Python使用If，Els根据输入切换显示

Question

我想使用IF/Else或Switch根据输入值显示打印文本。并让我知道如何使用下面的代码开关情况

 # OnButtonOK after clicking it, display the input value    
  def OnButtonOK(self):
    Input = self.entrytext.get()
    # self.text.insert(END, Input + '\n')
    # self.scroll.config(Input = self.text.yview)
    print Input
    useroption = atoi(Input)
    # self.OnButtonClick();
    if (useroption == 1):
            print "input is output"
        self.SubMenu1();
    else:
        print "Error:Invalid"

    return;

def SubMenu1(self):
        print 'SubMenu1'
    return;

def SubMenu2(self):
        print 'SubMenu2'
    return;

def SubMenu3(self):
        print 'SubMenu3'
    return;

我只能打印其他部分：

if (useroption == 1):
            print "input is output"
        self.SubMenu1();
    else:
        print "Error:Invalid"

让我知道我到底错在哪里

Answer 1

这是一个简单的初学者的错误，你正在缩进它：

if (useroption == 1):
        print "input is output"
    self.SubMenu1();
else:
    print "Error:Invalid"

应该是

if (useroption == 1):
    print "input is output" # You had an indent too many here
    self.SubMenu1();
else:
    print "Error:Invalid"

Python对缩进敏感；缩进太多或太少都会破坏代码