Print error to screen but continue code execution

Question

I have some code that iterates through a series of URLs. If there is an error in my code because one of the URLs does not contain a valid JSON body, I want the error generated to be printed to screen, but then the code moves onto the next iteration. A simple version of my code is:

for a in myurls:

    try:

        #mycode

    except Exception as exc:

        print traceback.format_exc()
        print exc
        pass

However this prints the error to screen and ends execution of the code. Is there a way I can get the error to continue execution by moving to the next iteration of my 'for' loop?

Solution

Just put try-except over the code for which you expect an exception to occur. That code basically lies inside the loop.

for a in myurls:
    try:
        #mycode

    except Exception as exc:

        print traceback.format_exc()
        print exc

Comments

[Marcus Müller]

@gdogg371 Animesh's answer is pretty basic python. The point is that your exception handling happens at exactly the level you specify the try/except.

[gdogg371]

@animesh sharma my apologies, I have amended my question as my code is currently structured the way animesh has answered. i typed my question incorrectly. any ideas then?

[Animesh Sharma]

I guess the code should work fine. The except block should not make the for loop break.

[Hugues]

@animesh What is the answer for python3?