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Python raw_input Example (Input From Keyboard)

I would like to read data from the user using Python. How do I read data using raw_input()?

Tutorial details
DifficultyEasy (rss)
Root privilegesNo
RequirementsPython
Estimated completion timeN/A

raw_input() function reads a line from input (i.e. the user) and returns a string by stripping a trailing newline. The syntax is:


mydata = raw_input('Prompt :')
print (mydata)

In the above example, a string called mydata stores users data. Please note that if you want to compare mydata, then convert mydata to a numeric variable using int().

Examples

In this example, read the user name using raw_input() and display back on the screen using print():

 
#!/usr/bin/python
name=raw_input('Enter your name : ')
print ("Hi %s, Let us be friends!" % name);
 

Sample outputs:

Enter your name : nixCraft
Hi nixCraft, Let us be friends!

In this following example, a string called choice converted to a numeric variable:

 
#!/usr/bin/python
# Version 1
## Show menu ##
print (30 * '-')
print ("   M A I N - M E N U")
print (30 * '-')
print ("1. Backup")
print ("2. User management")
print ("3. Reboot the server")
print (30 * '-')
 
## Get input ###
choice = raw_input('Enter your choice [1-3] : ')
 
### Convert string to int type ##
choice = int(choice)
 
### Take action as per selected menu-option ###
if choice == 1:
        print ("Starting backup...")
elif choice == 2:
        print ("Starting user management...")
elif choice == 3:
        print ("Rebooting the server...")
else:    ## default ##
        print ("Invalid number. Try again...")
 

OR

 
#!/usr/bin/python
# Version 2
print (30 * '-')
print ("   M A I N - M E N U")
print (30 * '-')
print ("1. Backup")
print ("2. User management")
print ("3. Reboot the server")
print (30 * '-')
 
###########################
## Robust error handling ##
## only accept int       ##
###########################
## Wait for valid input in while...not ###
is_valid=0
 
while not is_valid :
        try :
                choice = int ( raw_input('Enter your choice [1-3] : ') )
                is_valid = 1 ## set it to 1 to validate input and to terminate the while..not loop
        except ValueError, e :
                print ("'%s' is not a valid integer." % e.args[0].split(": ")[1])
 
### Take action as per selected menu-option ###
if choice == 1:
        print ("Starting backup...")
elif choice == 2:
        print ("Starting user management...")
elif choice == 3:
        print ("Rebooting the server...")
else:
        print ("Invalid number. Try again...")
 

Sample outputs (note down invalid inputs are detected on fly):

------------------------------
   M A I N - M E N U
------------------------------
1. Backup
2. User management
3. Reboot the server
------------------------------
Enter your choice [1-3] : x
''x'' is not a valid integer.
Enter your choice [1-3] :
'''' is not a valid integer.
Enter your choice [1-3] : 1
Starting backup...
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{ 8 comments… add one }

  • Kevin February 25, 2013, 10:34 pm

    How to print “Hello World” in python?

    • Joshua Carr June 6, 2014, 6:40 pm

      print(“Hello World”)

    • sivaji January 24, 2015, 11:04 am

      print ‘Hello World’

  • cuturrr February 26, 2013, 8:29 am

    nice questione

    print “Hello World”

    • Tolli March 5, 2013, 5:06 am

      If you are using python 3, you have to do:
      print (“Hello World!”)
      In python 3 print is a built-in function, not a keyword like in older versions.

  • Igor October 7, 2013, 11:03 am

    My vVersion _:D ::

    print "Welcome to Russian Roulette!!"
    time.sleep(2)
    number =  raw_input('Enter a number from 1 to 6 ')
    print ('you entered %s !' % number)
    q =  raw_input ('ARE you ready  ?! y or n   ' )
    if q == 'y':
        x = random.randrange(1,6)
        for i in range(20):
            print "BE READY !!"
        time.sleep(4)
        print ("my number is %x" % x)
        if x != int(number):
            print 'you are lucky'
        else:
            print ' BAAAAAAAAAAAAANGG !!!!!you`re dead! '
    
  • Nigel Haslam September 14, 2014, 4:16 am

    Dear nixCraft,

    Thanks for a great scripting example. I’m a novice working on a simple Turtle drawing exercise and I’ve been struggling for hours to convert some raw_input into a variable that I could pass as an argument and your code worked for me when I was about to quit.

    Encouraged by my success ( via your help) I tried to improve my implementation by adding another exception so that non valid, or out of range integers would give the user another chance to input a number. like the exception for non integers does, instead of quitting the program with an error but I’m stumped.

    I tried making a list
    myintegers = [1,2,3]

    and adding this extra code

    choice = int for int in myintegers ( raw_input(‘Enter your choice [1-3] : ‘) )

    but I found that lists are made up of strings, not integers :(

    So I tried to cast my list of strings into a new list of integers :

    new_integers = [];
    for n in myintegers:
    new_integers.append(int(n));

    and tried
    choice = int for int in new_integers ( raw_input(‘Enter your choice [1-3] : ‘) )

    I suspect that it’s my way of checking if the integer is in the list that’s falling down.

    Any pointers?

    Thanks for sharing.

    • Robb Tolliver September 22, 2014, 3:41 am

      How about this?

      valid_choices = [1,2,3]
      choice = None
      while choice not in valid_choices:
      choice = int(raw_input(‘Enter your choice [1-3] : ‘))

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