PHP MySQL表单选择返回

I am struggling to get this code to work, what I want from it is to show the item (that is already in the database) to be selected in the selection form.

<label>Server Ports:</label>
<select multiple class="form-control" name="select[]">
<?php 
// Get Server Information
    $query = "SELECT port_no FROM _servers WHERE (server_id = '$servid') ";
    $result = mysql_query($query) or die ('Unable to run query:'.mysql_error());
    while($row = mysql_fetch_assoc($result)){
        $no = $row['port_no'];
    }                               
    $query = "SELECT id, name, port_no, unique_id FROM ports ORDER BY name ASC";
    $result = mysql_query($query) or die ('Unable to run query:'.mysql_error());
    while($row = mysql_fetch_assoc($result))
    {
        $port_no = $row['port_no'];
        $port_name = $row['name'];
        $p_unique = $row['unique_id'];
    }
?>
<option value="<?php echo $p_unique;?>"
<?php
    if ($p_unique == $no) {
        $check = 'selected';
    } else {
        $check = '';
    }
    echo $check;
?>
>
<?php
    echo $row['name'];
?>
(
<?php
    echo $row['port_no'];
?>
)</option>
<?php
    }
?>
</select>

Are you sure your test must be if ($p_unique == $no) {} and not if ($port_no == $no) {} ?

If yes, try to checks your variables values :

Do a var_dump() of $no :

var_dump($no);

In your while() loop, you can also check values of $p_unique and $no like that :

var_dump($p_unique.'/'.$no);

Also, here is a simplest way to test for selected :

<option value="<?php echo $p_unique;?>" <?php if ($p_unique == $no) echo 'selected="selected"'; ?>><?php echo $row['name'] ;?> (<?php echo $row['port_no'] ;?>)</option>