为什么我的动态下拉列表不起作用？

Question

The drop down list is suppose to fetch records from a column in another table but currently there are no records appearing. Also, i would need an option in the drop down list that says "others" for the users to key in if they cannot find what they want in the list. Here are my codes:

<script type="text/javascript">
    function showfield(name){
    if(name=='Other')document.getElementById('div1').innerHTML='Please Specify: <input type="text" name="other" />';
    else document.getElementById('div1').innerHTML='';
    }
    </script>   
<label for="issue_type">Issue Type</label>
<?php 
include ("../db/dbConn.php"); 
$sql = "SELECT issue_type FROM issue where deleted =0"; 
$result=mysql_query($sql);
echo '<select class ="form-control" type="text" name="issue_type" id="issue_type" onchange="showfield(this.options[this.selectedIndex].value)" >';
 while ($row = mysql_fetch_array($result)) 
{ 
 echo "<option value='".$row['issue_type']."'>".$row['issue_type']."      </option>"; 
    } 
   echo "</select>";

?>
<div id="div1"></div>

Answer 1

assign html section to a variable, then echo it. try

<script type="text/javascript">
function showfield(name){
if(name=='Other')document.getElementById('div1').innerHTML='Please Specify: <input type="text" name="other" />';
else document.getElementById('div1').innerHTML='';
}
</script>   
<label for="issue_type">Issue Type</label>
<?php 
include ("../db/dbConn.php"); 
$sql = "SELECT issue_type FROM issue where deleted =0"; 
$result=mysql_query($sql);

$htm = '';
$htm .='<select class ="form-control" type="text" name="issue_type" id="issue_type" onchange="showfield(this.options[this.selectedIndex].value)" >';
while ($row = mysql_fetch_array($result)) 
{ 
$htm .="<option value='".$row['issue_type']."'>".$row['issue_type']."      </option>"; 
} 
$htm .= "<option value='Other'>Other</option>"; // add other option
$htm .="</select>";
echo $htm;
?>
<div id="div1"></div>