How to return to beginning of program from inside of if statement? - for-loop

Question

I'm practicing some basic coding, I'm running a simple math program running in the terminal on Visual Studio Code.
How do I create an option to return to the beginning of the program, or exit the program after getting caught in an if statement?
Example:
#beginning of program
user_input=input('Please select "this" or "that": ')
findings=user_input
If findings == this:
print(this)
# How can I redirect back to first user input question, instead
# of just ending here?
if findings == that:
print (that)
# Again, How do I redirect back to first user input, instead of
# the program ending here?
# Can I setup a Play_again here with options to return to user_input,
# or exit program? And then have all other If statements
# redirect here after completion? How would I do that? with
# another If? or with a for loop?
#end program

Answer 1

You can try wrapping the whole program in a while loop like this:
while(True):
user_input=input('Please select "this" or "that": ')
this = 'foo'
if user_input == this:
print(this)
continue
if user_input == this:
print(this)
continue

Answer 2

Unfortunately, that 'another technique' I thought of using didn't work.
Here's the code (the first example modified):
import sys
def main():
# Lots of setup code here.
def start_over():
return #Do nothing and continue from the next line
condition = 1==1 #Just a sample condition, replace it with "check(condition)".
float_condition = condition
def play(*just_define:bool):
if not just_define:
play_again = input('play again? "y" or "n"')
if play_again == 'y':
start_over() #Jump to the beginning of the prohram.
if play_again == 'n':
sys.exit() #Exit the program
while True:
if float_condition == True:
# print(float_condition)
play() #skip to play_again from here?
if float_condition == False:
#print(float_condition)
play() #skip to play_again from here?
#I removed the extra "main()" which was here because it'd cause an infinite loop-like error.
main()
Output:
play again? "y" or "n"y
play again? "y" or "n"n
Process finished with exit code 0
The * in the play(*just_define:bool) function makes the just_define parameter optional. Use the parameter if you want to only tell Python to search for this function, so it doesn't throw a ReferenceError and not execute anything that's after the line if not just_define:. How to call like so: play(just_define=True).
I've used nested functions. I've defined play so that you can call it from other places in your code.

Answer 3

Thanks to #Hack3r - I was finally able to choose to return back to the beginning of the program or exit out. But it resulted in a new issue. Now my print(results) are printing 4 or 5 times...
Here is the actual code I built and am working with:
def main():
math_Options=['Addition +','Subtraction -','Multiplication *','Division /']
math_func=['+','-','*','/']
for options in math_Options:
print(options)
print('Lets do some Math! What math function would you like to use? ')
while True:
my_Math_Function = input('Please make your choice from list above using the function symbol: ')
my_Number1=input('Please select your first number: ')
x=float(my_Number1)
print('Your 1st # is: ', x)
my_Number2=input('Please select your Second Number: ')
y=float(my_Number2)
print('Your 2nd # is: ', y)
z=float()
print('')
for Math_function in math_func:
if my_Math_Function == math_func[0]:
z=x+y
if my_Math_Function == math_func[1]:
z=x-y
if my_Math_Function == math_func[2]:
z=x*y
if my_Math_Function == math_func[3]:
z=x/y
if (z % 2) == 0 and z>0:
print(z, ' Is an EVEN POSITIVE Number')
if (z % 2) == 1 and z>0:
print(z, ' IS a ODD POSTIVE Number')
if (z % 2) == 0 and z<0:
print(z, ' Is an EVEN NEGATIVE Number')
if (z % 2) ==1 and z<0:
print(z, ' IS a ODD NEGATIVE Number')
if z==0:
print(z, 'Is is Equal to Zero')
print('')
play_again=input('Would you like to play again? "y" or "n" ')
if play_again == 'y':
continue
if play_again == 'n':
break
main()
main()