This is my MySQL table:
_______________________________________________
Id | Name| Employee id | Date | Attendance
_______________________________________________
1 | xyz | 196 | 2013-04-01 | present
2 | xyz | 196 | 2013-04-02 | present
3 | xyz | 196 | 2013-04-03 | present
4 | xyz | 196 | 2013-04-04 | absent
5 | xyz | 196 | 2013-04-05 | present
6 | abc | 197 | 2013-04-01 | present
7 | abc | 197 | 2013-04-02 | present
8 | abc | 197 | 2013-04-03 | present
9 | abc | 197 | 2013-04-04 | present
10 | abc | 197 | 2013-04-05 | present
_______________________________________________
I want to count a days who employee are mostly present, and I want like this result in PHP:
___________________________________________________________
Name| Employee id| Attendance | Best OR NOT
___________________________________________________________
xyz | 196 | 4 Present days |
abc | 197 | 5 Present days | This is best employees of the year
__________________________________________________________________________
How can i do this?
Try this.
$query = mysql_query("select name ,`Employee id` as emplyee_id, CONCAT(count(`Employee id`),' Present days') as Attendance
from Table1
where `Attendance` = 'present' group by `Employee id` ");
while($row = mysql_fetch_array($query)) {
echo $row['name']. " - " .$row['emplyee_id']. " - " .$row['Attendance']."</ br>";
}