How to prevent try catching every possible line in python?

Question

I got many lines in a row which may throw an exception, but no matter what, it should still continue the next line. How to do this without individually try catching every si

Answer 1

I ran into something similar, and asked a question on SO here. The accepted answer handles logging, and watching for only a specific exception. I ended up with a modified version:

class Suppressor:
    def __init__(self, exception_type, l=None):
        self._exception_type = exception_type
        self.logger = logging.getLogger('Suppressor')
        if l:
            self.l = l
        else:
            self.l = {}
    def __call__(self, expression):
        try:
            exec expression in self.l
        except self._exception_type as e:
            self.logger.debug('Suppressor: suppressed exception %s with content \'%s\'' % (type(self._exception_type), e))

Usable like so:

s = Suppressor(yourError, locals()) 
s(cmdString)

So you could set up a list of commands and use map with the suppressor to run across all of them.