Find previously selected option in while loop - php

Question

I'm trying to get a drop down menu to keep its selected value when the user hits submit, but it fails due to errors on the form.
I have a while loop returning values from a database to build the options for the drop down, but how do I echo "selected" on the right option?
I have tried if($district == $row["name"]) { echo "selected";} as you see below, but it doesn't work.
<?php
$result = mysql_query("SELECT dist.name FROM districts AS dist JOIN int_bd AS ibd ON dist.id = ibd.districts_id WHERE banners_id = 6 GROUP BY dist.id ORDER BY dist.id ASC", $connection);
if (!result) {
die("Database query failed: " . mysql_error());
}
while ($row = mysql_fetch_array($result)) {
echo '<option value="{$row["name"]}"'; if($district == $row["name"]) { echo "selected";} ; echo '>' . $row["name"] . "</option>";
}
?>
Sorry for the delay. None of the suggested answers worked for me. Any other ideas?

Answer 1

Can you try this,
<?php
$result = mysql_query("SELECT dist.name FROM districts AS dist JOIN int_bd AS ibd ON dist.id = ibd.districts_id WHERE banners_id = 6 GROUP BY dist.id ORDER BY dist.id ASC", $connection);
if (!result) {
die("Database query failed: " . mysql_error());
}
$district = $_REQUEST['name']; // You need pass the value you have been submitted
while ($row = mysql_fetch_array($result)) {
$selected ="";
if(trim($district) == trim($row["name"])) { $selected = "selected";}
echo '<option value="{$row["name"]}" '.$selected.' >' . $row["name"] . "</option>";
}
?>

Answer 2

Try this..
if($district == $row["name"])
{
echo "<option value='$district' selected>$district</option>";
}

Answer 3

I just found the answer. This is what I did:
while ($row = mysql_fetch_array($result)) {
echo '<option value="' . $row["name"] . '"';
if($row["name"] == $district) { echo 'selected';} ;
echo '>' . $row["name"] . '</option>';
}
It seems to have been this line
echo '<option value="{$row["name"]}"';
that was causing the problem.