)
if __name__ == '__main__':
import sys
app = QtGui.QApplication(sys.argv)
window = Window()
window.show()
sys.exit(app.exec_())https://stackoverflow.com/questions/12116073
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