为什么我的elif语句不会执行，但我的if语句将在Python中执行？

Question

I wrote a simple age verification in python to where if the user enters a year less than 2000, it will say welcome. However, if the user enters a year greater than 2000, it should redirect them to another site.

My code works but only executes only the if statement not the elif no matter what year I input.

Here is my code:

from Tkinter import *
import webbrowser
import tkMessageBox


url = 'google.com'
root = Tk()
frame = Frame(root, width=100, height=100)
frame.pack()
L1 = Label(root, text = "Month")
L1.pack(side = LEFT)
E1 = Entry(root,bd=5)
E1.pack(side=LEFT)
L2 = Label(root, text = "Day")
L2.pack(side =LEFT)
E2 = Entry(root, bd= 5)
E2.pack(side = LEFT)
L3 = Label(root, text = "Year")
L3.pack(side = LEFT)
E3 = Entry(root, bd = 5)
E3.pack(side = LEFT)

def getdate():
    tkMessageBox.showinfo(title ="Results", message = E1.get() + " "+ E2.get() + " " + E3.get())
    getage()
    root.destroy()
    #tkMessageBox.showinfo(E2.get())
    #tkMessageBox.showinfo(E3.get())
def getage():

    if E3 < 2000:

        tkMessageBox.showinfo(message= "Welcome! ")
    elif E3 > 2000:
        tkMessageBox.showinfo(message="You will be redirected")
        webbrowser.open_new(url)


b1 = Button(root, text = "Submit", width = 25, command = getdate)
b1.pack()


root.mainloop()

Answer 1

You are comparing the Entry() object to an integer. In Python 2, numbers are always sorted before other types, so they always register as smaller :

>>> object() > 2000
True

You want to get the value from the entry box and convert to an integer before testing:

entry = int(E3.get())
if entry > 2000:
     # ...
else:
     # ...

There is no need to do elif here, unless you want the value 2000 to be ignored altogether (your test only works for numbers either greater than, or smaller than, not equal).

Answer 2

Entry is the object type. You want to call the get() function. Like this:

EntryBox = Entry(root, bd = 5)
E3 = int(EntryBox.get())

Also, note that the elif is completely unnecessary. Just use else instead.

if E3 < 2000:
    tkMessageBox.showinfo(message= "Welcome! ")
else:
    tkMessageBox.showinfo(message="You will be redirected")
    webbrowser.open_new(url)

Answer 3

You compare an Entry object to an integer ,the if statement will never be reached because integers are always sorted before objects . Look:

>>> a = ['0', 9999, {}, [], False, (), object()]
>>> sorted(a)
[False, 9999, {}, [], <object object at 0xb74fa558>, '0', ()]
>>> 

Use get method to get text from Entry then convert it with int :

if int(E3.get()) < 2000:

You can use IntVar() if you want to get integer from Entry without converting:

...
v = IntVar()
E3 = Entry(root, bd = 5, textvariable=v)
...
if E3.get() < 2000:
    tkMessageBox.showinfo(message= "Welcome! ")
else:
    tkMessageBox.showinfo(message="You will be redirected")
    webbrowser.open_new(url)