上下文管理器中捕獲異常__enter __（）

Question

即使在__enter__()中存在異常，是否可以確保調用__exit__()方法？

>>> class TstContx(object): 
... def __enter__(self): 
...  raise Exception('Oops in __enter__') 
... 
... def __exit__(self, e_typ, e_val, trcbak): 
...  print "This isn't running" 
... 
>>> with TstContx(): 
...  pass 
... 
Traceback (most recent call last): 
    File "<stdin>", line 1, in <module> 
    File "<stdin>", line 3, in __enter__ 
Exception: Oops in __enter__ 
>>> 

編輯

這是接近我能得到...

class TstContx(object): 
    def __enter__(self): 
     try: 
      # __enter__ code 
     except Exception as e 
      self.init_exc = e 

     return self 

    def __exit__(self, e_typ, e_val, trcbak): 
     if all((e_typ, e_val, trcbak)): 
      raise e_typ, e_val, trcbak 

     # __exit__ code 


with TstContx() as tc: 
    if hasattr(tc, 'init_exc'): raise tc.init_exc 

    # code in context 

在後的視線，上下文經理或許不會

Answer 1

像這樣：

import sys 

class Context(object): 
    def __enter__(self): 
     try: 
      raise Exception("Oops in __enter__") 
     except: 
      # Swallow exception if __exit__ returns a True value 
      if self.__exit__(*sys.exc_info()): 
       pass 
      else: 
       raise 


    def __exit__(self, e_typ, e_val, trcbak): 
     print "Now it's running" 


with Context(): 
    pass 

，讓程序繼續得意揚揚地繼續不執行，你需要檢查上下文塊內的上下文對象，僅做了重要的東西，如果__enter__成功的上下文塊。

class Context(object): 
    def __init__(self): 
     self.enter_ok = True 

    def __enter__(self): 
     try: 
      raise Exception("Oops in __enter__") 
     except: 
      if self.__exit__(*sys.exc_info()): 
       self.enter_ok = False 
      else: 
       raise 
     return self 

    def __exit__(self, e_typ, e_val, trcbak): 
     print "Now this runs twice" 
     return True 


with Context() as c: 
    if c.enter_ok: 
     print "Only runs if enter succeeded" 

print "Execution continues" 

據我所知，你不能完全跳過with-block。並且請注意，此上下文現在吞併了全部例外。如果您希望在__enter__成功時不想吞下例外，請在__exit__和return False中檢查self.enter_ok，如果它是True。

Answer 2

最好的設計決策

不可以。如果在__enter__()有可能發生異常，那麼您需要自己動手，並打電話給幫手包含清理代碼的函數。

Answer 3

你可以使用contextlib.ExitStack（未測試）：

with ExitStack() as stack: 
    cm = TstContx() 
    stack.push(cm) # ensure __exit__ is called 
    with ctx: 
     stack.pop_all() # __enter__ succeeded, don't call __exit__ callback 

或者從the docs一個例子：

stack = ExitStack() 
try: 
    x = stack.enter_context(cm) 
except Exception: 
    # handle __enter__ exception 
else: 
    with stack: 
     # Handle normal case 

見contextlib2 on Python <3.3。

Answer 4

如果不需要繼承或複雜的子程序，你可以使用更短的方式：

from contextlib import contextmanager 

@contextmanager 
def test_cm(): 
    try: 
     # dangerous code 
     yield 
    except Exception, err 
     pass # do something 

Answer 5

class MyContext: 
    def __enter__(self): 
     try: 
      pass 
      # exception-raising code 
     except Exception as e: 
      self.__exit__(e) 

    def __exit__(self, *args): 
     # clean up code ... 
     if args[0]: 
      raise 

我已經做到了這個樣子。它以錯誤作爲參數調用__exit __（）。如果args [0]包含錯誤，它會在執行清理代碼後重新渲染異常。