I have a large dataset, and I have to do 3 joins, one of which is a sub-query. I chose to use a subquery instead of WHERE (IN or FIND_IN_SET) so that I don't lose values in my left, or base table. I need all of the data in the left column. Overall, I'm matching 11 million values with 900,000 values, so I expect this to take long, but it took ~20 seconds on a set of 200.
The engine is innoDB, each table has a primary key (IDvar).
I use the sub-query because I have to many values that I need to select from ( val1, val2,..., val100
) and I want to avoid using the 'AND' command with a clause for every 'val'.
The query I am using is:
SELECT *
FROM table1
LEFT JOIN (SELECT * FROM table2 WHERE table2.var IN(val1, val2,..., val100)) AS t
USING (IDvar)
LEFT JOIN table3
USING (IDvar);
The query looks fine to me. You'd want the following indexes:
create index idx_t1 on table1(idvar);
create index idx_t2 on table2(var, idvar);
create index idx_t3 on table3(idvar);
(Maybe it's just the second one that's missing.)
May I clarify why you do not use ON
statement?
In general, when I do joins, I do the following
SELECT *
FROM table1 JOIN table2 ON table1.common_var = table2.common_var
JOIN table3 ON table1.common_var2 = table3.common_var2
WHERE ...;
so that there is no need to load the whole huge table.
If there is a need to get every possible combination of the two table, we can get the two table separately and programatically get the combinations.
SELECT * FROM table1;
SELECT * FROM table2;
... the rest in another program ...
Won't this do the same task? And possibly be more efficient?
SELECT t1.*, t2.*, t3.*
FROM table1 AS t1
LEFT JOIN table2 AS t2 USING (IDvar)
LEFT JOIN table3 AS t3 USING (IDvar)
WHERE t2.var IN(val1, val2,..., val100);
Indexes needed:
t2: (IDvar, var) -- in this order
t3: (IDvar)
No index on t1
will be used.
Having LEFT
or not having it -- there is a big difference in this query.
The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.