从日期获取数据

Question

我在数据库中有一些记录。现在我想从最新wise.Please看到我的桌子的下面格式的记录..

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 

我希望得到“阿朗”从2012-08-01到2012年7月8日这样

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  

请帮我看看这个报告。是新的MySQL ..

Answer 1

为了显示丢失的日期，你需要一个临时表。 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 DEMO HERE.

Answer 2

我认为这是你在找什么：

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; 

Answer 3

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'