A switch case program in python

Question

i am struggling to get a switch case program in python,when i run the script i want to display three options

Enter your choice :

1.Insert Records

2.Update Records

3.Display Records

and after seeing these,the user should be able to enter his choice.Since am new to python i googled and found that there is no switch case in python.

def main():
    print("Enter your choice : ")
    print("1.Insert Records \n2.Update Records \n3.Display Records")
    choice = sys.argv[1]
    if(choice == 1):
        print 1
    if(choice == 2):
        print 2
    if(choice == 3):
        print 3
    else:
        print("You entered a wrong choice")

if __name__ == "__main__":
    main()

This is what i tried but its of no use, because it needs to enter choice at the time of running the script(eg. python abc.py 1 )

Answer 1

Instead of switch/case, you can use a dict, like this:

from sys import argv

def insertRecord(args):
  ...

def updateRecord(args):
  ...

def displayRecords(args):
  ...

{
  1 : insertRecords,
  2 : updateRecords,
  3 : displayRecords
}[argv[1]](argv[2])

If you want to catch a default case, you can add something like this:

from sys import argv

def insertRecord(args):
  ...

def updateRecord(args):
  ...

def displayRecords(args):
  ...

def printHelp():
  ...

try:
  {
    1 : insertRecords,
    2 : updateRecords,
    3 : displayRecords
  }[argv[1]](argv[2])    
except KeyError:
  printHelp()

Hope this helps.

Answer 2

sys.argv is a list of strings . You are trying to compare one of those to integers, and that won't work without conversion.

In this case, you don't need to convert, just compare to strings instead:

if choice == '1':
    print 1
elif choice == '2':
    print 2
elif choice == '3':
    print 3
else:
    print("You entered a wrong choice")

You can still convert the string to an integer first, allowing you to do more complex checks:

try:
    choice = int(sys.argv[1])
except ValueError:
    print("You entered a wrong choice")
else:
    if not 1 <= choice <= 3:
        print("You entered a wrong choice")
    else:
        print(choice)

If you are building an interactive menu, then perhaps the command line options are not the best choice here. Use the input() function instead so you print prompts and use a series of exchanges. See Asking the user for input until they give a valid response .

Answer 3

Look at the description of sys.argv . It is not used for input in runtime. Rather it is used to pass arguments to script when it is called. So this script:

import sys

print sys.argv

When called with python script.py one two three would print ['one', 'two', 'three']

If you want to get user input on executiopn of the program you should use input . Look carefully at the details such as returned value etc.

The next issue as makred by Martijn is that python does not have switch-case. You should use if-elif-else constrcution.

Answer 4

It sounds like you want to prompt the user for input. In python 2.x, use raw_input for that. Users enter strings, so I changed what you are looking for. And used elif to end the compares after you find a match.

def main():
    print("Enter your choice : ")
    choice = raw_input("1.Insert Records \n2.Update Records \n3.Display Records")
    if(choice == '1'):
        print 1
    elif(choice == '2'):
        print 2
    elif(choice == '3'):
        print 3
    else:
        print("You entered a wrong choice")

if __name__ == "__main__":
    main()

Answer 5

You are able to pass the choice as argument OR it will ask. I prefer the lambda-version in "switch-cases" to be sure it will only be executed when accessed. This way you can replace print(1) etc. with a performance intense task and it won't run for every entry in the dict.

Python3 example:

import sys
def get_input():
    print("Enter your choice : ")
    return input("1.Insert Records \n2.Update Records \n3.Display Records")

def main():
    choice = sys.argv[1] if len(sys.argv)>1 else get_input()
    {
        '1': lambda: print(1),
        '2': lambda: print(2),
        '3': lambda: print(3),
        '4': lambda: print(4),
        }.get(choice, lambda: print("You entered a wrong choice"))()

if __name__ == "__main__":
    main()

Same in python2:

import sys
def get_input():
    print "Enter your choice : "
    return raw_input("1.Insert Records \n2.Update Records \n3.Display Records")

def print_val(val):
    print val

def main():
    choice = sys.argv[1] if len(sys.argv)>1 else get_input()
    {
        '1': lambda: print_val(1),
        '2': lambda: print_val(2),
        '3': lambda: print_val(3),
        '4': lambda: print_val(4),
        }.get(choice, lambda: print_val("You entered a wrong choice"))()

if __name__ == "__main__":
    main()