获取当月的所有日期

Question

i have table

userID | date | time
===================
1 | 2015-02-08 | 06:32
1 | 2015-02-08 | 05:36
1 | 2015-02-08 | 17:43
1 | 2015-02-08 | 18:00
1 | 2015-02-09 | 06:36
1 | 2015-02-09 | 15:43
1 | 2015-02-09 | 19:00
1 | 2015-02-10 | 05:36
1 | 2015-02-10 | 17:43
1 | 2015-02-10 | 18:00
2 | 2015-02-08 | 06:32
2 | 2015-02-08 | 05:36
2 | 2015-02-08 | 17:43
2 | 2015-02-08 | 18:00
2 | 2015-02-09 | 06:36
2 | 2015-02-09 | 15:43
2 | 2015-02-09 | 19:00
2 | 2015-02-10 | 05:36
2 | 2015-02-10 | 17:43
2 | 2015-02-10 | 18:00

But i want the number of records returned to be exactly the same as the number of days of the current month and get min time for in and max time for the out. if the current month has 28 days and only had two records it should bring:

userID | date | in | out
========================
1 | 2015-02-01 | |
1 | 2015-02-02 | |
1 | 2015-02-03 | |
1 | 2015-02-04 | |
1 | 2015-02-05 | |
1 | 2015-02-06 | |
1 | 2015-02-07 | |
1 | 2015-02-08 | 06:32 | 18:00
1 | 2015-02-09 | 06:36 | 19:00
1 | 2015-02-10 | 05:36 | 18:00
1 | 2015-02-11 | |
1 | 2015-02-12 | |
1 | 2015-02-13 | |
1 | 2015-02-14 | |
1 | 2015-02-15 | |
1 | 2015-02-16 | |
1 | 2015-02-17 | |
1 | 2015-02-18 | |
1 | 2015-02-19 | |
1 | 2015-02-20 | |
1 | 2015-02-21 | |
1 | 2015-02-22 | |
1 | 2015-02-23 | |
1 | 2015-02-24 | |
1 | 2015-02-25 | |
1 | 2015-02-26 | |
1 | 2015-02-27 | |
1 | 2015-02-28 | |

How can i modify my query to achieve the above result?
this is my query:

$sql = "SELECT 
                    colUserID, 
                    colDate, 
                    if(min(colJam) < '12:00:00',min(colJam), '') as in, 
                    if(max(colJam) > '12:00:00',max(colJam), '') as out
                FROM tb_kehadiran
                    WHERE colDate > DATE_ADD(MAKEDATE($tahun, 31),
                    INTERVAL($bulan-2) MONTH)
                    AND
                    colDate < DATE_ADD(MAKEDATE($tahun, 1),
                    INTERVAL($bulan) MONTH)
                    AND
                    colUserID = $user_id
        GROUP BY colUserID,colDate";

Answer 1

I had to think about this one. But probably the simpliest answer so far:

WITH AllMonthDays as (
    SELECT n = 1
    UNION ALL
    SELECT n + 1 FROM AllMonthDays WHERE n + 1 <= DAY(EOMONTH(GETDATE())) 
)

SELECT 
    DISTINCT datefromparts(YEAR(GETDATE()), MONTH(GETDATE()), n) As dates
    , MIN(d.time) as 'In'
    , MAX(d.time) as 'Out'
FROM AllMonthDays as A
    LEFT OUTER JOIN
    table as d on 
        DAY(d.date) = A.n
GROUP BY n,(d.date);

--- Test and tried in this environment: ---

use Example;
CREATE TABLE demo (
     ID int identity(1,1)
    ,date date
    ,time time
    );

INSERT INTO demo (date, time) VALUES 
    ('2015-12-08', '06:32'),
    ('2015-12-08', '05:36'),
    ('2015-12-08', '17:43'),
    ('2015-12-08', '18:00'),
    ('2015-12-09', '06:36'),
    ('2015-12-09', '15:43'),
    ('2015-12-09', '19:00'),
    ('2015-12-10', '05:36'),
    ('2015-12-10', '17:43'),
    ('2015-12-10', '18:00')
    ;

WITH AllMonthDays as (
    SELECT n = 1
    UNION ALL
    SELECT n + 1 FROM AllMonthDays WHERE n + 1 <= DAY(EOMONTH(GETDATE())) 
)

SELECT 
    DISTINCT datefromparts(YEAR(GETDATE()), MONTH(GETDATE()), n) As dates
    , MIN(d.time) as 'In'
    , MAX(d.time) as 'Out'
FROM AllMonthDays as A
    LEFT OUTER JOIN
    demo as d on 
        DAY(d.date) = A.n
GROUP BY n,(d.date);

DROP table demo;

Answer 2

The way I've approached this problem in the past is to have a date table that is pre-populated for some years in the future.

You could create such a table, possibly defining columns for year, month and date, with indexes on year and month.

You can then use this table with a JOIN on your data to ensure that all dates are present in your results.

Answer 3

You need three things:

• A list of dates.
• A left join
• Aggregation

So:

select d.dte, min(t.time), max(t.time)
from (select date('2015-02-01') as dte union all
      select date('2015-02-02') union all
      . . 
      select date('2015-02-28')
     ) d left join
     t
     on d.dte = t.date
group by d.dte
order by d.dte;

Answer 4

Try this

set @is_first_date = 0;
set @temp_start_date =  date('2015-02-01');
set @temp_end_date =  date('2015-02-28');

select my_dates.date,your_table_name.user_id, MIN(your_table_name.time), MAX(your_table_name.time)  from 
( select if(@is_first_date , @temp_start_date := DATE_ADD(@temp_start_date, interval 1 day), @temp_start_date) as date,@is_first_date:=@is_first_date+1    as start_date from information_schema.COLUMNS 
where @temp_start_date < @temp_end_date limit 0, 31
) my_dates left join your_table_name on 
my_dates.date = your_table_name.date
group by my_dates.date

Answer 5

Try This query

 SELECT `date`, MIN(`time`) as `IN`, MAX('time') AS `OUT` 
   FROM `table_name` WHERE month(current_date) = month(`date`)
   GROUP BY `date`;