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Akshay Homkar
NA
19
17.3k
Pivot Table or Cross Tab in PHP using MYSQL for attendance
Jan 15 2014 11:29 AM
Hello
I want to create attendance sheet on which date are printed as column and name of student/staff as column
database is as shown
CREATE TABLE IF NOT EXISTS `attendance` (
`date` varchar(500) DEFAULT NULL,
`time` varchar(1000) DEFAULT NULL,
`staffname` varchar(1000) DEFAULT NULL,
`id` int(11) DEFAULT NULL,
`role` varchar(1000) NOT NULL,
`status` varchar(1) DEFAULT NULL
)
but when I querying the table repeating the names of student and staff which not gives me report as expected here I attaching code also
<table align="letf" style="margin-left: 0px; border: 1px solid black; border-spacing: 0px;" width="8">
<th style="border: 1px solid black; text-align: center;">Date</th>
<?php
$sql133="select distinct date from attendance";
$sql_row133=mysqli_query($dbConn,$sql133);
while($sql_res133=mysqli_fetch_assoc($sql_row133)) {
$date=$sql_res133["date"];
?>
<th style="border: 1px solid black; text-align: center;">
<?php echo $date; ?>
</th>
<?php
$a=$date;
$sql13="
SELECT atten.date,atten.time,atten.staffname,atten.id, atten.status, supst.id, supst.staffname
FROM
(examcenter.attendance atten INNER JOIN examcenter.supportstaff supst ON
atten.id = supst.id) where atten.date='$a' group by supst.staffname,supst.id ORDER BY
atten.id ASC ";
$sql_row13=mysqli_query($dbConn,$sql13);
while($sql_res13=mysqli_fetch_assoc($sql_row13)) {
$staffname=$sql_res13["staffname"];
$status=$sql_res13["status"];
?>
<tr>
<td><?php echo $staffname; ?></td>
<td><?php echo $status; ?></td>
<?php
}
}
?>
</table>
please guide what do to create a report as expected
Name/Date
12-11-2013
13-11-2013
16-11-2013
Student name1
P
A
A
Staffname 1
P
P
A
awaiting valuable reply
Reply
Answers (
1
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