Обсуждение: basic stored proc/transaction question

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basic stored proc/transaction question

От
Ben
Дата:
My understanding is that a stored procedure does an implicit begin/commit when
it executes. Maybe my brain isn't working so well this morning, because I can't
figure out how I would do:

begin;
call stored proc;
call another stored proc;
commit;

It seems that the transaction would be committed after the first call.

Re: basic stored proc/transaction question

От
"Ian Harding"
Дата:
On 3/24/06, Ben <bench@silentmedia.com> wrote:
> My understanding is that a stored procedure does an implicit begin/commit when
> it executes. Maybe my brain isn't working so well this morning, because I can't
> figure out how I would do:
>
> begin;
> call stored proc;
> call another stored proc;
> commit;
>
> It seems that the transaction would be committed after the first call.
>
Nope.  Unless you use the new SAVEPOINT stuff, the explicit
transaction is the transaction.  Any error in any function will
rollback the whole thing.  The commit happens at the explicit commit.

Every SQL statement (such as calling a function) runs in an implicit
transaction.  Explicit transactions effectively "group" these implicit
transactions such that any one failure causes them all to fail.

- Ian

Re: basic stored proc/transaction question

От
Ben
Дата:
Well, that's awesome. Thanks!

On Fri, 24 Mar 2006, Ian Harding wrote:

> On 3/24/06, Ben <bench@silentmedia.com> wrote:
>> My understanding is that a stored procedure does an implicit begin/commit when
>> it executes. Maybe my brain isn't working so well this morning, because I can't
>> figure out how I would do:
>>
>> begin;
>> call stored proc;
>> call another stored proc;
>> commit;
>>
>> It seems that the transaction would be committed after the first call.
>>
> Nope.  Unless you use the new SAVEPOINT stuff, the explicit
> transaction is the transaction.  Any error in any function will
> rollback the whole thing.  The commit happens at the explicit commit.
>
> Every SQL statement (such as calling a function) runs in an implicit
> transaction.  Explicit transactions effectively "group" these implicit
> transactions such that any one failure causes them all to fail.
>
> - Ian
>