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
I believe an UPDATE FROM
with a JOIN
will help:
MS SQL
UPDATE
Sales_Import
SET
Sales_Import.AccountNumber = RAN.AccountNumber
FROM
Sales_Import SI
INNER JOIN
RetrieveAccountNumber RAN
ON
SI.LeadID = RAN.LeadID;
MySQL and MariaDB
UPDATE
Sales_Import SI,
RetrieveAccountNumber RAN
SET
SI.AccountNumber = RAN.AccountNumber
WHERE
SI.LeadID = RAN.LeadID;
Best Answer
This is an example of the
greatest-n-per-group
problem that has appeared regularly on StackOverflow.Here's how I usually recommend solving it:
Explanation: given a row
p1
, there should be no rowp2
with the same customer and a later date (or in the case of ties, a laterid
). When we find that to be true, thenp1
is the most recent purchase for that customer.Regarding indexes, I'd create a compound index in
purchase
over the columns (customer_id
,date
,id
). That may allow the outer join to be done using a covering index. Be sure to test on your platform, because optimization is implementation-dependent. Use the features of your RDBMS to analyze the optimization plan. E.g.EXPLAIN
on MySQL.Some people use subqueries instead of the solution I show above, but I find my solution makes it easier to resolve ties.