Question

I am writing a web app in PHP using mySQL that models an election.

I have three tables: Candidates, Elections, and Votes. Votes contains CandidateID, ElectionID and Count, which is the number of times that the given candidate was voted for in the given Election. Votes also contains TimeStamp which is the last time the row was modified which is used for breaking ties (the earlier vote wins). A candidate may have run in multiple elections. How do I find how many elections a given candidate has ever won?

All help greatly appreciated, thanks.

Some sample data:

CREATE TABLE IF NOT EXISTS `Votes` (
`ElectionID` int(11) unsigned NOT NULL,
`CandidateID` int(11) unsigned NOT NULL,
`Count` smallint(5) unsigned NOT NULL DEFAULT '0',
`stamp` int(11) unsigned NOT NULL,
PRIMARY KEY (`ElectionID`,`CandidateID`)
) ENGINE=MyISAM DEFAULT CHARSET=utf8;

INSERT INTO `Votes` (`ElectionID`, `CandidateID`, `Count`, `stamp`) 
VALUES
(1, 1, 3, 1332897534),
(4, 1, 3, 1333149930),
(4, 4, 2, 1333149947),
(4, 5, 3, 1333149947),
(1, 4, 4, 1333153373);

Desired output: One row, with one column, being the number of wins for a certain candidate

Answer 1

You can write:

SELECT COUNT(1)
  FROM Elections AS e
 INNER
  JOIN Votes AS v1 -- representing the candidate of interest
    ON v1.ElectionID = e.ID
   AND v1.CandidateID = ...
  LEFT
 OUTER
  JOIN Votes AS v2 -- representing a candidate who beat the candidate of interest
    ON v2.ElectionID = e.ID
   AND (    v2.Count > v1.Count
         OR (     v2.Count = v1.Count
              AND v2.stamp < v1.stamp
            )
       )
 WHERE v2.ElectionID IS NULL -- meaning that no candidate beat the candidate of interest
;

(It’s also possible to represent either or both of those joins with EXISTS and a correlated subquery; or the first join could be changed to IN with an uncorrelated subquery; but the above is the most likely to perform best, IMHO, and my experience on StackOverflow has been that people seem to like joins better than subqueries for some reason. If you’d prefer a subquery answer, let me know.)