how to find out missing date in mysql using php - php

Question

How to find out missing date in MySQL using PHP?
echo "<tr> <th>Username</th><th>Date</th><th>Check In</th><th>Check Out</th> </tr>";
// get results1 from database
$result1 = mysql_query("SELECT * FROM attend WHERE user_name='ali' AND date BETWEEN '2015-07-01' AND '2015-07-15' order by date");
while($row = mysql_fetch_array($result1))
{
// echo out the contents of each row into a table
echo "<tr>";
echo '<td>' . $row['user_name'] . '</td>';
echo '<td>' . $row['date'] . '</td>';
echo '<td>' . $row['checkin'] . '</td>';
echo '<td>' . $row['checkout'] . '</td>';
echo "</tr>";
}
echo "</table>";
currently result
Username Date Check In Check Out
ali 2015-07-01 11:30:34 17:23:47
ali 2015-07-02 10:11:34 17:15:15
ali 2015-07-03 09:32:34 18:16:27
ali 2015-07-06 10:41:34 16:56:13
ali 2015-07-07 08:51:34 17:36:01
ali 2015-07-08 05:61:34 17:16:26
ali 2015-07-09 04:11:34 17:14:12
ali 2015-07-10 02:81:34 17:25:25
ali 2015-07-13 11:71:34 17:02:29
ali 2015-07-14 10:81:34 17:04:20
ali 2015-07-15 09:31:34 17:00:43
and i want result like this
Username Date Check In Check Out
ali 2015-07-01 11:30:34 17:23:47
ali 2015-07-02 10:11:34 17:15:15
ali 2015-07-03 09:32:34 18:16:27
2015-07-04
2015-07-05
ali 2015-07-06 10:41:34 16:56:13
ali 2015-07-07 08:51:34 17:36:01
ali 2015-07-08 05:61:34 17:16:26
ali 2015-07-09 04:11:34 17:14:12
ali 2015-07-10 02:81:34 17:25:25
2015-07-11
2015-07-12
Ali 2015-07-13 11:71:34 17:02:29
ali 2015-07-14 10:81:34 17:04:20
ali 2015-07-15 09:31:34 17:00:43

Answer 1

All of your missing check ins are not assigned to any user. If you want to fetch records assigned to user ali or unassigned, between gives dates:
SELECT * FROM attend WHERE (user_name='ali' OR user_name = '') AND date BETWEEN '2015-07-01' AND '2015-07-15' order by date
It would be much cleaner if you show us your schema.

Answer 2

I'd use timestamp for that date column. If you do this, you can simply "check" which day is expected to come next. If its not that day, print that day like you wanted and then continue, if you got that expected date.
Do do so, you should save all your timestamps at a given time of that day. Like always 12:00:00. This makes handling those numbers much easier.
echo "<tr> <th>Username</th><th>Date</th><th>Check In</th><th>Check Out</th> </tr>";
// get results1 from database
$iLastTimestamp = 0
$result1 = mysql_query("SELECT * FROM attend WHERE user_name='ali' AND date BETWEEN '2015-07-01' AND '2015-07-15' order by date");
while($row = mysql_fetch_array($result1))
{
if ($iLastTimestamp != 0)
{
while ($iLastTimestamp != $row['date'])
{
// echo out the contents of each row into a table
echo "<tr>";
echo '<td> </td>';
echo '<td>' . date("Y-m-d", $iLastTimestamp) . '</td>';
echo '<td> </td>';
echo '<td> </td>';
echo "</tr>";
$iLastTimestamp = strtotime("+1 Day 12:00:00", $iLastTimestamp)
}
}
// echo out the contents of each row into a table
echo "<tr>";
echo '<td>' . $row['user_name'] . '</td>';
echo '<td>' . $row['date'] . '</td>';
echo '<td>' . $row['checkin'] . '</td>';
echo '<td>' . $row['checkout'] . '</td>';
echo "</tr>";
}
echo "</table>";
Another way is to convert your date into an timestamp first, before using the code I posted. You can use the "strtotime"-Func here, too. Just add the time, just to be sure. :)

Answer 3

You can do this with a range in a function, this will permit you to reuse your code:
function getAttendInInterval($begin,$end) {
$begin = new DateTime( $begin );
$end = new DateTime( $end );
$interval = DateInterval::createFromDateString('1 day');
$period = new DatePeriod($begin, $interval, $end);
foreach ( $period as $dt ) {
$result = mysql_query('SELECT * FROM attend WHERE user_name="ali" AND date = '".$dt->format( 'Y-m-d' )."' order by date');
...
}
This fonction permits to show the attend for each day in an interval.
and then:
getAttendInInterval('2015-07-01','2015-07-15');

Answer 4

you can use date in LOOP and if date is match in database then show all info. of database with date otherwise show only date.