Hi,
so...@la... wrote:
> I think S=E9bastien is busy right now, and I hope he
> find clear this point.
Alas, sorry for not being more reactive, but that's right, I'm really
busy these days.
> Ansering myself,
>=20
> I found the way to avoid this issue by using=20
> ec.lock() and ec.unlock() at each web app request.
> (not really usefull for a threaded web app !!)=20
>=20
>=20
> But i really think the doc, isn't clear about this=20
> point. In fact i think that all operation on a=20
> ec should be locked, even fetching ( !=3D the doc ).
> Cause if you don't, you will get some strange=20
> behaviours meanly in the database adapator, that
> cause a segfault in the mysql one.
You're right indeed. Here is the problem:
>>> import MySQLdb
>>> MySQLdb.threadsafety
1
Quoted from http://www.python.org/topics/database/DatabaseAPI-2.0.html:
threadsafety
Integer constant stating the level of thread safety the interface
supports. Possible values are:
0 =3D Threads may not share the module.
1 =3D Threads may share the module, but not connections.
2 =3D Threads may share the module and connections.
3 =3D Threads may share the module, connections and cursors.
Hence the problem, because the framework shares a single connection
among threads.
And you're right when you say that the documentation is not clear on
that point. In fact, I wrote it with psycopg in mind (threadsafety=3D=3D2)
and it is definitely *wrong* as far as MySQLdb is concerned; same for
pgdb and pypgsql, btw --all but psycopg in fact ):
However, as usual ;) the behaviour you reveal is a bug, not a
feature. The framework should either:
- automatically lock/unlock the shared connection --this solution
would be equivalent to your ec.lock/unlock(), but the period of
locking will be somehow smaller than when locking the ec
- or use a different connection for each thread -- I expect this to=20
be longer to implement than the first solution.
Could you please fill in a bug report (category: core) ? I'll try to
have a look at it and propose a solution in the next two weeks, but I
cannot promise since I've very little time at the moment.
In the meantime, your solution is correct, and it's the only one.
Cheers,
-- S=E9bastien.
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