合并具有唯一属性的两行

This is a sample output of my table(time_entries)

user_name| entry_type   | entry_datetime
 User1   |   Time In    | 28-JUL-13  16:40:40
 User1   |   Time Out   | 28-JUL-13  16:40:41
 User2   |   Time In    | 28-JUL-13  16:41:13
 User2   |   Time Out   | 28-JUL-13  16:41:15
 User1   |   Time In    | 27-JUL-13  16:42:30
 User1   |   Time Out   | 27-JUL-13  16:42:34
 User2   |   Time In    | 27-JUL-13  16:43:32
 User2   |   Time Out   | 27-JUL-13  16:43:35

Now I used this query

SELECT te.user_name, te.entry_name, MAX(te.entry_datetime) AS date
FROM time_entries AS te
GROUP BY te.entry_type, te.user_name

and the result is this

user_name| entry_type   | entry_datetime
 User1   |   Time In    | 28-JUL-13  16:40:40
 User1   |   Time Out   | 28-JUL-13  16:40:41
 User2   |   Time In    | 28-JUL-13  16:41:13
 User2   |   Time Out   | 28-JUL-13  16:41:15

QUESTION: Is there a way to merge username with the same name in 1 line? and have an output something like this.

user_name|  Date     | Timein   | Timeout
User1    | 28-JUL-13 | 16:40:40 | 16:40:41
User2    | 28-JUL-13 | 16:41:13 | 16:41:15

I'm a little confuse how to do it.

I would approach it in this manner:

1. Create a query that returns the "time in" time for each user and date.
2. Create a query like #1 that returns the "time out" time for each user and date.
3. Merge the results together using LEFT JOIN.

The end result:

SELECT te.user_name, LEFT(te.entry_datetime, 10) AS entry_date, cin.entry_datetime AS timein, cout.entry_datetime AS timeout
FROM time_entries te
LEFT JOIN (SELECT user_name, LEFT(entry_datetime, 10) AS entry_date, MIN(entry_datetime) AS entry_datetime
    FROM time_entries
    WHERE entry_type = 'Time In'
    GROUP BY user_name, entry_date
) cin ON cin.user_name = te.user_name AND cin.entry_date = LEFT(te.entry_datetime, 9)
LEFT JOIN (SELECT user_name, LEFT(entry_datetime, 10) AS entry_date, MAX(entry_datetime) AS entry_datetime
    FROM time_entries
    WHERE entry_type = 'Time Out'
    GROUP BY user_name, entry_date
) cout ON cout.user_name = te.user_name AND cout.entry_date = LEFT(te.entry_datetime, 10)
GROUP BY te.user_name, entry_date;

Demo

Btw, I'm using LEFT() here because your column doesn't contain an actual date format that MySQL understands.

What about this?

select
user_name,
date(entry_datetime) as date,
min(entry_datetime) as Timein,
max(entry_datetime) as Timeout
from (
  SELECT te.user_name, te.entry_name, MAX(te.entry_datetime) AS date
  FROM time_entries AS te
  GROUP BY te.entry_type, te.user_name) src
group by user_name, date

Here is a php side solution on how you can process the data.

function transformArray($data) {
   $final = array();
   foreach ($data as $key => $row) {
      if (!isset($final[$row['u']])) $final[$row['u']] = array('user' => $row['u']);
      $final[$row['u']][$row['t']] = date("H:i:s", strtotime($row['entrydatetime']));
      $final[$row['u']]['date'] = date("d:M:y", strtotime($row['entrydatetime']));
   }

   return $final;
}

You can pass it a data set that looks like this, you can adjust the field names as nessecery.

$data = array(
         array('u' => 'user1', 't' => 'TimeIn', 'entrydatetime' => '28-JUL-13  16:40:40'),
         array('u' => 'user1', 't' => 'TimeOut', 'entrydatetime' => '28-JUL-13  16:40:40'),
         array('u' => 'user2', 't' => 'TimeIn', 'entrydatetime' => '28-JUL-13  16:40:40'),
         array('u' => 'user2', 't' => 'TimeOut', 'entrydatetime' => '28-JUL-13  16:40:40')
);

This will do the job:

SELECT te.user_name,DATE(te.entry_datetime) as Date,te.entry_datetime AS TimeIn,te2.entry_datetime AS TimeOut
FROM time_entries te
JOIN time_entries te2 ON te.user_name=te2.user_name AND te2.entry_type="Time Out" AND
  te2.entry_datetime=
  (
    SELECT 
      entry_datetime
    FROM 
      time_entries
    WHERE entry_type='Time Out' AND user_name=te.user_name AND entry_datetime > te.entry_datetime
    ORDER BY 
      ABS(TIMESTAMPDIFF(SECOND, te.entry_datetime, `entry_datetime`))
    LIMIT 1
  )
WHERE te.entry_type="Time In";

Edit: a little explanation:

It selects the user_name, the date part and the entry_datetime for the TimeIn part. Then it joins on the same table to fetch the Time Out datetime part. It does this by selecting the entry with type "Time Out" with the same user_name and the date that is closest to the Time In datetime.

select
user_name,
dATE(entry_datetime) AS date1,
min(entry_datetime) as Timein,
max(entry_datetime) as Timeout
from time_entries
group by user_name, date1

SqlFiddleDemo

This Query will work appropriately when for every Time_In there is corresponding timeout. and each and every new day entry type should start with time_in