CODEWITHSUNDEEP
python
kitchen800
kitchen800

Reputation: 227

running a function after x seconds in python

I am trying to schedule to run very x seconds. I am getting mixed up with the simple logic.

import schedule
import time

names = ["1.jpg", "2.jpg", "3.jpg"]
i = 0

def printer():
    for item in names:
        print(names[i])
        i = i+1

schedule.every(5).seconds.do(printer)

while 1:
    schedule.run_pending()
    time.sleep(1)

what i am trying to get it to do is to print the name 1 then 5 seconds later print name 2 then another 5 seconds later print name 3. i et the error saying: UnboundLocalError: local variable 'i' referenced before assignment any ideas?

Upvotes: 0

Views: 694

Answers (2)

Samwise
Samwise

Reputation: 71562

Your printer function is broken. If you're trying to have this function print each item in names, get rid of i and just do:

names = ["1.jpg", "2.jpg", "3.jpg"]

def printer():
    for item in names:
        print(item)

If you want the function itself to have a delay between each item, and you only want to run through this once, the simplest thing is to not bother with schedule at all and just add a sleep to the loop:

from time import sleep

names = ["1.jpg", "2.jpg", "3.jpg"]

def printer():
    for item in names:
        print(item)
        sleep(5)

printer()

To make it even simpler, you can also get rid of names and printer:

from time import sleep

for i in range(1, 4):
    print(f"{i}.jpg")
    sleep(5)

Upvotes: 3

Bahae El Hmimdi
Bahae El Hmimdi

Reputation: 368

import schedule
import time

names = ["1.jpg", "2.jpg", "3.jpg"]
i = 0

def printer():
    global i # here is the fix
    for item in names:
        print(names[i])
        i = i+1 # you trying to change variable in function that already been initialised out of function so you must use global before using the variable 

schedule.every(5).seconds.do(printer)

while 1:
    schedule.run_pending()
    time.sleep(1)

Upvotes: 1

Related Questions