Question

我必须构建一个SQL SELECT查询。表结构如下：

[ID] [PHONE] [description] [DATE]
[1] [600898367] [main] [2016-01-23]
[2] [600898367] [] [2016-01-24]
[3] [600898367] [] [2016-01-26]
[4] [600898367] [] [2016-01-28]
[5] [662349093] [main] [2016-01-10]
[6] [662349093] [] [2016-01-21]
[7] [662349093] [] [2016-01-30]
[8] [662349093] [] [2016-01-31]

这里有不同的记录，分组在同一个电话号码中。第一个（最早的）出现标有[main]标志。 [main]标志没有两个相同的数字。我想选择每个[主要]记录，另外一个最年轻的同一个电话号码，所以结果应该给出记录1,4,5,8。

请帮忙。

Answer 1

使用main子句为您提供MAX标记的记录。使用JOIN获取最新记录，使用UNION ALL获取其他列。最后，执行-- Get the main records first SELECT * FROM tbl WHERE description = 'main' UNION ALL -- Get the most recent records SELECT b.* FROM ( SELECT t.PHONE, MAX(DATE) AS MaxDate FROM tbl t GROUP BY PHONE ) a INNER JOIN tbl b -- Do a JOIN to get the additional columns ON b.PHONE = a.PHONE AND b.DATE = a.MaxDate组合结果。

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Answer 2

试试这个;）

SQL Fiddle

MySQL 5.6架构：

CREATE TABLE table1
    (`ID` int, `PHONE` int, `description` varchar(4), `DATE` varchar(11))
;

INSERT INTO table1
    (`ID`, `PHONE`, `description`, `DATE`)
VALUES
    (1, 600898367, 'main', '2016-01-23'),
    (2, 600898367, NULL, '2016-01-24'),
    (3, 600898367, NULL, '2016-01-26'),
    (4, 600898367, NULL, '2016-01-28'),
    (5, 662349093, 'main', '2016-01-10'),
    (6, 662349093, NULL, '2016-01-21'),
    (7, 662349093, NULL, '2016-01-30'),
    (8, 662349093, NULL, '2016-01-31')
;

查询1 ：

select t.*
from table1 t
inner join (
    select `PHONE`, max(`DATE`) as `DATE` from table1 group by `PHONE`
) t1 on t.`PHONE` = t1.`PHONE` and (t.`DATE` = t1.`DATE` or t.`description` = 'main')
order by t.`ID`

<强> Results ：

| ID |     PHONE | description |       DATE |
|----|-----------|-------------|------------|
|  1 | 600898367 |        main | 2016-01-23 |
|  4 | 600898367 |      (null) | 2016-01-28 |
|  5 | 662349093 |        main | 2016-01-10 |
|  8 | 662349093 |      (null) | 2016-01-31 |

Answer 3

您可以使用以下查询：

SELECT t1.*, t3.*
FROM mytable AS t1
LEFT JOIN (
   SELECT PHONE, MAX(date) AS max_date
   FROM mytable 
   GROUP BY PHONE
) AS t2 ON t1.PHONE = t2.PHONE
LEFT JOIN mytable AS t3 ON t1.PHONE = t3.PHONE AND t2.max_date = t3.`date`
WHERE t1.description = 'main'

Answer 4

使用config.vm.synced_folder "host/dir", "/guset/home/dir"，我们会先在GROUP BY和PHONE进行分组，从而获得4行。

接下来，我们将使用GROUP_CONCAT创建逗号分隔集。可以采用description条款，按日期订购电话号码。

最后我们想要从集合中获取第一个项目，我们可以使用SUBSTRING_INDEX执行此操作。

ORDER BY

See Fiddle

Answer 5

尝试以下查询..

SELECT 
  A.* 
FROM
  `old_young` A 
  INNER JOIN 
    (SELECT 
      MIN(`DATE`) AS Res 
    FROM
      old_young 
    WHERE description = 'main' 
    GROUP BY PHONE 
    UNION
    ALL 
    SELECT 
      MAX(`DATE`) AS Res 
    FROM
      old_young 
    WHERE description = '' 
    GROUP BY PHONE) B 
    ON A.DATE = B.Res ;

检查FIDDLE

MySQL - SELECT查询 - 最早和最年轻的日期

5 个答案:

See Fiddle