如何从单个字段中查找超时和超时并将其值提取到SQL中的另一个表

Question

I have 2 tables in PHP MyAdmin, first is tb_data_log that stores all information from data records. I need to find time-in and time-out from first table and then extract the value to second table that is tb_attendance . Look at image bellow:

tb_data_log

uid date time 1 28/01/2017 07.12 1 28/01/2017 16.02 2 28/01/2017 07.05 2 28/01/2017 16.23 3 28/01/2017 07.00 3 28/01/2017 16.16 1 29/01/2017 07.24 1 29/01/2017 16.11 2 29/01/2017 07.09 2 29/01/2017 16.45 3 29/01/2017 07.12 3 29/01/2017 16.02 1 30/01/2017 07.12 1 30/01/2017 16.02 2 30/01/2017 07.29 2 30/01/2017 16.19 3 30/01/2017 07.22 3 30/01/2017 16.56

I need my table to look like this bellow:

tb_attendance

uid date time_in time_out 1 28/01/2017 07.12 16.02 2 28/01/2017 07.05 16.23 3 28/01/2017 07.00 16.16 1 29/01/2017 07.24 16.11 2 29/01/2017 07.09 16.45 3 29/01/2017 07.12 16.02 1 30/01/2017 07.12 16.02 2 30/01/2017 07.29 16.19 3 30/01/2017 07.22 16.56

I think it is basically simple, but I have no idea how to write the codes in MySQL. How am I supposed to do that? Thanks!

Answer 1

This SQL should do the task:

INSERT INTO `tb_attendance` 
SELECT `uid`, 
       `date`, 
       Min(`time`) AS 'time_in', 
       Max(`time`) AS 'time_out' 
FROM   `tb_data_log` 
GROUP  BY `date`, 
          `uid`;

Provided that you have the following field types:

`uid`  INT, 
`date` DATE, 
`time` TIME 

Answer 2

insert into tb_attendance 
    select a.uid,
          a.date,
          a.time as 'time_in',
          (select b.time from tb_data_log b where a.uid = b.uid and b.time >      
          a.time and a.date = b.date limit 1) as 'time_out' 
    from tb_data_log a group by a.uid,a.date order by a.date,a.uid;