Question

class A(object):
    def __init__(self):
        self.db = create_db_object()

    def change_Db_a(self):
         self.db.change_something()
         self.db.save()

    def change_db_b(self):
         self.db.change_anotherthing()
         self.db.save()

I am getting object from database, I changing it in multiple function and saving it back.
which is slow because it hits database on every function call. is there anything like deconstructor where I can save the database object so that I don’t have to save it for every function call and not waste time.

Answer 1

Don’t rely on the __del__ method for saving your object. For details, see this blog post.

You can use use the context management protocol by defining __enter__ and __exit__ methods:

class A(object):
    def __enter__(self):
        print 'enter'
        # create database object here (or in __init__)
        pass

    def __exit__(self, exc_type, exc_val, exc_tb):
        print 'exit'
        # save database object here

    # other methods

Then use the with statement when you create your object:

with A() as myobj:
    print 'inside with block'
    myobj.do_something()

When you enter the with block, the A.__enter__ method will be called. When you exit the with block the __exit__ method will be called. For example, with the code above you should see the following output:

enter

inside with block

exit

Here’s more information on the with statement:

• Understanding Python’s “with” statement