Yes. Sure... the data is modified in the DB.
On Friday, April 1, 2016 at 12:14:51 PM UTC-3, Niphlod wrote: > > sure the other process has commit()ed the changes ? that's what > transactions are for: until a change is commit()ed, no changes are visible. > > > BTW: mysql has the weirdest default setting of not allowing to see changes > from OTHER processes until the process which reads commit()s itself (which > may be your case). > Try to call db.commit() on the "shell" process even if you didn't issue > any update() statement yet. The restriction explained earlier still > applies: if the other process doesn't commit(), your shell will still see > no changes at all. > > On Friday, April 1, 2016 at 5:07:52 PM UTC+2, Marcello wrote: >> >> I tried it. >> It gives me the old value. >> I have to leave the shell and enter again to refresh it... >> >> This is my problem >> >> >> >> On Friday, April 1, 2016 at 11:53:52 AM UTC-3, Niphlod wrote: >>> >>> yep, you need to re-select it ... >>> >>> On Friday, April 1, 2016 at 3:26:39 PM UTC+2, Marcello wrote: >>>> >>>> Hello, >>>> >>>> I have a function that I call in a shell... >>>> I load a record from a database and do some stuff.. >>>> >>>> Problem is that in the meantime the record may be modified in the >>>> server... >>>> >>>> Is there a way to reload the record from the database to track possible >>>> changes ?? >>>> (I'm using mysql) >>>> >>>> >>>> Thanks, >>>> >>>> Marcello >>>> >>> -- Resources: - http://web2py.com - http://web2py.com/book (Documentation) - http://github.com/web2py/web2py (Source code) - https://code.google.com/p/web2py/issues/list (Report Issues) --- You received this message because you are subscribed to the Google Groups "web2py-users" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. For more options, visit https://groups.google.com/d/optout.
