对此错误的任何修复---> TypeError: 'StringVar' object is not callable

Question

我正在尝试创建这个 GUI 计算器，但似乎“StringVar”对象不可调用。我的其他运算符（例如 -、*、/、= 等）仍然存在此问题。（我刚接触编码和 python 世界），

从 tkinter 导入 Tk、Label、StringVar、Entry、Button、W、E

类计算器： 操作数：StringVar 运营商：无 结果：浮动

def __init__(self):
    master = Tk()        
    self.operator = None
    self.operand = StringVar()
    self.result = 0
    
         
    Label(master,text = "Calculator").grid(row = 0, column=0, columnspan=2)
    
    UserInputEntry = master.register(self.operandValid)
    Entry(master, validate="key", validatecommand=(UserInputEntry,'%P'), textvariable=self.operand).grid(row=0, column=0, columnspan=3)
                                     
    addAction = lambda: self.operand(lambda a, b: a+b)
    Button(master,text = "+",command=addAction).grid(row=1, column=3, sticky=W+E)
    
    subAction = lambda: self.operand(lambda a, b: a-b)
    Button(master,text = "-",command=subAction).grid(row=2, column=3, sticky=W+E)
    
    mulAction = lambda: self.operand(lambda a, b: a*b)
    Button(master,text = "X",command=mulAction).grid(row=3, column=3, sticky=W+E)
    
    divAction = lambda: self.operand(lambda a, b: a/b)
    Button(master,text = "/",command=divAction).grid(row=4, column=3, sticky=W+E)
    
    equal = lambda: self.operand(None)
    Button(master,text = "=",command=equal).grid(row=4, column=2, sticky=W+E)       
    
    enter1 = lambda: self.enterNumber("1")
    Button(master,text = "1",command=enter1).grid(row=1, column=0, sticky=W+E)
    
    
    enter2 = lambda: self.enterNumber("2")
    Button(master,text = "2",command=enter2).grid(row=1, column=1, sticky=W+E)
    
    enter3 = lambda: self.enterNumber("3")
    Button(master,text = "3",command=enter3).grid(row=1, column=2, sticky=W+E)
    
    
    enter4 = lambda: self.enterNumber("4")
    Button(master,text = "4",command=enter4).grid(row=2, column=0, sticky=W+E)
    
    enter5 = lambda: self.enterNumber("5")
    Button(master,text = "5",command=enter5).grid(row=2, column=1, sticky=W+E)
    
    enter6 = lambda: self.enterNumber("6")
    Button(master,text = "6",command=enter6).grid(row=2, column=2, sticky=W+E)
    
    enter7 = lambda: self.enterNumber("7")
    Button(master,text = "7",command=enter7).grid(row=3, column=0, sticky=W+E)
    
    enter8 = lambda: self.enterNumber("8")
    Button(master,text = "8",command=enter8).grid(row=3, column=1, sticky=W+E)
    
    enter9 = lambda: self.enterNumber("9")
    Button(master,text = "9",command=enter9).grid(row=3, column=2, sticky=W+E)
    
    enter0 = lambda: self.enterNumber("0")
    Button(master,text = "0", command=enter0).grid(row=4, column=0, sticky=W+E)
    
    enterDecimal = lambda: self.enterNumber(".")
    Button(master,text=".",command=enterDecimal).grid(row=4, column=1, sticky=W+E)
    
    master.mainloop()
    
    
def operation(self, op):
    print("The previous operation was {self.operation}")
    self.result = self.operator(self.result, float(self.operand))
    self.operator = op

def operandValid(self, operand):
    try: 
        operand = float(operand)
        return True
    except:
        return False
def enterNumber(self,character): 
            
    #if self.resultDisplayed: 
        #self.resultDisplayed = False
        #self.operand.set("")#reset the Entry for new input
    
    if(self.operandValid(self.operand.get()+character)):
        self.operand.set(self.operand.get()+character)

计算器()

addAction = lambda: self.operand(lambda a, b: a+b) TypeError: 'StringVar' 对象不可调用

Answer 1

那是因为您正在执行与StringVar()(lambda a, b: a+b) 等效的操作。您应该将 lambda 留在 self.operand 中，并保留其他所有内容。