Mysql 根据另一列的值选择“最小”、“最大”对应列

我有下表：

Date        Open        High        Low         Close
1-Oct-19    225.070007  228.220001  224.199997  224.589996
2-Oct-19    223.059998  223.580002  217.929993  218.960007
3-Oct-19    218.429993  220.960007  215.130005  220.820007
4-Oct-19    225.639999  227.490005  223.889999  227.009995
7-Oct-19    226.270004  229.929993  225.839996  227.059998
8-Oct-19    225.820007  228.059998  224.330002  224.399994
9-Oct-19    227.029999  227.789993  225.639999  227.029999
10-Oct-19   227.929993  230.440002  227.300003  230.089996
11-Oct-19   232.949997  237.639999  232.309998  236.210007
14-Oct-19   234.899994  238.130005  234.669998  235.869995
15-Oct-19   236.389999  237.649994  234.880005  235.320007
16-Oct-19   233.369995  235.240005  233.199997  234.369995
17-Oct-19   235.089996  236.149994  233.520004  235.279999
18-Oct-19   234.589996  237.580002  234.289993  236.410004
21-Oct-19   237.520004  240.990005  237.320007  240.509995
22-Oct-19   241.160004  242.199997  239.619995  239.960007
23-Oct-19   242.100006  243.240005  241.220001  243.179993
24-Oct-19   244.509995  244.800003  241.809998  243.580002
25-Oct-19   243.160004  246.729996  242.880005  246.580002
28-Oct-19   247.419998  249.25  246.720001  249.050003
29-Oct-19   248.970001  249.75  242.570007  243.289993
30-Oct-19   244.759995  245.300003  241.210007  243.259995
31-Oct-19   247.240005  249.169998  237.259995  248.759995

对于给定的日期范围，比如说从03Oct19到10Oct19，我想得到高列的最大值及其对应的日期值，以及低列的最小值及其对应的日期。 在上面的示例中，预期结果应该是

|  MAX(High)  | High-Date |  MIN(Low)  | Min-Date  |
+-------------+-----------+------------+-----------+
|  230.44002  | 10-Oct-19 | 215.130005 | 03-Oct-19 |

我正在尝试以下查询：

SELECT max(high)
       , Date
       , min(low)
       , Date 
FROM tbl1 
where date>='2019-10-03' 
and date<='2019-10-10' 
group by date

但是，上面的代码缺少一些内容，因为它只是返回所有行

你知道还应该用什么吗？还是嵌套查询

谢谢你可以试试这个：

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首先需要找到指定日期范围的高值和低值，然后与主表联接以获得相应的日期。结果将在同一日期具有一行最高值和最低值，或者在两个日期具有两行最高值和最低值，或者可能在多个日期具有多行相同的最高值或最低值。使用条件聚合来优化结果：

SELECT MIN(CASE WHEN t.High = v.High THEN t.High END) AS High     -- MIN/MAX does not matter here
     , MIN(CASE WHEN t.High = v.High THEN t.Date END) AS HighDate -- MIN for first date, MAX for last date
     , MIN(CASE WHEN t.Low  = v.Low  THEN t.Low  END) AS Low      -- MIN/MAX does not matter here
     , MIN(CASE WHEN t.Low  = v.Low  THEN t.Date END) AS LowDate  -- MIN for first date, MAX for last date
FROM (
    SELECT MAX(High) AS High
         , MIN(Low) AS Low
    FROM t
    WHERE Date BETWEEN '2019-10-03' AND '2019-10-10' 
) AS v
JOIN t ON t.High = v.High OR t.Low = v.Low
WHERE Date BETWEEN '2019-10-03' AND '2019-10-10'

---

试着这样做：

select * from 
(SELECT high as "Max(high)"
       , Date_c as "High-Date"
FROM tbl1 
where Date_c >= '2019-10-03' 
and Date_c <= '2019-10-10' 
And high = (select max(high) from tbl1)) a
cross join
(SELECT low as "MIN(Low)"
       , Date_c as "Min-Date"
FROM tbl1 
where Date_c >= '2019-10-03' 
and Date_c<= '2019-10-10' 
And low = (select min(low) from tbl1)) b

