我正在根据列X对结果进行分组,我希望返回组中列Y值最高的行。
SELECT *
FROM mytable
GROUP BY col1
HAVING col2 >= (SELECT MAX(col2)
FROM mytable AS mytable2
WHERE mytable2.col1 = mytable.col1 GROUP BY mytable2.col1)
我想优化上面的查询。没有子查询是否可行?
我找到了解决方案,它比你想象的更简单:
SELECT * FROM (SELECT * FROM mytable ORDER BY col2 DESC) temp GROUP BY col1
在20,000行上运行5毫秒。
答案 0 :(得分:13)
使用JOIN的派生表/内联视图:
SELECT x.*
FROM mytable x
JOIN (SELECT t.col1,
MAX(t.col2) AS max_col2
FROM MYTABLE t
GROUP BY t.col1) y ON y.col1 = x.col1
AND y.max_col2 >= x.col2
请注意,如果有多个相关的x
记录,则会重复y
个记录。要删除重复项,请使用DISTINCT
:
SELECT DISTINCT x.*
FROM mytable x
JOIN (SELECT t.col1,
MAX(t.col2) AS max_col2
FROM MYTABLE t
GROUP BY t.col1) y ON y.col1 = x.col1
AND y.max_col2 >= x.col2
以下是未经测试的,但不会返回重复项(假设有效):
SELECT x.*
FROM mytable x
WHERE EXISTS (SELECT NULL
FROM MYTABLE y
WHERE y.col1 = x.col1
GROUP BY y.col1
HAVING MAX(y.col2) >= x.col2)
答案 1 :(得分:1)
你的Col2永远不会>那么MAX(col2)所以我建议使用col2 = MAX(col2)
所以HERE是QUERY
SELECT * FROM mytable GROUP BY col1 HAVING col2 = MAX( col2 )