3

我在数据库中有一些记录。现在我想从日期获得记录。请参阅我的表格的以下格式。

 User_id    name    date         Present
  001      Jone    01-08-2012     yes
  002      arun    01-08-2012     yes
  001      Jone    02-08-2012     yes
  002      arun    02-08-2012     yes
  001      Jone    03-08-2012     yes
  001      Jone    04-08-2012     yes
  002      arun    05-08-2012     yes
  001      Jone    05-08-2012     yes
  001      Jone    06-08-2012     yes
  001      arun    06-08-2012     yes

我想像这样从 01-08-2012 到 07-08-2012 获得“arun”报告

 User_id    name    date         Present
  002      arun    01-08-2012     yes
  002      arun    02-08-2012     yes
  002      arun    03-08-2012    
  002      arun    04-08-2012     
  002      arun    06-08-2012     yes
  002      arun    07-08-2012
4

3 回答 3

1

为了显示丢失的日期，您需要一个临时表。临时表：

CREATE TABLE numbers (number INTEGER);
INSERT INTO numbers VALUES (0), (1), (2), (3), (4), (5), (6), (7), (8), (9);

sql如下：

SELECT CASE WHEN u.id IS NOT NULL 
       THEN u.id
       ELSE u2.id
END AS id ,
CASE WHEN u.name IS NOT NULL 
       THEN u.name
       ELSE u2.name
END AS name ,
cal.date, 
CASE WHEN u.present IS NOT NULL 
       THEN u.present
       ELSE ''
END AS present 
FROM (
    SELECT '2012-08-01' + INTERVAL a.number * 10 + b.number DAY as date
    FROM numbers a JOIN numbers b
    ORDER BY a.number * 10 + b.number
) cal LEFT JOIN (select * from users where name='arun') u 
ON DATE_FORMAT(STR_TO_DATE(u.date1, '%d-%m-%Y'), '%Y-%m-%d') = cal.date
left join (select id,name from users  group by id)u2 on u.id is null and u2.name='arun'
WHERE date BETWEEN '2012-08-01' AND '2012-08-07';

SQL 演示在这里。

于 2012-08-07T07:30:25.163 回答
0

我认为这就是你要找的：

SELECT User_id, name, `date`,
       IF(MAX(Present) = 1, 'yes', 'no')  AS Present
FROM
(
 SELECT User_id, name, `date`, IF('yes', 1, 0) Present
 FROM table_name
 WHERE `date` BETWEEN '2012-08-01' AND '2012-08-07' AND
       name = 'arun'

 UNION ALL

 SELECT '002' AS User_id, 'arun' AS name, a.`date`, 0 AS Present
 FROM  (
        SELECT CURDATE() - INTERVAL (a.a + (10 * b.a) + (100 * c.a)) DAY AS date
        FROM (SELECT 0 AS a UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9) AS a
        CROSS JOIN (SELECT 0 AS a UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9) AS b
        CROSS JOIN (SELECT 0 AS a UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9) AS c
       ) a
 WHERE a.date BETWEEN '2012-06-05' AND '2012-06-20';
) a
GROUP BY name, `date`
ORDER BY `date` ASC;
于 2012-08-07T06:31:51.407 回答
0 
select * from (

    select id, name, date, present
      FROM your_table 
     WHERE your_table.name = 'arun'
     union all
    select (select id from your_table where name = 'arun'), 'arun', all_dates.date, ''
      from (select distinct date from your_table where your_table.name != 'arun')
      left join your_table
        ON all_dates.date = your_table.date
       AND your_table.name = 'arun'
     WHERE your_table.id IS NULL

) subquery
order by date desc
where date between '01-08-2012' and '07-08-2012'
于 2012-08-07T06:47:05.227 回答