Using SET in mysqli query - php

Question

I've got this query to work in the editor but now that I transfer the Code into the table code I can't seem to get it to work...
Is there something I'm doing wrong in order for this to work in the php?
Having never used the SET in a mysql query before I assume it has something to do with this, I've tried moving the SET part to separate it to no avail but not sure whether it's just something really simple that I'm missing?
<table data-role="table" id="table-column" data-theme="f" data-mode="table" class="ui-responsive table-stroke table-stripe">
<thead>
<tr>
<th style="text-align: center" data-priority="persist">Date Recorded</th>
<th style="text-align: center" data-priority="1">Max Temp C°</th>
<th style="text-align: center" data-priority="1">rH%</th>
<th style="text-align: center" data-priority="2">Hours</th>
<th style="text-align: center" data-priority="2">Hours Running Total</th>
<th style="text-align: center" data-priority="persist">pH</th>
</tr>
</thead>
<tbody>
<tr>
<?php
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($link, "SET #runtot:= 0;
SELECT q1.dated, max_temp, max_rh, q1.c, (#runtot := #runtot + q1.c) AS rt
FROM
(SELECT datestamp AS dated,
max_temp,
max_rh,
sum(hours) AS c,
ph
FROM hours
WHERE hours > 0 AND username = '$_SESSION[USERNAME]' AND batch_no = '$_SESSION[batch_no]'
GROUP BY dated
ORDER BY dated) AS q1");
if (!$result) {
die("Query to show fields from table failed");
}
$fields_num = mysqli_num_fields($result);
for($i=0; $i<$fields_num; $i++)
{
$field = mysqli_fetch_field($result);
}
while($row = mysqli_fetch_row($result))
{
?>
<td style="text-align: center; vertical-align: middle"><?php echo "$row[0]"?></td>
<td style="text-align: center; vertical-align: middle"><?php echo "$row[1]"?></td>
<td style="text-align: center; vertical-align: middle"><?php echo "$row[2]"?></td>
<td style="text-align: center; vertical-align: middle"><?php echo "$row[3]"?></td>
<td style="text-align: center; vertical-align: middle"><?php echo "$row[4]"?></td>
<td style="text-align: center; vertical-align: middle"><?php echo "$row[5]"?></td>
</tr>
<?php
}
mysqli_free_result($result)
?>
</tbody>
</table>

Answer 1

Simply changing the mysql query worked wonders.
Removing the SET.... as the first line of the code and then as the last line of the mysql, the following:
ORDER BY d) AS q1, (SELECT #runtot:=0) AS n")