在PyQt中沒有直接提供左鍵雙擊的判斷方法,需要自己實現,其思路主要如下所示:
- 1、起動一個定時器,判斷在指定的時間之內,點擊次數超過2次,則視為雙擊(其主要思路判斷兩次點擊的時間差在預測的條件以內)
- 2、 起動一個定時器,判斷在指定的時間之內,點擊次數超過2次,另外再獲取鼠標點擊的坐標,如果前后兩次點擊的坐標位置,屬於同一個位置,滿足這兩個條件則判斷為雙擊(其主要思路判斷兩次點擊的時間差在預測的條件以內,且點擊的坐標在預設的坐標之內,允許存在一定的偏差)
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):
# 左鍵按下
if e.buttons() == QtCore.Qt.LeftButton:
QTimer.singleShot(0, lambda: self.judgeClick(e))
# 右鍵按下
elif e.buttons() == QtCore.Qt.RightButton:
self.mouse = "右"
# 中鍵按下
elif e.buttons() == QtCore.Qt.MidButton:
self.mouse = '中'
# 左右鍵同時按下
elif e.buttons() == QtCore.Qt.LeftButton | QtCore.Qt.RightButton:
self.mouse = '左右'
# 左中鍵同時按下
elif e.buttons() == QtCore.Qt.LeftButton | QtCore.Qt.MidButton:
self.mouse = '左中'
# 右中鍵同時按下
elif e.buttons() == QtCore.Qt.MidButton | QtCore.Qt.RightButton:
self.mouse = '右中'
# 左中右鍵同時按下
elif e.buttons() == QtCore.Qt.LeftButton | QtCore.Qt.MidButton | QtCore.Qt.RightButton:
self.mouse = '左中右'
def mouseDoubleClickEvent(self,e):
# 雙擊
self.mouse = "雙擊"
self.isDoubleClick=True
def judgeClick(self,e):
if self.isDoubleClick== False:
self.mouse="左"
else:
self.isDoubleClick=False
self.mouse = "雙擊"
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="雙擊"
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)
