Question

嘿大家喔我有一个表，其中包含一个名为id的自动增量值，一个用户ID，一个日期（称为日期）和一个用于多个注册的整数（称为多重注册）
由于多注册可以有多个日期，我需要选择具有最低日期的多注册您可以通过获取每个多注册的最小日期来比较它，我需要获取每个多注册的（id）。然后应该按最小日期

对它们进行排序

一些示例数据

 id --- user-id ---   date     --- multiregistration
 1   |    2      |  2010-02-01  |  1
 2   |    3      |  2010-02-01  |  2
 3   |    4      |  2010-01-01  |  2
 4   |    2      |  2010-02-03  |  1
 5   |    4      |  2010-02-03  |  3
 6   |    1      |  2010-02-02  |  3

预期输出

 multiregistration
  2
  1
  3

您可以将其与

进行比较

 SELECT `multiregistration`
 FROM `registrations`
 ORDER BY `date`

但我想删除重复项

这是我现在使用的PHP代码

 // Create the query
 $query  = 'SELECT `multiregistration` AS `multireg` FROM `registrations` WHERE `multiregistration`>0 GROUP BY `multiregistration`';

 // Execute the query and set the result into a variable
 $idsTo2ndQuery = arrayExecuteLocalQueryGetResults($query);

 // Create the second Query
 $query2 = 'SELECT `multiregistration` AS `multireg` FROM `registrations` WHERE `multiregistration`='.implode(' OR `multiregistration`=', $idsTo2ndQuery).' ORDER BY `date` ASC';

 // An array of IDs (sorted)
 $ids = array();

 // Execute the query and set the result into a variable
 $idsArray = arrayExecuteLocalQueryGetResults($query2);

 // For each entry
 foreach ($idsArray AS $idArray)
      if (!in_array($idArray, $ids))
           $ids[] = $idArray;

 // Return the IDs
 return $ids;

Answer 1

如果按多重注册进行分组，则每个不同的多重注册只能获得一行。这将为您提供首先按多重注册排序的所有行的集合，然后是日期。

SELECT multiregistration, date 
FROM registrations 
ORDER BY multiregistration, date

更新：

这将为您提供一组ID ...每个不同的多重注册，其中日期为分钟（日期）...

SELECT id  
FROM registrations 
GROUP BY multiregistration
ORDER BY multiregistration, date

更新：

这将为您提供一个多注册记录...其中日期是表格中的第一个日期......

SELECT top 1 multiregistration  
FROM registrations     
ORDER BY date

更新：（样本数据？）

ID-----USER-ID-----DATE------------MULTIREGISTRATION
1        101            1/2/2011         3
2        101            1/1/2011         3 
3        102            1/5/2011         2
4        102            1/7/2011         2
5        103            1/1/2011         4

Minumum date is 1/1/2011 multiregistration（3）将是1/1/2011 - id 2 ... 这将返回一个有序（按升序日期）结果集，其中多重注册值为3

SELECT * 
FROM registrations   
WHERE ID in (SELECT min(date) FROM registrations)   
ORDER BY date

对于列表中的两者（实际上在这种情况下都是3 - ＃3,2和4）...以下将返回结果集（如下所示......）

SELECT * 
FROM registrations          
ORDER BY multiregistration, date

返回结果集，如

3        102            1/5/2011         2
4        102            1/7/2011         2
2        101            1/1/2011         3 
1        101            1/2/2011         3
5        103            1/1/2011         4

Answer 2

以下是我将如何在T-SQL中执行此操作。它声明了一个与数据一起使用的变量表，并且查询产生了所请求的确切结果。

DECLARE @A AS TABLE(ID INT, USERID INT, SOMEDATE DATE, MULTIREG INT)

INSERT @A
SELECT 1,2,'20100201',1 UNION ALL
SELECT 2,3,'20100201',2 UNION ALL
SELECT 3,4,'20100101',2 UNION ALL
SELECT 4,2,'20100302',1 UNION ALL
SELECT 5,4,'20100203',3 UNION ALL
SELECT 6,1,'20100202',3

SELECT * FROM @A AS A
WHERE A.SOMEDATE = (SELECT MIN(SOMEDATE) FROM @A AS B WHERE B.MULTIREG = A.MULTIREG)
ORDER BY USERID DESC

ID  USERID  SOMEDATE     MULTIREG
3    4     2010-01-01       2
1    2     2010-02-01       1
6    1     2010-02-02       3

Answer 3

SELECT MIN(date), `multiregistration` 
FROM `registrations` 
GROUP BY `multiregistration` 
ORDER by `date`

Answer 4

SELECT multiregistration FROM registrations 
GROUP BY multiregistration ORDER BY date

它正在做这项工作吗？

MySQL从具有group by的表中选择最小值

4 个答案: