从sql中获取信息并将其放在表单中

Question

因此，我需要从sql中获取信息并将其放在我的表单中的下拉列表中。这就是我所拥有的......我非常迷茫......信息已被预填充到sql中。我认为最重要的部分是相对正确的，然后我不知道如何在表格中引用它。

     <div class="form-group"> 
        <label class='col-xs-4 control-label'>What school did you go to for your undergrad? </label> 
        <div class='col-xs-8'> 
         <select class="form-control background" id='dropdown'> 
          <option>"<?php print $schoolname ?>"</option> 
          <option>"<?php print $schoolname ?>"</option> 
          <option>"<?php print $schoolname ?>"</option> 
          <option>"<?php print $schoolname ?>"</option> 
          <option value="bing">"<?php print $schoolname ?>"</option> 
         </select> 
         <input type="hidden" name="id" id='id' value="<?php print $id ?>"> 
         <input type="hidden" name="editMode" value="edit"> 
        </div> 
       </div 

Answer 1

在这里你去

<?php 
if(!isset($_GET['id']]){ 
    echo 'id= not present in URL. Exiting.'; 
    return false; 
} 
$id = intval($_GET['id']); 

$conn = mysql_connect("localhost", "root", "MIS42520!$") or die (mysql_error()); 

mysql_select_db("assignment 3", $conn); 

$sql = "select * FROM schooltable WHERE id='" . mysql_real_escape_string($id) . "'"; 

$result = mysql_query($sql, $conn) or die(mysql_error()); 

$schools = array(); 
while ($row = mysql_fetch_assoc($result)) { 
    $schools[] = $row; 
} 
?> 

<div class="form-group"> 
    <label class='col-xs-4 control-label'>What school did you go to for your undergrad? </label> 
    <div class='col-xs-8'> 
     <select class="form-control background" id='dropdown'> 
      <?php foreach($schools as $school){?> 
       <option value="<?php echo $school['schoolname'];?>"><?php echo $school['schoolname'];?></option> 
      <?php } ?> 
     </select> 
     <input type="hidden" name="id" id='id' value="<?php echo $id ?>"> 
     <input type="hidden" name="editMode" value="edit"> 
    </div> 
</div