Assuming you're joining on columns with no duplicates, which is a very common case:
An inner join of A and B gives the result of A intersect B, i.e. the inner part of a Venn diagram intersection.
An outer join of A and B gives the results of A union B, i.e. the outer parts of a Venn diagram union.
Examples
Suppose you have two tables, with a single column each, and data as follows:
A B
- -
1 3
2 4
3 5
4 6
Note that (1,2) are unique to A, (3,4) are common, and (5,6) are unique to B.
Inner join
An inner join using either of the equivalent queries gives the intersection of the two tables, i.e. the two rows they have in common.
select * from a INNER JOIN b on a.a = b.b;
select a.*, b.* from a,b where a.a = b.b;
a | b
--+--
3 | 3
4 | 4
Left outer join
A left outer join will give all rows in A, plus any common rows in B.
select * from a LEFT OUTER JOIN b on a.a = b.b;
select a.*, b.* from a,b where a.a = b.b(+);
a | b
--+-----
1 | null
2 | null
3 | 3
4 | 4
Right outer join
A right outer join will give all rows in B, plus any common rows in A.
select * from a RIGHT OUTER JOIN b on a.a = b.b;
select a.*, b.* from a,b where a.a(+) = b.b;
a | b
-----+----
3 | 3
4 | 4
null | 5
null | 6
Full outer join
A full outer join will give you the union of A and B, i.e. all the rows in A and all the rows in B. If something in A doesn't have a corresponding datum in B, then the B portion is null, and vice versa.
select * from a FULL OUTER JOIN b on a.a = b.b;
a | b
-----+-----
1 | null
2 | null
3 | 3
4 | 4
null | 6
null | 5
Correct; you cannot truncate a table which has an FK constraint on it.
Typically my process for this is:
- Drop the constraints
- Trunc the table
- Recreate the constraints.
(All in a transaction, of course.)
Of course, this only applies if the child has already been truncated. Otherwise I go a different route, dependent entirely on what my data looks like. (Too many variables to get into here.)
The original poster determined WHY this is the case; see this answer for more details.
Best Answer
PostgreSQL automatically creates indexes on primary keys and unique constraints, but not on the referencing side of foreign key relationships.
When Pg creates an implicit index it will emit a
NOTICE
-level message that you can see inpsql
and/or the system logs, so you can see when it happens. Automatically created indexes are visible in\d
output for a table, too.The documentation on unique indexes says:
and the documentation on constraints says:
Therefore you have to create indexes on foreign-keys yourself if you want them.
Note that if you use primary-foreign-keys, like 2 FK's as a PK in a M-to-N table, you will have an index on the PK and probably don't need to create any extra indexes.
While it's usually a good idea to create an index on (or including) your referencing-side foreign key columns, it isn't required. Each index you add slows DML operations down slightly, so you pay a performance cost on every
INSERT
,UPDATE
orDELETE
. If the index is rarely used it may not be worth having.