I'd just like to exit out of a with statement under certain conditions:
with open(path) as f:
print 'before condition'
if <condition>: break #syntax error!
print 'after condition'
Of course, the above doesn't work. Is there a way to do this? (I know that I can invert the condition: if not <condition>: print 'after condition' -- any way that is like above?)
with giving you trouble? Throw more with-able objects at the problem!
class fragile(object):
class Break(Exception):
"""Break out of the with statement"""
def __init__(self, value):
self.value = value
def __enter__(self):
return self.value.__enter__()
def __exit__(self, etype, value, traceback):
error = self.value.__exit__(etype, value, traceback)
if etype == self.Break:
return True
return error
Just wrap the expression you're going to with with fragile, and raise fragile.Break to break out at any point!
with fragile(open(path)) as f:
print 'before condition'
if condition:
raise fragile.Break
print 'after condition'
with and just the with; doesn't wrap your function in a semantically misleading one-run 'loop' or a narrowly specialized function, and doesn't force you to do any extra error handling after the with.Nestable!
with fragile(open(path1)) as f:
with fragile(open(path2)) as g:
print f.read()
print g.read()
raise fragile.Break
print "This wont happen"
print "This will though!"
This way, you don't have to create a new function to wrap the outer with if you want both to break.
fragile and you're good to go!
The best way would be to encapsulate it in a function and use return:
def do_it():
with open(path) as f:
print 'before condition'
if <condition>:
return
print 'after condition'
This is an ancient question, but this is an application for the handy "breakable scope" idiom. Just imbed your with statement inside:
for _ in (True,):
with open(path) as f:
print 'before condition'
if <condition>: break
print 'after condition'
This idiom creates a "loop", always executed exactly once, for the sole purpose of enclosing a block of code inside a scope that can be broken out of conditionally. In OP's case, it was a context manager invocation to be enclosed, but it could be any bounded sequence of statements that may require conditional escape.
The accepted answer is fine, but this technique does the same thing without needing to create a function, which is not always convenient or desired.
I think you should just restructure the logic:
with open(path) as f:
print 'before condition checked'
if not <condition>:
print 'after condition checked'
Since break can only occur inside a loop your options are somewhat limited inside the with to:
Having a function and using return is probably the cleanest and easiest solution here if you can isolate the with and the associated statements (and nothing else) inside a function.
Otherwise generate an exception inside the with when needed, catch immediately below/outside the with to continue the rest of the code.
Update: As OP suggests in the comments below (perhaps tonuge in cheek?) one could also wrap the with statement inside a loop to get the break to work - though that would be semantically misleading. So while a working solution, probably not something that would be recommend).
This question was asked before Python 3.4 existed but with 3.4 you can use contextlib.supress,
suppressing your own personal exception.
See that this (runnable as is) code
from contextlib import suppress
class InterruptWithBlock(UserWarning):
"""To be used to interrupt the march of a with"""
condition = True
with suppress(InterruptWithBlock):
print('before condition')
if condition: raise InterruptWithBlock()
print('after condition')
# Will not print 'after condition` if condition is True.
So with the code in the question, you'd do:
with suppress(InterruptWithBlock) as _, open(path) as f:
print('before condition')
if <condition>: raise InterruptWithBlock()
print('after condition')
Note: If you're (still) before 3.4, you can still make your own suppress context manager easily.
f = open("somefile","r")
for line in f.readlines():
if somecondition: break;
f.close()
I dont think you can break out of the with... you need to use a loop...
[edit] or just do the function method others mentioned
as shorthand snippet:
class a:
def __enter__(self):
print 'enter'
def __exit__(self ,type, value, traceback):
print 'exit'
for i in [1]:
with a():
print("before")
break
print("after")
...
enter
before
exit
There exists a function __exit__() for this purpose. Syntax is as follows:
with VAR = EXPR:
try:
BLOCK
finally:
VAR.__exit__()
Use while True:
while True:
with open(path) as f:
print 'before condition'
if <condition>:
break
print 'after condition n'
break
You could put everything inside a function, and when the condition is true call a return.
Here is another way of doing it using try and except, even though inverting the condition is probably more convenient.
class BreakOut(Exception): pass
try:
with open(path) as f:
print('before condition')
if <condition>:
raise BreakOut #syntax error!
print('after condition')
except BreakOut:
pass
Change from "break" to "f.close"
with open(path) as f:
print('before condition')
if <condition>: f.close()
print('after condition')
