PHP calendar with dates --> distinct weeks and merging rows - php

Question

I am currently creating a calendar where you can see all users that are saved in the database for that day and the possibility to add an user per date.
I want to work with dates because I think it is easier to work with in the future.
So what I am doing: I save a date into the database combined to an userid.
While I am fetching the data from database into a table I am converting the dates to days and weeks and years. Just to see which day is at 2015/10/05.
So I get $day = Monday, $week = 41, $year = 2015
In the table I want to put the users to the matching day.
The calendar looks like this:
I am calling an if statement:
if ($day == 'Mon') {
// SELECT * FROM users WHERE datum=$date
// Fetch and echo the users name into that day
}
The database looks like this:
The weeks showed here in the database, maybe they can be used for a proper display? But I don't know I tried several things.
I am using this code for putting the database rows into a table:
$sql = "SELECT distinct(week),datum, van , tot FROM roosternew ORDER BY datum ASC";
$run_sql = mysqli_query($conn , $sql);
?>
<table class="table">
<thead>
<tr class="primary">
<th width="5%"><center>Week</center></th>
<th width="10%">Maandag</th>
<th width="10%">Dinsdag</th>
<th width="10%">Woensdag</th>
<th width="10%">Donderdag</th>
<th width="10%">Vrijdag</th>
<th width="10%">Zaterdag</th>
<th width="10%">Zondag</th>
</tr>
</thead>
<tbody>
<?php while ($row = mysqli_fetch_assoc($run_sql)) {
echo $row['datum'].'test';
$datum = $row['datum'];
$van = $row['van'];
$tot = $row['tot'];
$tijd = $van.' - '.$tot;
$date = new DateTime($datum);
$week = $date->format("W");
$dag = $date->format("D");
$jaar = $date->format("Y");
$newdate_ = new DateTime($datum);
$newdate = $newdate_->format('Y-m-d');
$week_start = new DateTime();
$week_start->setISODate($jaar,$week);
$thisdate_ = $week_start->format('d-m');
//Create plus button
$plus_button = ' <i class="fa fa-plus-square"></i><br/>';
//Plus button
$thisdate = '<div class="datetable"><div class="datetable">'.$thisdate_.'</div></div><br/>';
?>
<tr>
<td class="primary"><center><strong><?= $week; ?></strong><br/><?php echo $thisdate.'<h4><small>'.$jaar.'</small></h4>'; ?></center></td>
<td>
<?php
createPlusButton(0, $jaar,$week);
if ($dag == 'Mon') {
getScheduleAdmin($newdate);
}
?>
</td>
<td>
<?php
createPlusButton(1, $jaar,$week);
if ($dag == 'Tue') {
getScheduleAdmin($newdate);
}
?>
</td>
<td>
<?php
createPlusButton(2, $jaar,$week);
if ($dag == 'Wed') {
getScheduleAdmin($newdate);
}
?>
</td>
<td>
<?php
createPlusButton(3, $jaar,$week);
if ($dag == 'Thu') {
getScheduleAdmin($newdate);
}
?>
</td>
<td>
<?php
createPlusButton(4, $jaar,$week);
if ($dag == 'Fri') {
getScheduleAdmin($newdate);
}
?>
</td>
<td>
<?php
createPlusButton(5, $jaar,$week);
if ($dag == 'Sat') {
getScheduleAdmin($newdate);
}
?>
</td>
<td>
<?php
createPlusButton(6, $jaar,$week);
if ($dag == 'Sun') {
getScheduleAdmin($newdate);
}
?>
</td>
</tr>
<? } ?>
</tbody>
</table>
The function getScheduleAdmin($newdate):
function getScheduleAdmin($newdate) {
require('connect.php');
$sql2 = "SELECT * FROM roosternew WHERE datum='$newdate'";
$run_sql2 = mysqli_query($conn , $sql2);
while ($row2 = mysqli_fetch_assoc($run_sql2)) {
$id_rooster_users = $row2['idusers'];
$sql= mysqli_query($conn,"SELECT * FROM users WHERE idusers='$id_rooster_users'");
$row = mysqli_fetch_array($sql);
$name = $row['name'];
$color = $row['color'];
$sql3 = "SELECT van,tot FROM roosternew WHERE idusers= '$id_rooster_users'";
$sql3_run = mysqli_query($conn,$sql3);
$sql3_array = mysqli_fetch_array($sql3_run);
$van_user = $sql3_array['van'];
$tot_user = $sql3_array['tot'];
$tijd_user = $van_user.' - '.$tot_user;
echo ''.$name.'<br/>'.'('.$tijd_user.')';
echo '<br/>';
}
}
The problem is described in the first image (the calendar), it will show me more than once the week and they won't merge.
I dont know how to merge the weeks so that it will be :
Something like that.
And not like this (situation right now):
if anyone knows a better way of doing this, please tell me.