python catch exception and continue try block

Question

Can I return to executing try-block after exception occurs? (The goal is to write less) For Example:

try:
    do_smth1()
except:
    pass

try:
    do_smth2(         


        

Answer 1

Depending on where and how often you need to do this, you could also write a function that does it for you:

def live_dangerously(fn, *args, **kw):
    try:
        return fn(*args, **kw)
    except Exception:
        pass

live_dangerously(do_smth1)
live_dangerously(do_smth2)

But as other answers have noted, having a null except is generally a sign something else is wrong with your code.

Answer 2

You could iterate through your methods...

for m in [do_smth1, do_smth2]:
    try:
        m()
    except:
        pass

Answer 3

'continue' is allowed within an 'except' or 'finally' only if the try block is in a loop. 'continue' will cause the next iteration of the loop to start.

So you can try put your two or more functions in a list and use loop to call your function.

Like this:

funcs = [f,g]
for func in funcs:
    try: func()
    except: continue

For full information you can go to this link

Answer 4

No, you cannot do that. That's just the way Python has its syntax. Once you exit a try-block because of an exception, there is no way back in.

What about a for-loop though?

funcs = do_smth1, do_smth2

for func in funcs:
    try:
        func()
    except Exception:
        pass  # or you could use 'continue'

Note however that it is considered a bad practice to have a bare except. You should catch for a specific exception instead. I captured for Exception because that's as good as I can do without knowing what exceptions the methods might throw.

Answer 5

While the other answers and the accepted one are correct and should be followed in real code, just for completeness and humor, you can try the fuckitpy ( https://github.com/ajalt/fuckitpy ) module.

Your code can be changed to the following:

@fuckitpy
def myfunc():
    do_smth1()
    do_smth2()

Then calling myfunc() would call do_smth2() even if there is an exception in do_smth1())

Note: Please do not try it in any real code, it is blasphemy

Answer 6

one way you could handle this is with a generator. Instead of calling the function, yield it; then whatever is consuming the generator can send the result of calling it back into the generator, or a sentinel if the generator failed: The trampoline that accomplishes the above might look like so:

def consume_exceptions(gen):
    action = next(gen)
    while True:
        try:
            result = action()
        except Exception:
            # if the action fails, send a sentinel
            result = None

        try:
            action = gen.send(result)
        except StopIteration:
            # if the generator is all used up, result is the return value.
            return result

a generator that would be compatible with this would look like this:

def do_smth1():
    1 / 0

def do_smth2():
    print "YAY"

def do_many_things():
    a = yield do_smth1
    b = yield do_smth2
    yield "Done"

>>> consume_exceptions(do_many_things())
YAY

Note that do_many_things() does not call do_smth*, it just yields them, and consume_exceptions calls them on its behalf