python catch exception and continue try block

Question

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.

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

You can achieve what you want, but with a different syntax. You can use a "finally" block after the try/except. Doing this way, python will execute the block of code regardless the exception was thrown, or not.

Like this:

try:
    do_smth1()
except:
    pass
finally:
    do_smth2()

But, if you want to execute do_smth2() only if the exception was not thrown, use a "else" block:

try:
    do_smth1()
except:
    pass
else:
    do_smth2()

You can mix them too, in a try/except/else/finally clause. Have fun!

You could iterate through your methods...

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