How to continue with next line in a Python's try block?

Question

eg

try:
    foo()
    bar()
except: 
    pass

When foo function raise an exception, how to skip to the next line (bar) and execute it?

Answer 1

Take bar() out of the try block:

try:
    foo()
except: 
    pass
bar()

Btw., watch out with catch-all except clauses. Prefer to selectively catch the exceptions that you know you can handle/ignore.

Answer 2

Can't be done if the call to bar is inside the try -block. Either you have to put the call outside of the try-except block, or use the else :

try:
    foo()
except:
    pass
else:
    bar()

If bar might throw an exception as well, you have to use a separate try block for bar .

Answer 3

That is not the intended way for try/except blocks to be used. If bar() should be executed even if foo() fails, you should put each in its own try/except block:

try:
  foo()
except:
  pass # or whatever

try:
  bar()
except:
  pass # or whatever

Answer 4

If you have just two functions, foo() bar(), check the other solutions. If you need to run A LOT of lines, try something like this example:

def foo():
    raise Exception('foo_message')
def bar():
    print'bar is ok'
def foobar():
    raise  Exception('foobar_message')

functions_to_run = [
     foo,
     bar,
     foobar,
]

for f in functions_to_run:
    try:
        f()
    except Exception as e:
        print '[Warning] in [{}]: [{}]'.format(f.__name__,e)

Result:

[Warning] in [foo]: [foo_message]
bar is ok
[Warning] in [foobar]: [foobar_message]

Answer 5

If you want exceptions from both functions to be handled by the same except clause, then use an inner try/finally block:

try:
    try:
        foo()
    finally:
        bar()
except Exception:
    print 'error'

If there is an exception in foo() , first bar() will be executed, then the except clause.

However, it's generally good practice to put the minimum amount of code inside a try block, so a separate exception handler for each function might be best.