这是解决这个问题的方法

这是@SalmanACheers的代码！忠告:

select A.high as "Max(high)"
       , A.Date_c "High-Date" 
       , B.low as "MIN(Low)"
       , B.Date_c as "Min-Date"
from 
(SELECT High, Date_c 
FROM tbl1
where Date_c between '2019-10-03' and '2019-10-10'
ORDER BY High DESC LIMIT 1) A
CROSS JOIN 
(SELECT Low, Date_c 
FROM tbl1 
where Date_c between '2019-10-03' and '2019-10-10'
ORDER BY Low LIMIT 1
) B;

---

和用于此的。您可以使用此查询获得所需的结果。它查找感兴趣期间的MAXhigh和MINlow值，然后将这些值连接回原始表以查找相应的日期。注：我们在日期值上使用最小值，以避免在一天以上出现高/低值的情况；这给出了它发生的最早日期。如果要获取高/低值出现的最新日期，请将其更改为“最大”

输出：

high    low     high_date   low_date
230.44  215.13  2019-10-10  2019-10-03

---

MySQL的哪个版本？MySQL版本：5.6.42-84.2我详细研究了@VBokšić和Nick的答案，Ersin的答案与VBokšić的类似，而Salman的答案与Nick的类似。谢谢你们四位的帮助。在我的表上分别运行VBokšić和Nick的查询10次，平均而言，VBokšić的SQL在执行时间上略优于Nick的SQL。这就是将VBokšić的答案作为答案的原因，尽管尼克的答案也很完美，厄辛和萨尔曼也是如此。向所有四个人投票@用户6337701到目前为止，我还没有在S.O.上看到如此恭敬的解释。谢谢！很高兴有帮助。@user6337701感谢您的反馈。看到基准测试结果总是很好的。可能在“2019-10-03”和“2019-10-10”之间的日期必须重复两次才能加入。前提是历史上从未重复过高/低。是的，我看到了问题。修正。嘿@VBokšić，我尝试了你提供的第二种解决方案。但它忽略了您在查询的最后一部分中指定的日期范围，其中a.date_c>='2019-10-03'和a.date_cHi@user6337701，您是100%正确的。我会改变代码，我会把新的演示。检查它。如果您使用交叉连接方法，则更简单的做法是选择高，按高描述限制从t订单开始日期1交叉连接选择低，按低限制从t订单开始日期1。@SalmanA。。。很好，在VBokšić的答案中列出的第三个解决方案甚至更好，因为它是基于排序的，不太复杂+1.

select A.high as "Max(high)"
       , A.Date_c "High-Date" 
       , B.low as "MIN(Low)"
       , B.Date_c as "Min-Date"
from 
(SELECT High, Date_c 
FROM tbl1
where Date_c between '2019-10-03' and '2019-10-10'
ORDER BY High DESC LIMIT 1) A
CROSS JOIN 
(SELECT Low, Date_c 
FROM tbl1 
where Date_c between '2019-10-03' and '2019-10-10'
ORDER BY Low LIMIT 1
) B;

SELECT hl.high, 
       MIN(t1.date) AS high_date, 
       hl.low, 
       MIN(t2.date) AS low_date
FROM (SELECT MAX(high) AS high, MIN(low) AS low
      FROM tbl1
      WHERE `Date` BETWEEN '2019-10-03' AND '2019-10-10') hl
JOIN tbl1 t1 ON t1.high = hl.high AND t1.date BETWEEN '2019-10-03' AND '2019-10-10'
JOIN tbl1 t2 ON t2.low = hl.low AND t2.date BETWEEN '2019-10-03' AND '2019-10-10'
GROUP BY hl.high, hl.low

high    low     high_date   low_date
230.44  215.13  2019-10-10  2019-10-03