På torsdag 12. juni 2014 kl. 21:49:59, skrev David G Johnston <
david(dot)g(dot)johnston(at)gmail(dot)com <mailto:david(dot)g(dot)johnston(at)gmail(dot)com>>: SELECT
coalesce(apl.company_name, bns.company_name, pns.company_name,
prs.company_name) AS company_name,
coalesce(apl.project_name, bns.project_name, pns.project_name,
prs.project_name) AS project_name,
apl.qty AS
num_apples,
bns.qty AS
num_bananas,
pns.qty AS
num_pineapples,
prs.qty AS
num_pears FROM company_apples AS apl FULL OUTER JOIN
company_bananas AS bns
ON apl.company_id = bns.company_id
AND (apl.project_id IS NULL AND bns.project_id IS NULL OR
apl.project_id = bns.project_id) FULL OUTER JOIN company_pineapples AS pns
ON apl.company_id = pns.company_id
AND (apl.project_id IS NULL AND pns.project_id IS NULL OR
apl.project_id = pns.project_id) FULL OUTER JOIN company_pears AS prs
ON apl.company_id = prs.company_id
AND (apl.project_id IS NULL AND prs.project_id IS NULL OR
apl.project_id = prs.project_id) ORDER BY company_name ASC, project_name ASC
NULLS FIRST; This gives the result: # company_name project_name num_apples
num_bananas num_pineapples num_pears 1 C1 NULL 2 2 10 NULL 2 C1 P1 5 12 NULL
NULL 3 C1 P2 2 NULL NULL NULL 4 C2 NULL 3 NULL 10 NULL 5 C2 P1 3 NULL NULL NULL
6 C2 P2 3 NULL NULL NULL 7 C3 NULL NULL 8 NULL NULL 8 C3 NULL NULL NULL NULL 7
As you see, there are two rows for C3, which should have been 1, with
num_bananas=8 and num_pears=7. There has to be something wrong with my FULL
OUTER JOINs but I don't know what.... Row #7 is from the right side of
the outer join between apples and bananas Row #8 is from the right side of the
outer join between apples and pears Since you are only using apple as the
source of valid company-project pairs only those records will be "correct" per
your definition. Even if you do not use a CROSS JOIN between company/product
you have to obtain a master list of valid company-project pairs from ALL of
the target tables. You can then LEFT JOIN that master against each of the
target tables and be sure that you have a valid master record to attach to.
The direct way to do this is: SELECT DISTINCT company_id, project_id FROM
apples UNION DISTINCT SELECT DISTINCT company_id, project_id FROM pears [and so
forth] The problem with retreiving a list of companies/projects is that the
actual sub-queries are pretty complex and don't run instantly, so I'll end up
querying lots of things twice. -- Andreas Jospeh Krogh CTO / Partner - Visena
AS Mobile: +47 909 56 963 andreas(at)visena(dot)com <mailto:andreas(at)visena(dot)com>
www.visena.com <https://www.visena.com> <https://www.visena.com>