Joris Dobbelsteen writes:
Did you really check your list thoroughly.
SELECT distinct export_messages.export_id as id
FROM export_messages
LEFT OUTER JOIN exports ON
(export_messages.export_id = exports.export_id);
Take any value from "SELECT export_id FROM exports"
Does it not exist in your list?
Correct.
I thought of that.. and the outerjoin as I showed.. only shows values that
are in export_messages but are not in exports.
I went over nearly 100 values and that select only had the right values.
Try this:
SELECT distinct export_messages.export_id as id,
exports.export_id as exports_export_id
FROM export_messages
LEFT OUTER JOIN exports ON
(export_messages.export_id = exports.export_id)
WHERE exports.export_id IS NOT NULL;
Thanks will try it.
The LEFT OUTER join will at all times return ALL rows from
export_messages.
What is very, very strange is that it didn't return all values.
In effect, you generate a list with ALL export_messages.expor
we must conclude that for every row you are trying to delete, the
condition must evaluate to true.
This is what was driving me crazy.. when I did the select by itself the list
was correct.
That's the trade-off: effects of a mistake * chance of a mistake against
the cost to prevent these.
Thanks much. Will try your query.
doing all this within a transaction so I can double check the results.. that
is the primary reason i would rather get it done from within psql.
If I do it in a program I will have no easy way to tell if I am doing the
right thing... Small tests.. and print statements will helpfully help, but
once I believe the program is working.. and run it.. the only solution is a
restore (I do a backup before doing any changes of course).
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