Question

如何查看每天出现多次的所有记录？

我有这张桌子：

ID   Name     Date
1    John     27.03.2010 18:17:00
2    Mike     27.03.2010 16:38:00
3    Sonny    28.03.2010 20:23:00
4    Anna     29.03.2010 13:51:00
5    Maria    29.03.2010 21:59:00
6    Penny    29.03.2010 17:25:00
7    Alba     30.03.2010 09:36:00
8    Huston   31.03.2010 10:19:00

我想得到：

1    John     27.03.2010 18:17:00
2    Mike     27.03.2010 16:38:00
4    Anna     29.03.2010 13:51:00
5    Maria    29.03.2010 21:59:00
6    Penny    29.03.2010 17:25:00

Answer 1

假设您正在使用MySQL，应工作。

SELECT *
FROM `thetable`
GROUP BY DATE(`Date`)
HAVING COUNT(*) > 1

我没有对此进行过测试，这只是我能想到的第一件事。

日期函数只占用DateTime的日期部分（我假设你正在使用它，因为也显示了一个时间组件。Reference。我还在反引号中包含Date字段名，因为Date是保留的MySQL中的单词（并且为了一致性而对表名执行相同的操作）。

请记住，不同的RDBMS可能会有不同的功能来实现这一目标。

Answer 2

我正在调用表Mytable，并将Date的名称更改为somedate以不使用关键字：

--create table mytable(ID int,Name varchar(32), somedate datetime)
select *
from mytable
where id in (
select id
from mytable
group by convert(varchar(10), somedate, 101), id
having count(1) > 1
)

Answer 3

您的意思是“我如何选择包含多条记录的天数的所有记录？”

select *
from your_table
where trunc(date) in ( select trunc(date)
                       from your_table
                       group by trunc(date)
                       having count(*) > 1)
/

修改

哦，你在SQL Server上。我使用了ORACLE的TRUNC()函数，该函数采用日期时间并删除时间元素。显然，SQL Server没有确切的等价物，但有some workarounds。

Answer 4

更新：使用SQL Server：

CREATE TABLE t1 (id int IDENTITY PRIMARY KEY, name varchar(20), date datetime);

INSERT INTO t1 VALUES('John',   '2010-03-27 18:17:00');
INSERT INTO t1 VALUES('Mike',   '2010-03-27 16:38:00');
INSERT INTO t1 VALUES('Sonny',  '2010-03-28 20:23:00');
INSERT INTO t1 VALUES('Anna',   '2010-03-29 13:51:00');
INSERT INTO t1 VALUES('Maria',  '2010-03-29 21:59:00');
INSERT INTO t1 VALUES('Penny',  '2010-03-29 17:25:00');
INSERT INTO t1 VALUES('Alba',   '2010-03-30 09:36:00');
INSERT INTO t1 VALUES('Huston', '2010-03-31 10:19:00');

SELECT  t1.id, t1.name, sub_t.date
FROM    t1
JOIN    (SELECT   DATEADD(dd, DATEDIFF(dd,0, date), 0) as date
         FROM     t1 
         GROUP BY DATEADD(dd, DATEDIFF(dd,0, date), 0) 
         HAVING   COUNT(id) > 1) sub_t ON 
                  (sub_t.date = DATEADD(dd, DATEDIFF(dd,0, t1.date), 0));

返回：

+----+-------+---------------------+
| id | name  | date                |
+----+-------+---------------------+
|  1 | John  | 2010-03-27 00:00:00 |
|  2 | Mike  | 2010-03-27 00:00:00 |
|  4 | Anna  | 2010-03-29 00:00:00 |
|  5 | Maria | 2010-03-29 00:00:00 |
|  6 | Penny | 2010-03-29 00:00:00 |
+----+-------+---------------------+

以前的答案假定MySQL：

加入子查询将是一个选项：

CREATE TABLE t1 (id int AUTO_INCREMENT PRIMARY KEY, 
                 name varchar(20), 
                 date datetime);

INSERT INTO t1 VALUES(NULL, 'John',   '2010-03-27 18:17:00');
INSERT INTO t1 VALUES(NULL, 'Mike',   '2010-03-27 16:38:00');
INSERT INTO t1 VALUES(NULL, 'Sonny',  '2010-03-28 20:23:00');
INSERT INTO t1 VALUES(NULL, 'Anna',   '2010-03-29 13:51:00');
INSERT INTO t1 VALUES(NULL, 'Maria',  '2010-03-29 21:59:00');
INSERT INTO t1 VALUES(NULL, 'Penny',  '2010-03-29 17:25:00');
INSERT INTO t1 VALUES(NULL, 'Alba',   '2010-03-30 09:36:00');
INSERT INTO t1 VALUES(NULL, 'Huston', '2010-03-31 10:19:00');

SELECT  t1.id, t1.name, sub_t.date
FROM    t1
JOIN    (SELECT   DATE(date) as date
         FROM     t1 
         GROUP BY DATE(date) 
         HAVING   COUNT(id) > 1) sub_t ON (sub_t.date = DATE(t1.date));

返回：

+----+-------+------------+
| id | name  | date       |
+----+-------+------------+
|  1 | John  | 2010-03-27 |
|  2 | Mike  | 2010-03-27 |
|  4 | Anna  | 2010-03-29 |
|  5 | Maria | 2010-03-29 |
|  6 | Penny | 2010-03-29 |
+----+-------+------------+
5 rows in set (0.02 sec)

Answer 5

  Select *  
    From YourTable  
   Where ADate in (Select ADate  
                     From YourTable  
                    Group By ADate  
                   Having Count(Distinct Id) > 1  
                 )

Answer 6

您可以按照以下方式执行此操作：

SELECT * 
FROM   theTable t1 
WHERE  t1.created_at IN 
       (SELECT   t2.created_at 
        FROM     theTable t2 
        GROUP BY created_at 
        HAVING COUNT(*) > 1)

Answer 7

您可以在日期变量的日期部分进行连接，这将排除该日期只有一行的任何位置（因为它没有任何内容可以加入）。

此示例位于t-sql：

select  distinct l.*
from    @table l
join 
    @table r
on  convert(varchar, l.[date], 102) = convert(varchar, r.[date],102)
and l.id != r.id

Answer 8

这是最终版本：

SELECT * 
FROM   table 
WHERE  convert(varchar, table.DataInput, 102) IN 
       (SELECT convert(varchar, table.DataInput, 102) 
        FROM table 
        GROUP BY convert(varchar, table.DataInput, 102) 
        HAVING COUNT(*) > 2)

谢谢大家的投入！：）

获取更多出现多次的记录

8 个答案: