Display unlimited row by random ID selected by user (created by php) - php

Question

I need to display row from different table with random id that what i make it php generate it when i add new car or new driver
For example :
In this image table record's for same driver and different Car
and in this image the records for random id for the car
I need to show every records by date for same driver and Which car he use
Note: In this case its show like this
and I need it show like this
and sorry for arabic database records tell me if there anything not understood
my code:
<?php
require 'config.php';
$query = "SELECT * FROM `driversuser`";
$result1 = mysqli_query($con, $query);
?>
<div class="container">
<div class="addingcar">
<form dir="rtl" action="#" method="post">
<div class="">
<b>عرض وردية سائق</b>
</div>
<br/>
<label>
<b>اختر السائق</b>
<select name="insdname" id="framework" class="align-right selectpicker" data-live-search="true">
<option style="text-align: right" value="">اختر السائق ...</option>
<?php while($row1 = mysqli_fetch_array($result1)):; ?>
<option style="text-align: right" value="<?php echo $row1[10]; ?>"><?php echo $row1[1]; ?></option>
<?php endwhile; ?>
</select>
</label>
<br />
<label>
<b>من</b>
<input type="date" name="datefrom">
</label>
<br />
<label>
<b>الي</b>
<input type="date" name="dateto">
</label>
<br /><br />
<input type="hidden" name="hidden_framework" id="hidden_framework" />
<input type="submit" name="submit" class="btn btn-info" value="عــــرض" />
</form>
<?php
if(isset($_POST['submit'])){
$selected_val = $_POST['insdname'];
$date_val = $_POST['datefrom'];
$dateto_val = $_POST['dateto'];
$report = mysqli_query($con, "SELECT * FROM `reports` WHERE Datefrom BETWEEN '$date_val' AND '$dateto_val' AND DriverCode = '$selected_val' ORDER BY Datefrom");
while ($datareport = mysqli_fetch_array($report)) {
echo $datareport['Datefrom']." ".$datareport['DateTo']." ".$datareport['Price']." ".$datareport['PriceTaken']." ";
$report2 = mysqli_query($con, "SELECT * FROM `reports` WHERE DriverCode = '$selected_val' GROUP BY CarCode");
while ($datareport2 = mysqli_fetch_array($report2)) {
$carcode = $datareport2['CarCode'];
}
$report3 = mysqli_query($con, "SELECT * FROM `cars` WHERE Car_ID = '$carcode'");
while ($datareport3 = mysqli_fetch_array($report3)) {
echo $datareport3['CarName']." ".$datareport3['CarModel']." ".$datareport3['CarNumber']." ".$datareport3['CarColor']." ";
echo '<br/>';
}};
}
?>
</div>
</div>

Answer 1

I don't have your database schema, but I tried to create one for testing. So, try this:
SELECT * FROM reports r1 INNER JOIN cars c1 ON r1.carcode = c1.car_id ORDER BY r1.date
r1 is a shortcut for reports table ( change it to whatever you like ).
c1 is a shortcut for cars table ( also, you can change it ).
You can read more about INNER JOIN.