显示从mysql表中检索到的下拉列表中的正确值

Question

I have a dropdown list which should be displaying the correct value of the record retrieved from mysql table. Below is what I have tried so far:

   <strong>Authority Id: *</stong><select name="authid">
   <?php

            $authid = $row['AuthorityId'];
    $selectedId = array(
    5, 6, 7);
    $selection = array(
            5 => "Admin",
            6 => "Employee",
            7 => "Student" );
    foreach($selection as $value){
        $text = $value;
        echo '<option value="'.$selectedId.'" selected="'.$authid.'">'.$text.'</option>';
    }
  ?>
  </select>

But it's not displaying the correct value. Can someone help me figure out what went wrong here? Thanks.

Answer 1

<strong>Authority Id: *</strong><select name="authid">
<?php

$authid = $row['AuthorityId']; // Get $authid from database
$selection = array( // Create Index Of AuthIDs and AuthNames
    5 => "Admin",
    6 => "Employee",
    7 => "Student" );

foreach($selection as $key => $value) // Loop Through $selection, Where $key is AuthID and $value is AuthName
{
    echo '<option value="' . $key . '"'; // Start Menu Item
    if ($authid == $key) // Check If AuthID from $selection equals $authid from database
        echo ' selected="selected"'; // Select The Menu Item If They Match
    echo '>' . $value . '</option>'; // End Menu Item
}
?>
</select>