Question

这是我的数据库表结构：

-----------------------------------
|  id |     title    |    date    |
-----------------------------------
| 001 | my event OO1 | 04/02/2014 |
| 002 | my event OO2 | 04/02/2014 |
| 003 | my event OO3 | 04/02/2014 |
| 004 | my event OO4 | 05/02/2014 |
| 005 | my event OO5 | 05/02/2014 |
| 006 | my event OO6 | 05/02/2014 |
| 007 | my event OO7 | 05/02/2014 |
| 008 | my event OO8 | 06/02/2014 |
| 009 | my event OO9 | 06/02/2014 |
-----------------------------------

我开始像它一样显示它：

04/02/2014
- my event OO1
- my event OO2
- my event OO3
05/02/2014
- my event OO4
- my event OO5
- my event OO6
- my event OO7
06/02/2014
- my event OO8
- my event OO9

但我希望每个日期随机显示1个结果（事件），如：

04/02/2014
- my event OO2
05/02/2014
- my event OO6
06/02/2014
- my event OO8

在我的sql结果中，我尝试了ORDER BY RAND() LIMIT 1，但当然我没有按日期排序所以结果是随机排序的

那么，如何对日期进行排序并获得随机结果？

Answer 1

使用：

SELECT d1.date, (
    SELECT d2.title 
    FROM table_1 d2 
    WHERE d2.date = d1.date 
    ORDER BY RAND()
    LIMIT 1
) AS title
FROM table_1 d1 
GROUP BY d1.date
HAVING COUNT(d1.date)>=1
ORDER BY d1.date ASC

您唯一需要更改的是表格的名称。请确保您只在table_1个子句中将FROM列为表格的实际名称。

您的想法是检索日期，然后执行子查询以在该日期内检索随机事件，方法是从同一个表中选择并加入日期。

Answer 2

SELECT distinct t1.date, 
    (SELECT t2.ID, t2.title 
    FROM mytable as t2 
    WHERE t2.date = t1.date 
    ORDER BY RAND()
    LIMIT 1) AS event_title
FROM mytable as t1 
ORDER BY t1.date;

Answer 3

一种方法

SELECT e.* 
  FROM
(
  SELECT
  (
    SELECT id
      FROM events
     WHERE date = d.date
     ORDER BY RAND()
     LIMIT 1
  ) id
    FROM
  (
    SELECT DISTINCT date
      FROM events
  ) d
) q JOIN events e 
    ON q.id = e.id
 ORDER BY date

示例输出：

| ID |        TITLE |                         DATE |
|----|--------------|------------------------------|
|  2 | my event OO2 | April, 02 2014 00:00:00+0000 |
|  4 | my event OO4 |   May, 02 2014 00:00:00+0000 |
|  8 | my event OO8 |  June, 02 2014 00:00:00+0000 |

这是 SQLFiddle 演示

显示按日期排序的随机结果

3 个答案: