Hi, On 2023-07-31 21:25:06 +0000, José Neves wrote: > Ok, if I understood you correctly, I start to see where my logic is faulty. > Just to make sure that I got it right, taking the following example again: > > T-1 > INSERT LSN1-1000 > UPDATE LSN2-2000 > UPDATE LSN3-3000 > COMMIT LSN4-4000 > > T-2 > INSERT LSN1-500 > UPDATE LSN2-1500 > UPDATE LSN3-2500 > COMMIT LSN4-5500 > > Where data will arrive in this order: > > INSERT LSN1-500 > INSERT LSN1-1000 > UPDATE LSN2-1500 > UPDATE LSN2-2000 > UPDATE LSN3-2500 > UPDATE LSN3-3000 > COMMIT LSN4-4000 > COMMIT LSN4-5500
No, they won't arrive in that order. They will arive as BEGIN INSERT LSN1-1000 UPDATE LSN2-2000 UPDATE LSN3-3000 COMMIT LSN4-4000 BEGIN INSERT LSN1-500 UPDATE LSN2-1500 UPDATE LSN3-2500 COMMIT LSN4-5500 Because T1 committed before T2. Changes are only streamed out at commit / prepare transaction (*). Within a transaction, they however *will* be ordered by LSN. (*) Unless you use streaming mode, in which case it'll all be more complicated, as you'll also receive changes for transactions that might still abort. > You are saying that the LSN3-3000 will never be missing, either the entire > connection will fail at that point, or all should be received in the > expected order (which is different from the "numeric order" of LSNs). I'm not quite sure what you mean with the "different from the numeric order" bit... > If the connection is down, upon restart, I will receive the entire T-1 > transaction again (well, all example data again). Yes, unless you already acknowledged receipt up to LSN4-4000 and/or are only asking for newer transactions when reconnecting. > In addition to that, if I commit LSN4-4000, even tho that LSN has a "bigger > numeric value" than the ones representing INSERT and UPDATE events on T-2, I > will be receiving the entire T-2 transaction again, as the LSN4-5500 is > still uncommitted. I don't quite know what you mean with "commit LSN4-4000" here. > This makes sense to me, but just to be extra clear, I will never receive a > transaction commit before receiving all other events for that transaction. Correct. Greetings, Andres Freund
