als ik het met firebug controlleer staat er dit:
<img src="Images/elearning.png,Images/expertise.png">
Hoe kan ik dit oplossen?
dit is de code: overzicht-typetraining_soort_all
<?php
require_once 'class/all_training_type.php';
$alle_training_type = new type_training();
$alle_type_trainingen = $alle_training_type->select_training_type();
while($a=mysql_fetch_assoc($alle_type_trainingen))
{
echo "<div id='container'><td>";
echo "<div id='left3'>";
echo '<td><b>' . $a['Training'] . '</b></td><br>';
$plaatje = $a['Plaatjes'];
echo '<td>' . $a['Beschrijving'] . '</td><br>';
echo '<br>';
echo '<td>Type training: <b>' . $a['Type'] . '</b></td><br>';
echo "</div>";
echo "<div id='right5'>";
echo "<div class='overzicht-plaatje'> <img src='$plaatje'/> </div> <br>" ;
{
echo "</div>";
echo"</tr></div>";
}
}
?>
all_training_type.php
<?php
require_once 'Singleton.php';
class type_training {
protected $connect;
protected $sql;
public function __construct() {
$this->connect = Singleton::getconnect();
}
public function __destruct() {
$this->connect = null;
}
public function select_training_type() {
$all_type = 'SELECT DISTINCT ( train.naam_training) AS Training, (train.beschrijving) AS Beschrijving, GROUP_CONCAT( typetrain.type_naam ) AS Type , GROUP_CONCAT( typetrain.type_plaatje ) AS Plaatjes
FROM training train
JOIN training_type trainingtype ON trainingtype.training_id = train.id
JOIN type_training typetrain ON typetrain.id = trainingtype.type_id
ORDER BY train.naam_training
';
return $resultaat_all_type=mysql_query($all_type);
}
}
?>
