Working CODE :)
Is this the right way to do this? or am i missing some more security code??
What went wrong? the first if condition was: if (isset($_POST['submit']) {
The code..
}
i don't know how but it wouln'd work liek that, so i changed it to the code below and now it works!! :)
<?php
if (isset($_POST['username']) || isset($_POST['locatie']) || isset($_POST['geslacht']) || isset($_POST['online'])) {
$name = mysqli_real_escape_string($server, $_POST['username']);
$waar = mysqli_real_escape_string($server, $_POST['locatie']);
$sex = mysqli_real_escape_string($server, $_POST['geslacht']);
$status = mysqli_real_escape_string($server, $_POST['online']);
$sql= "SELECT * FROM users WHERE 1=1";
if (isset($_POST['username'])) {
$name = $_POST['username'];
$sql .= " and username LIKE '%$name%'";
}
if (isset($_POST['locatie'])) {
$name = $_POST['locatie'];
$sql .= " and locatie LIKE '%$waar%'";
}
if (isset($_POST['geslacht'])) {
$name = $_POST['geslacht'];
$sql .= " and geslacht LIKE '%$sex%'";
}
if (isset($_POST['online'])) {
$name = $_POST['online'];
$sql .= " and online LIKE '%$status%'";
}
$sql .= " ORDER BY RAND()";
$result_set=mysqli_query($server,$sql) or die(mysqli_error($server));
//echo $sql;
echo '<div class="col-sm-12">';
while($row=mysqli_fetch_array($result_set)) {
echo '<div class="col-sm-2">';
echo '<center><img class="img-vrienden" src=' . $row['prof_pic'] . ' /><br>' . $row['username'].'</center>';
echo '</div>';
}
echo '</div>';
}
?>