Question

简化，但对于像：

这样的表格

 id time distance price
 1  20   500      8 
 2  50   500      10 
 3  90   500      12 
 4  80   1000     17 
 5  170  1000     11 
 6  180  1000     13 
 7  19   800      12

我希望获得距离为500和1000的最快时间的行，即

 id time distance price
 1  20   500      8 
 4  80   1000     17

如果我这样做

select min(time) from table

可以很好地找到价格，但我无法获得ID和价格 - 只有所有ID /价格的最大/最小/平均/第一个值。

我可以通过多次查找来实现 - 例如

select * from table where distance = 500 and time = 20 
select * from table where distance = 1000 and time = 80

但是有一种更好的方法，不涉及1 +（距离数）查询（或者至少提供一个结果集，即使在内部使用该数量的查询）

Answer 1

您需要使用内部选择：

SELECT t.id, t.time, t.distance, t.price
FROM table t
JOIN (SELECT MIN(time) as min_time, distance
        FROM table
        GROUP BY distance) as tmp
      ON (t.distance = tmp.distance AND t.time = tmp.min_time)
WHERE t.distance IN (500, 1000)

Answer 2

您需要将min内容放入having子句中，因此您的查询将为select * from table group by distance having min(distance); （另）或者您可以使用subquerys找出： select * from table where distance = (select distance from table where min(time)) and time = select min(time) from table)（也未经测试：））

Answer 3

这个怎么样......准确地说你正在寻找什么（测试）

select *  from Table1 where time in (select min(time)  from table1 where distance = 500 or distance = 1000 group by distance) and (distance = 500 or distance = 1000)

Answer 4

只是订购和限制。那么你在1次查询中获得最快的500距离。

select * from thetable where distance = 500 ORDER BY time ASC LIMIT 1

Answer 5

试试这个 -

SELECT t1.* FROM table1 t1
  JOIN (SELECT distance, MIN(time) min_time FROM table11 WHERE distance = 500 OR distance = 1000 GROUP BY distance) t2
    ON t1.distance = t2.distance AND t1.time = t2.min_time;

Answer 6

SELECT * FROM tblData INNER JOIN (SELECT MIN(TIME) AS minTime, distance FROM tblData WHERE distance IN (500,1000) GROUP BY distance) AS subQuery ON tblData.distance = subQuery.distance AND tblData.time = subQuery.minTime

SQL Group By query - 获取聚合函数的相关字段

6 个答案: