1 reply to this topic

[rj_jr]

hi i am new to programing so please help if you can. ok this is what im doing,

Write a computer program in the python programming lanuage that:

1. Asks the user to enter up to three mobile telephone numbers in a single screen.

2. For those numbered entered, displays the list of numbers entered and clearly displays the mobile network belongs to alongside. You program should also indicate if a number entered does not belong to a valid mobile network.

3. Continues to run until the user indicates that they have finished entering phone numbers.

Number Prefix Mobile Network
083 3 Ireland
085 Meteor
086 O2
087 Vodafone

and this is what i have done.

import easygui


RunProgram = "Yes"


prompt = "Enter mobile phone numbers: "

Title = "Mobile phone networks"


options = ["Mobile Number 1: ","Mobile Number 2: ","Mobile Number 3: "]

userinput = []


t

while RunProgram == "Yes":



userinput = easygui.multenterbox(prompt,Title,options)


mobOne = userinput[0]

mobTwo = userinput[1]

mobThree = userinput[2]



posOne = mobOne.find('8')

posTwo = mobTwo.find('8')

posThree = mobTwo.find('8')



prefixOne = mobOne[posOne+1:3]

prefixTwo = mobTwo[posTwo+1:3]

prefixThree = mobThree[posThree+1:3]


networks = ['3','5','6','7']


three = "3 Ireland Network"

meteor = "Meteor Network"

Otwo = "O2 Network"

Vod = "Vodafone Network"



if prefixOne == networks[0]:

easygui.msgbox(mobOne + " is on the " + three)

elif prefixOne == networks[1]:

easygui.msgbox(mobOne + " is on the " + meteor)

elif prefixOne == networks[2]:

easygui.msgbox(mobOne + " is on the " + Otwo)

elif prefixOne == networks[3]:

easygui.msgbox(mobOne + " is on the " + Vod)

elif prefixOne != networks[0] or prefixOne != networks[1] or prefixOne != networks[2] or prefixOne != networks[3]:

easygui.msgbox(mobOne + " is not a valid network")



if prefixTwo == networks[0]:

easygui.msgbox(mobTwo + " is on the " + three)

elif prefixTwo == networks[1]:

easygui.msgbox(mobTwo + " is on the " + meteor)

elif prefixTwo == networks[2]:

easygui.msgbox(mobTwo + " is on the " + Otwo)

elif prefixTwo == networks[3]:

easygui.msgbox(mobTwo + " is on the " + Vod)

elif prefixTwo != networks[0] or prefixTwo != networks[1] or prefixTwo != networks[2] or prefixTwo != networks[3]:

easygui.msgbox(mobTwo + " is not a valid network")



if prefixThree == networks[0]:

easygui.msgbox(mobThree + " is on the " + three)

elif prefixThree == networks[1]:

easygui.msgbox(mobThree + " is on the " + meteor)

elif prefixThree == networks[2]:

easygui.msgbox(mobThree + " is on the " + Otwo)

elif prefixThree == networks[3]:

easygui.msgbox(mobThree + " is on the " + Vod)

elif prefixThree != networks[0] or prefixThree != networks[1] or prefixThree != networks[2] or prefixThree != networks[3]:

easygui.msgbox(mobThree + " is not a valid network")




RunProgram = easygui.enterbox("Would you like to enter more mobile numbers? Yes/No")

i have the programme working, but i want it shorter is there anyway 2 put the if statements into forloops, and to display the message in a single easygui.msg.box?

Edited by Roger, 20 January 2011 - 04:28 PM.

[manux]

As a generaly tip, instead of having mobOne,mobTwo,mobThree, just use a list called mob, then you can easily loop on it:

You could write most of this in probably one or two nested loops:

for i in range(3):
    for j in range(3):
        if prefix[i]==networks[j]:
            gui.msg(mob[i]+" bla bla "+networkNames[j]

Edited by manux, 20 January 2011 - 03:33 PM.
grammar