I am using MySQL. Here is my schema:
Suppliers(sid: integer, sname: string, address string)
Parts(pid: integer, pname: string, color: string)
Catalog(sid: integer, pid: integer, cost: real)
(primary keys are bolded)
I am trying to write a query to select all parts that are made by at least two suppliers:
-- Find the pids of parts supplied by at least two different suppliers.
SELECT c1.pid -- select the pid
FROM Catalog AS c1 -- from the Catalog table
WHERE c1.pid IN ( -- where that pid is in the set:
SELECT c2.pid -- of pids
FROM Catalog AS c2 -- from catalog
WHERE c2.pid = c1.pid AND COUNT(c2.sid) >= 2 -- where there are at least two corre开发者_如何学Csponding sids
);
First off, am I even going about this the right way?
Secondly, I get this error:
1111 - Invalid use of group function
What am I doing wrong?
You need to use HAVING
, not WHERE
.
The difference is: the WHERE
clause filters which rows MySQL selects. Then MySQL groups the rows together and aggregates the numbers for your COUNT
function.
HAVING
is like WHERE
, only it happens after the COUNT
value has been computed, so it'll work as you expect. Rewrite your subquery as:
( -- where that pid is in the set:
SELECT c2.pid -- of pids
FROM Catalog AS c2 -- from catalog
WHERE c2.pid = c1.pid
HAVING COUNT(c2.sid) >= 2)
First, the error you're getting is due to where you're using the COUNT
function -- you can't use an aggregate (or group) function in the WHERE
clause.
Second, instead of using a subquery, simply join the table to itself:
SELECT a.pid
FROM Catalog as a LEFT JOIN Catalog as b USING( pid )
WHERE a.sid != b.sid
GROUP BY a.pid
Which I believe should return only rows where at least two rows exist with the same pid
but there is are at least 2 sid
s. To make sure you get back only one row per pid
I've applied a grouping clause.
If you don't have an aggregate function in your where clause, another possible source of the 1111 - Invalid use of group function
error is if you have nested aggregate functions:
select sum(avg(close)) from prices;
(1111, 'Invalid use of group function')
You can get around this by breaking up the problem into two steps:
- Save the inner aggregation into a variable
select @avg:=avg(close) from prices;
- Run the outer aggregation against the variable
select sum(@avg) from prices;
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