How can i go back to the if statement in Python 3?

Question

I am making a Python program where I can work with files from any part of my computer. It's not done yet but I ran into a problem. This is my code:

import os
from os.path import join
import subprocess
def opening_file(lookfor):
    global store_1
    for root, dirs, files in os.walk('/home/'):
        if lookfor in files:
           file = join(root, lookfor)
           store_1.append(join(root, lookfor))
    if len(store_1) <= 0:
        ask_1 = str(input("Your file was not found. Do you want to try again(Y/n)?:"))
        #This is where I have the problem
    elif len(store_1) > 0:
        print("Your file was found")
        subprocess.call(["xdg-open", file])
    #print(store_1)
store_1 = []
print("Welcome to our program for working with files.")
print("Press O for opening and editing a file, C for making a copy of a file. M for moving a file and R for renaming a file. If you are done working with the file, press F to end the program.")
choice = str(input("Your choice:"))
if choice == "O" or choice == "o":
   lookfor = input("File name(make sure to include the extension as well):")
   opening_file(lookfor)

I want to know how can i go back to the if statement the user entered with his/her input when the file is not being found.

Is there any way I can do this? I have googled but I cannot find a solution to my problem. My OS is Ubuntu 16.04.

None of the duplicates accepted answers solutions are using recursion. — Sayse, Sep 10 '18 at 10:29
@mahir The solution given in the link isn't using recursion. — Abhishek Keshri, Sep 10 '18 at 10:29
I'm willing to bet a good amount of money that you haven't even tried any of those answers that "don't solve your problem". — JJJ, Sep 10 '18 at 10:39
@JJJ, but I did try the ones that seemed most likely to work. I tried many different ways before asking this question. I asked this question because none of the solutions in any of the questions I entered solved my problem, except the one I just marked. Please understand, no hate — Mahir Islam, Sep 10 '18 at 10:44
The one you marked is practically identical to the answers in the duplicate. Next time, show what you have tried and you'll avoid the downvotes. — JJJ, Sep 10 '18 at 11:11

Ochmar · Accepted Answer · 2018-09-10T10:29:36.380

Simply use while?

import os
from os.path import join
import subprocess
def opening_file(lookfor):
    global store_1
    for root, dirs, files in os.walk('/home/'):
        if lookfor in files:
           file = join(root, lookfor)
           store_1.append(join(root, lookfor))
    if len(store_1) <= 0:
        ask_1 = str(input("Your file was not found. Do you want to try again(Y/n)?:"))
        #This is where I have the problem
    elif len(store_1) > 0:
        print("Your file was found")
        subprocess.call(["xdg-open", file])
    #print(store_1)
store_1 = []
print("Welcome to our program for working with files.")
choice = ""
while(choice.lower() != "f"):
    print("Press O for opening and editing a file, C for making a copy of a file. M for moving a file and R for renaming a file. If you are done working with the file, press F to end the program.")
    choice = str(input("Your choice:"))
    if choice == "O" or choice == "o":
       lookfor = input("File name(make sure to include the extension as well):")
       opening_file(lookfor)

You will need to define `choice` before the while loop, e.g. `choice = ""` — Martin Evans, Sep 10 '18 at 10:27

score 1 · Answer 2 · answered Sep 10 '18 at 10:40

import os
from os.path import join
import subprocess
def opening_file(lookfor):
    store_1 = []
    for root, dirs, files in os.walk('/home/'):
        if lookfor in files:
           file = join(root, lookfor)
           store_1.append(join(root, lookfor))
    if len(store_1) <= 0:
        ask_1 = str(input("Your file was not found. Do you want to try again(Y/n)?:"))
        #This is where I have the problem
        if (ask_1.lower() == 'y':
            return 'y'
        else:
            return 'n'
    else:
        print("Your file was found")
        subprocess.call(["xdg-open", file])
        return store_1
        #print(store_1)

def show_menu():
    choice = str(input("Your choice:"))
    if choice == "O" or choice == "o":
        lookfor = input("File name(make sure to include the extension as well):")
        result = opening_file(lookfor)
        if result == 'y':
            show_menu()
        elif result == 'n':
            #your code for no
        else:
            # you can manage the result list 'store_1'
            print(result)

print("Welcome to our program for working with files.")
print("Press O for opening and editing a file, C for making a copy of a file. 
M for moving a file and R for renaming a file. If you are done working with the file, press F to end the program.")
show_menu()

How can i go back to the if statement in Python 3?

2 Answers2