在PyQt中没有直接提供左键双击的判断方法,需要自己实现,其思路主要如下所示:
- 1、起动一个定时器,判断在指定的时间之内,点击次数超过2次,则视为双击(其主要思路判断两次点击的时间差在预测的条件以内)
- 2、 起动一个定时器,判断在指定的时间之内,点击次数超过2次,另外再获取鼠标点击的坐标,如果前后两次点击的坐标位置,属于同一个位置,满足这两个条件则判断为双击(其主要思路判断两次点击的时间差在预测的条件以内,且点击的坐标在预设的坐标之内,允许存在一定的偏差)
from PyQt5.QtCore import QTimer
from PyQt5 import QtCore, QtGui, QtWidgets
class myWidgets(QtWidgets.QTableWidget):
def __init__(self, parent=None):
super(myWidgets, self).__init__(parent)
self.isDoubleClick = False
self.mouse = ""
def mousePressEvent(self, e):
# 左键按下
if e.buttons() == QtCore.Qt.LeftButton:
QTimer.singleShot(0, lambda: self.judgeClick(e))
# 右键按下
elif e.buttons() == QtCore.Qt.RightButton:
self.mouse = "右"
# 中键按下
elif e.buttons() == QtCore.Qt.MidButton:
self.mouse = '中'
# 左右键同时按下
elif e.buttons() == QtCore.Qt.LeftButton | QtCore.Qt.RightButton:
self.mouse = '左右'
# 左中键同时按下
elif e.buttons() == QtCore.Qt.LeftButton | QtCore.Qt.MidButton:
self.mouse = '左中'
# 右中键同时按下
elif e.buttons() == QtCore.Qt.MidButton | QtCore.Qt.RightButton:
self.mouse = '右中'
# 左中右键同时按下
elif e.buttons() == QtCore.Qt.LeftButton | QtCore.Qt.MidButton | QtCore.Qt.RightButton:
self.mouse = '左中右'
def mouseDoubleClickEvent(self,e):
# 双击
self.mouse = "双击"
self.isDoubleClick=True
def judgeClick(self,e):
if self.isDoubleClick== False:
self.mouse="左"
else:
self.isDoubleClick=False
self.mouse = "双击"
或
from PyQt5.QtCore import QTimer
from PyQt5 import QtCore, QtGui, QtWidgets
class myWidgets(QtWidgets.QTableWidget):
def __init__(self, parent=None):
super(myWidgets, self).__init__(parent)
self.mouse = ""
self.timer=QTimer(self)
self.timer.timeout.connect(self.singleClicked)
def singleClicked(self):
if self.timer.isActive():
self.timer.stop()
self.mouse="左"
def mouseDoubleClickEvent(self,e):
if self.timer.isActive() and e.buttons() ==QtCore.Qt.LeftButton:
self.timer.stop()
self.mouse="双击"
super(myWidgets,self).mouseDoubleClickEvent(e)
def mousePressEvent(self,e):
if e.buttons()== QtCore.Qt.LeftButton:
self.timer.start(1000)
elif e.buttons()== QtCore.Qt.RightButton:
self.mouse="右"
super(myWidgets,self).mousePressEvent(e)