Here is what I need to do:
Create an IF branching statement that will take the
users input from 1-10 and return the English word for
the number. (1 is , 2 is Two, etc.) If the user
enters a value outside of the range of 1-10, display
an error message, and ask the user to enter a valid
selection.
Here is what I have so far:
# Print the word for a number 1-10 entered by the user
# If there is an incorrect value entered print an
error message
# and have them type in a correct value
print "\nThis program prints the English word for a
number entered between 1 and 10."
print "\nIf an incorrect value is entered you will get
an error message and be asked to enter a correct
value."
raw_input("\n\nPlease press enter to continue.")
number = int(raw_input("Please enter a number between
1 and 10: "))
if number < 1:
print "That is an incorrect number. Please try
again."
raw_input("Please enter a number between 1 and 10:
")
if number 10:
print "That is an incorrect number. Please try
again."
raw_input("Please enter a number between 1 and 10:
")
elif number == 1:
print ""
elif number == 2:
print "Two"
elif number == 3:
print "Three"
elif number == 4:
print "Four"
elif number == 5:
print "Five"
elif number == 6:
print "Six"
elif number == 7:
print "Seven"
elif number == 8:
print "Eight"
elif number == 9:
print "Nine"
elif number == 10:
print "Ten"
raw_input("\n\nPress the enter key to exit.")
How do I get it to work after an incorrect number is
entered? I am stuck on this and would appreciate and
help or suggestions?
Thanks,
Josh
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Tutor maillist - Tutor (AT) python (DOT) org

How do I get it to work after an incorrect number is
entered? I am stuck on this and would appreciate and
help or suggestions?

Use exception

try:
int("hoge")
except(ValueError):
print('incorrect')

incorrect

elif number == 1:
print ""
elif number == 2:
print "Two"
elif

Using table is more smart way.

NUMBER_STRINGS = ['Zero', '', 'Two', 'Three', ]
print NUMBER_STRINGS[number]

Tutor maillist - Tutor (AT) python (DOT) org

cimjls wrote:
Here is what I need to do:

Create an IF branching statement that will take the
users input from 1-10 and return the English word for
the number. (1 is , 2 is Two, etc.) If the user
enters a value outside of the range of 1-10, display
an error message, and ask the user to enter a valid
selection.

Here is what I have so far:

# Print the word for a number 1-10 entered by the user
# If there is an incorrect value entered print an
error message
# and have them type in a correct value

print "\nThis program prints the English word for a
number entered between 1 and 10."
print "\nIf an incorrect value is entered you will get
an error message and be asked to enter a correct
value."
raw_input("\n\nPlease press enter to continue.")

number = int(raw_input("Please enter a number between
1 and 10: "))
if number < 1:
print "That is an incorrect number. Please try
again."
raw_input("Please enter a number between 1 and 10:
")
if number 10:
print "That is an incorrect number. Please try
again."
raw_input("Please enter a number between 1 and 10:
")
elif number == 1:
print ""
elif number == 2:
print "Two"
elif number == 3:
print "Three"
elif number == 4:
print "Four"
elif number == 5:
print "Five"
elif number == 6:
print "Six"
elif number == 7:
print "Seven"
elif number == 8:
print "Eight"
elif number == 9:
print "Nine"
elif number == 10:
print "Ten"

raw_input("\n\nPress the enter key to exit.")

How do I get it to work after an incorrect number is
entered? I am stuck on this and would appreciate and
help or suggestions?

What happens when you try an incorrect number? You will generally get
better help on this list when you are very specific about what happens,
instead of saying it doesn't work.

This looks like homework so I will just give some hints.

There is a difference between how you use the first raw_input() call and
the ones you make when the data is bad; can you see it?

Do you know about while loops yet? If so, what happens if the user
enters a bad number the second time? Can you fix it with a while loop?

Kent

Tutor maillist - Tutor (AT) python (DOT) org