holding value in dropdown after submitting or refreshing - php

Question

<?php
$sql=mysqli_query($mysqli,"select * from calls where eid='$aid'")or die(mysqli_error($mysqli));
while($row2=mysqli_fetch_assoc($sql))
{
?><select name='city'>
<option value="">Select City</option>
<?php
$c=explode(',', $row2['city']);
foreach ($c as $c1)
{
?>
<option value='$c1' <?php if(isset($_POST['search']) || isset($_POST['next'])|| isset($_POST['submit']))
{
if(($c==$c1))
{ echo 'Selected';}
}?>><?php echo $c1;?></option>"
<?php }
echo"</select>";
}
?>
calls table city is stored as comma separated array. I have successfully get it into drop-down box. Now I want that whenever I submit or page refresh it holds se selected drop-down value.

Answer 1

You should check for isset($_POST["city"]) && $_POST["city"] == $city
foreach ($cities as $city)
{
$selection = (isset($_POST["city"]) && $_POST["city"] == $city ? 'selected':'' );
echo ' <option value="'.$city.'" '.$selection.' >'.$city.'</option>';
}

Answer 2

You only need to compare the value of $c1 against its previous value returned by the form submit, after checking that this execution is the result of the form being posted.
<?php
$sql = "select * from calls where eid=?";
$stmt = $mysqli->prepare($sql);
$stmt->bind_param('s', $aid)
$stmt->execute();
$result = $stmt->get_result();
while($row2 = $result->fetch_assoc()) {
?>
<select name='city'>
<option value="">Select City</option>
<?php
$c=explode(',', $row2['city']);
foreach ($c as $c1) {
$sel = ''; // init value
if ( $_SERVER['REQUEST_METHOD'] == 'POST' && $c1 == $_POST['city'] ) {
$sel = "selected='selected'";
}
echo "<option $sel value='$c1' >$c1</option>";
}
echo "</select>";
}
?>