User only get three chances to enter a valid value

Question

Adding a loop such that the user gets three chances to enter a
valid value. If the user enters an invalid value more than three times in a row, the program
should issue an error message and exit

In python please
I know how to do the while loop, but how do you make it to only three times

What I have:

miles= eval(input("Enter a value for miles "))
kilometers= miles * 1.6 
while miles < 0: 
    print("Error, you can't enter a negative value for miles")
    miles = eval(input("Enter the correct value for miles "))

Asked By: DesiBoy

||

Source

Answer 1

I’m not a python programmer but I can help you regarding to the logic behind it.

first declare a global variable for example integer. something like.

int invalid= 0;

Then do the loop using a do while statement or likewise loop.

while (true)
{

}

inside your while statement, put a validation on it or your logic and also the checking of number of iteration (not valid). something like

while (true)
{
 bool valid = isthisValid(); // your method
 if (valid == false)
  {
     invalid = invalid+ 1;
     if (invalid >= 3)
     {
      // Throw your error here!
     }
  }
}

Answered By: Chris

Answer 2

I would fix your code first…you’re evaluating “miles < 0” after you convert it to kilometers, so whether or not the user corrects the miles variable, you already have the resulting kilometers variable.

miles= eval(input("Enter a value for miles ")) 
kilometers= miles * 1.6 
while miles < 0: 
   print("Error, you can't enter a negative value for miles") 
   miles = eval(input("Enter the correct value for miles "))

change to:

miles = eval(input("Enter a value for miles ")) 
while miles < 0: 
    print("Error, you can't enter a negative value for miles") 
    miles = eval(input("Enter the correct value for miles "))
kilometers= miles * 1.6 
print(kilometers)

You can add a counter to your while loop to take note of iterations, and stop when it goes over:

counter = 0
miles = eval(input("Enter a value for miles ")) 
while miles < 0: 
    print("Error, you can't enter a negative value for miles") 
    miles = eval(input("Enter the correct value for miles "))
    counter+=1
    if counter > 2:
       break
kilometers= miles * 1.6 
print(kilometers)

And make it so that the kilometer variable only displays if the counter doesn’t exceed:

counter = 0
miles = eval(input("Enter a value for miles ")) 
while miles < 0: 
    print("Error, you can't enter a negative value for miles") 
    miles = eval(input("Enter the correct value for miles "))
    counter+=1
    if counter > 2:
       break
if counter <= 2:
  kilometers= miles * 1.6 
  print(kilometers)
else:
  print("Exceeded error count")

Answered By: Tyress

Answer 3

There are several ways to achive this. Here are two:

Using a for loop. And two check conditions:

import sys

def isvalid(input):
    if input == 1:
        return True
    else:
        return False

for n in xrange(3):
    check = isvalid(input(">>"))
    if check == True:
        break
    elif check == False and n == 2:
        sys.exit("Exceed tries")
    else:
        print "Wrong input, try again"

print "Right"

Using a while loops and a counter: ( I prefere this )

import sys

def isvalid(input):
    if input == 1:
        return True
    else:
        return False

mistakes_counter = 0
while True:
    check = isvalid(input(">>"))
    if check == True:
        break
    else:
        mistakes_counter += 1
        if mistakes_counter == 3:
           sys.exit("Exceed tries")
        print "Wrong input, try again"

print "Right"

Answered By: f.rodrigues

Answer 4

I understand your problem (because i already faced up a long time ago)

I find my solution myself… don’t give up just face it …i done this same….
Ohh sorry we are taking about your problem,not mine
…

Here a solution….`

for i in range(3):
   pwd = float(input('enter password '))
    if pwd == 123456:
        #your work 
        #my recommended (function)
    else :
        i += 1
        print("wrong password")
        if i == 3:
            print("wrong password limit exeded")`

I done for password u can edit that line to input

Answered By: Vedant more

User only get three chances to enter a valid value

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