Python if error in a loop continue loop from next iteration

Question

I have such code:

for i in range(0,5):
  try:
      print(f"opened some_file_{i}")
  except Exception as e:
      print(f"error opening some_file_{i}")
      return
  print(f"i = {i}")

After exception, I need contuinue the loop from the next iteration, but not continue code (print(f"i = {i}")

How can I do it?

I tried to use a break, continue, pass statements and return

Please provide a minimal reproducible example. The expected behavior and difference from the provided code is hard to understand. — Michael Szczesny
– Michael Szczesny, Commented Oct 27, 2022 at 7:10

Myrkytyn · Accepted Answer · 2022-10-27 07:19:52Z

1

Sorry, continue works

for i in range(0,5):
  try:
      print(f"opened some_file_{i}")
  except Exception as e:
      print(f"error opening some_file_{i}")
      continue
  print(f"i = {i}")

So if I had an exception, continue will skip this loop iteration

edited Oct 27, 2022 at 7:19

answered Oct 27, 2022 at 7:02

Myrkytyn

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3 Comments

Epsi95 Over a year ago

why do you need continue, once you catch the exception, it will go to next iteration anyway

Michael Szczesny Over a year ago

@Epsi95 - I think the commented parts shouldn't be executed, so continue is necessary. But the question is hardly understandable.

Epsi95 Over a year ago

yes, then, they should be put in try block, I think.

jazz99 · Accepted Answer · 2022-10-27 08:14:05Z

0

If you have handled the exception properly it should continue on to the next iteration. You do not require break, continue, pass or return statements. When you try to break or return the control will exit out of the loop/ function. Example:

    for i in range(0,5):
  try:
      print("Hello {}".format(i))
      if i == 3:
          raise Exception("Wrong file path error")
  except Exception as e:
      print(e)

Output

answered Oct 27, 2022 at 8:14

jazz99

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