SQL most popular

Question

I have a mysql table with items in relation to their order.

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CREATE DATABASE IF NOT EXISTS `sqltest`;
USE `sqltest`;

DROP TABLE IF EXISTS `testdata`;
CREATE TABLE `testdata` (
  `orderID` varchar(10) DEFAULT NULL,
  `itemID` varchar(10) DEFAULT NULL,
  `qtyOrdered` int(10) DEFAULT NULL,
  `sellingPrice` decimal(10,2) DEFAULT NULL
)

INSERT INTO `testdata`(`orderID`,`itemID`,`qtyOrdered`,`sellingPrice`) 
values ('1','a',1,'7.00'),('1','b',2,'8.00'),('1','c',3,'3.00'),('2','a',1,'7.00'),('2','c',4,'3.00');

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Intended Result:

A = (1+1)2

B = 2

C = (2+4)6 <- most popular

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How do I add up all the qty's for each item and result the highest one?

It should be fairly strait forward but I'm new to SQL and I can't work this one out :S

Solution needs to be mysql and or php.

I guess there needs to be some sort of temporary tally variable for each item ID, but that seems like it could get messy with too many items.

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ANSWER:

(thanks nuqqsa)

SELECT itemID, SUM(qtyOrdered) AS total FROM testdata GROUP BY itemID ORDER BY total DESC LIMIT 1;

Answer 1

select count(qtyOrdered), qtyOrdered from testdata group by qtyOrdered

Answer 2

How about this:

SELECT itemID, SUM(qtyOrdered) AS total FROM testdata GROUP BY itemID ORDER BY total DESC;

Answer 3

SELECT itemID, SUM(qtyOrdered) as blah FROM sqltest GROUP BY itemID ORDER BY blah DESC should do it

Answer 4

SELECT      *
FROM        testdata
ORDER BY    SUM(gtyOrdered) DESC
GROUP BY    itemID

Answer 5

SELECT SUM( qtyOrdered ) AS sum_ordered, itemID
FROM testdata
GROUP BY itemID
ORDER BY sum_ordered