1

最初のMYSQLクエリ

-- selects the latest records by unique member_id
SELECT * FROM table t
JOIN (SELECT MAX( id ) AS id
FROM table
GROUP BY member_id) t2 ON t2.id = t.id
WHERE `t`.`timestamp` >= DATE_SUB(NOW(), INTERVAL 5 MINUTE)
ORDER BY t.id DESC
LIMIT 10

2番。

-- sums the item_qt column by unique member_id
SELECT SUM(item_qt) AS sum_item_qt FROM table t
WHERE `t`.`timestamp` >= DATE_SUB(NOW(), INTERVAL 5 MINUTE)
GROUP BY member_id
ORDER BY t.id DESC
LIMIT 10

これらの2つのクエリを組み合わせて、sum_item_qtがmember_idに結合されるようにする方法はありますか？

4

2 に答える 2

3

このクエリは、あなたが探している答えを与えるはずだと思います:

SELECT *
FROM table1 t
INNER JOIN
(SELECT MAX(id) AS id, SUM(item_qt) AS sum_item_qt
 FROM table1
 WHERE timestamp >= DATE_SUB(NOW(), INTERVAL 5 MINUTE)
 GROUP BY member_id) AS t2
ON t2.id = t.id
ORDER BY t.id DESC
LIMIT 10
于 2012-12-17T13:55:17.540 に答える
1 
SELECT  a.*, c.*
FROM    tableName a
        INNER JOIN
        (
            SELECT member_ID, max(ID) maxID
            FROM tableName
            GROUP BY member_ID
        ) b ON  a.member_ID = b.member_ID AND
                a.ID = b.ID
        INNER JOIN
        (
            SELECT  member_ID, SUM(item_qt) sum_item_qt 
            FROM    tableName
            WHERE   timestamp >= DATE_SUB(NOW(), INTERVAL 5 MINUTE)
            GROUP BY member_id
        ) c ON  a.member_ID = c.member_ID
-- WHERE
-- ORDER BY
-- LIMIT
于 2012-12-16T14:13:38.850 に答える