[Tutor] Interactive Menu Woes
Evert Rol
evert.rol at gmail.com
Wed Nov 14 13:28:45 CET 2007
> I am trying to build a menu for the following script to make it
> more "user friendly". Nothing fancy, just a simple add data and
> look up the entered results.
>
> The problem is that when I run my modified version with this
> snippet (please see attachment for original and my modified versions):
>
> [code]
> class MenuInput:
>
> # ask user to input data and store to be passed to manageable
> variables.
> def addName(self):
> print "To add a name, please input he following
> information: "
> self.enter = phoneentry()
> self.name = raw_input("Name: ")
> self.number = raw_input("Number: ")
> self.get = int(raw_input("What type of number is
> this? (choose one): \n 1. Home:\n 2. Work:\n 3. Cell:\n : "))
> def numberType(self, get = 1):
> if self.get == 1:
> self.returnType = HOME
> #return self.getType
> elif self.gete == 2:
> self.returnType = WORK
> #return self.getType
> elif self.get == 3:
> self.returnType = FAX
> #return self.getType
> return self.returnType
>
> self.type = numberType(self.get)
> self.enter(self.name, self.number, self.returnType)
>
Bypassing the whole 'function within function' problem, consider
using a dictionary instead: somedict = {1: HOME, 2: WORK, 3: FAX},
and self.type = somedict[self.get]. This also avoids endless if-elif-
elif-elif sequences (looking at showtype() in your original script,
where you can apply the same technique); it's often clearer, and
certainly shorter.
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