MySQL Forums
Forum List  »  InnoDB

Re: NOT Solved:: Bug in HAVING COUNT?
Posted by: John Viescas
Date: June 13, 2016 02:33PM

Variation on the theme - this is still a bug in 5.7.

CREATE VIEW CustStyles AS
SELECT C.CustomerID, C.CustFirstName,
C.CustLastName, MS.StyleName
FROM Customers AS C INNER JOIN Musical_Preferences AS MP
ON C.CustomerID = MP.CustomerID
INNER JOIN Musical_Styles AS MS
ON MP.StyleID = MS.StyleID;

CREATE VIEW EntStyles AS
SELECT E.EntertainerID, E.EntStageName, MS.StyleName
FROM Entertainers AS E INNER JOIN Entertainer_Styles AS ES
ON E.EntertainerID = ES.EntertainerID
INNER JOIN Musical_Styles AS MS
ON ES.StyleID = MS.StyleID;

-- NOTE: The following returns the WRONG answer
-- because of a bug in MySQL
SELECT CustStyles.CustomerID, CustStyles.CustFirstName,
CustStyles.CustLastName, EntStyles.EntStageName
FROM CustStyles INNER JOIN EntStyles
ON CustStyles.StyleName = EntStyles.StyleName
GROUP BY CustStyles.CustomerID, CustStyles.CustFirstName,
CustStyles.CustLastName, EntStyles.EntStageName
HAVING COUNT(EntStyles.StyleName) =
(SELECT COUNT(StyleName)
FROM CustStyles AS CS1
WHERE CS1.CustomerID = CustStyles.CustomerID)
ORDER BY CustStyles.CustomerID;

-- By forcing the count in the outer SELECT clause, we
-- get the correct answer:
SELECT CustStyles.CustomerID, CustStyles.CustFirstName,
CustStyles.CustLastName, EntStyles.EntStageName, COUNT(EntStyles.StyleName) As EntCount,
(SELECT COUNT(StyleName) FROM CustStyles AS CS2 WHERE CS2.CustomerID = CustStyles.CustomerID) AS CustCount
FROM CustStyles INNER JOIN EntStyles
ON CustStyles.StyleName = EntStyles.StyleName
GROUP BY CustStyles.CustomerID, CustStyles.CustFirstName,
CustStyles.CustLastName, EntStyles.EntStageName
HAVING COUNT(EntStyles.StyleName) = CustCount
ORDER BY CustStyles.CustomerID;

BUG!

Options: ReplyQuote


Subject
Views
Written By
Posted
1288
June 07, 2016 04:56AM
489
June 07, 2016 05:19AM
Re: NOT Solved:: Bug in HAVING COUNT?
484
June 13, 2016 02:33PM


Sorry, you can't reply to this topic. It has been closed.

Content reproduced on this site is the property of the respective copyright holders. It is not reviewed in advance by Oracle and does not necessarily represent the opinion of Oracle or any other party.