CODEWITHSUNDEEP
python
JohnTheBadge
JohnTheBadge

Reputation: 67

"if choice" is calling further code in program when I want it to stop

my code is if choice == "1" prints code further than I want it too. I cant use break as I want the user to be able to enter the correct password. There's some more code previously that shows an enter screen and includes the choice = input("choice: ")

This is what I have

# password
if choice == "1":
    print ("Requires Password")

#open the entire text file 
if choice == "password": 
    fp = open('answers.txt')
    while 1:
        line = fp.readline()
        if not line:
            break
        print (line)

 #quiz programming
 #generate a random keyword
elif choice == "2":
     input("You Ready? Press Enter Then")
print ('''

 Here is your Keyword''')
import random
with open('keywords.txt') as f:
       a = random.choice(list(f)).strip() #.strip cleans the line \n problem
       print ("    ")
       print ("------", a)

With this code, I hope that when the user enter 1, "requires password prints" however this is what I get.

choice: 1
Incorrect option
Requires Password


     Here is your Keyword

------ haemoglobin

     Now press enter for your definitions

How do I stop at requires password and allow the user to enter their password. Also the Incorrect option, further in the code is showing and I can't rid of it.

Upvotes: 0

Views: 196

Answers (4)

Joël
Joël

Reputation: 2822

First, reading your explanation shows that a part at higher level is missing in your code sample; would you please provide more details?

Second, a remark on your code: when working on an item that returns elements, Python best way is to use for.

With that in mind, your code:

fp = open('answers.txt')
while 1:
    line = fp.readline()
    if not line:
        break
    print (line)

becomes a much clearer and leaner

for line in open('answers.txt'):  # this is an iterator
    print line,

This is explained in the Python Tutorial.

Upvotes: 1

YXD
YXD

Reputation: 32511

I suspect further back in your code you should have used raw_input instead of input when getting choice

choice = raw_input("Choose an option").strip()

Upvotes: 2

tmj
tmj

Reputation: 1858

# password
if choice == "1":
    print ("Requires Password")

#open the entire text file 
if choice == "password": 
    fp = open('answers.txt')
    while 1:
        line = fp.readline()
        if not line:
            break
        print (line)

 #quiz programming
 #generate a random keyword
elif choice == "2":
     input("You Ready? Press Enter Then")

What happens is your first if executes, printing Requires Password then your control passs over to the second if-elif ladder. None matches, which is expected as choice is already equal to "1".

Then the rest of the code, not being under any if/elif statement executes no matter what. The presence of the if-elif block doesn't affect the execution of this block.

So what you need to do is either put the code of the block below, under some if-elif (whichever condition needs to be met), else what you can do is, put another condition before this block can be entered to check if your desired conditions have been met.

Upvotes: 1

PearsonArtPhoto
PearsonArtPhoto

Reputation: 39698

To answer your second question, you want to use a series of if/elif/else statements, as you've done later on, such as is included below.

if choice =='1':
   #do stuff here
elif choice=='2':
   #do stuff here
else:
   print("Invalid choice")

As far as the first one goes, I would recommend calling a subroutine that allows for entering the password. Here's the start of such a subroutine:

def enterPassword():
   #Code to enter password here
   if password =='password':
      print 'You entered the right password!
      return True
   else:
      print 'Wrong password'
      return False

if choice =='1':
    if enterPassword():
        #Authenticated, do stuff that requires password
    else:
        print('Wrong password')

Upvotes: 0

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