[RESOLVED] Using Array To Populate Formfield

[obrienkev]

Hi all,

I have a Database Query that I want to use to populate a HTML Form Select Menu.

How do I get each Database Row to populate each Option Value?

Here's my code so far...

	 $query = "SELECT fund_ID, client_ID, fund_name ".
	 					"FROM fund_name ".
						"WHERE client_ID = $client";
	 $result = mssql_query($query) or die('Select Error');
	 $num_results = mssql_num_rows($result);

 $options = array(); // declare array


 while($row = mssql_fetch_array($result)) // SET RESULTS
 {
     $options[] = $row;	
 } // END WHILE

 foreach ($options as $option) // LOOP THROUGH ARRAY
   {
	     $option_list = array($option['fund_ID'],$option['fund_name']);
	 } // END FOREACH

HTML:

<select name="fund_name" id="fund_name">
<option value="0">Select Fund Name</option>
			<?php
echo '<option value="'.$option_list[0].'">'.$option_list[1].'</option>';
			?>	 
</select>

[ClarkF1]

 $query = "SELECT fund_ID, client_ID, fund_name ".
                         "FROM fund_name ".
                        "WHERE client_ID = $client";
     $result = mssql_query($query) or die('Select Error');
     $num_results = mssql_num_rows($result);

?>
<select name="fund_name" id="fund_name">
<option value="0">Select Fund Name</option>
<?php
     while($row = mssql_fetch_array($result)) // SET RESULTS
     {
echo '<option value="'.$row['fund_ID'].'">'.$row['fund_name'].'</option>';          

     } 
?></select>    

[konrad]

If you need HTML output only, try something like this:

<?php
     $select = '<select name="fund_name" id="fund_name">
                      <option value="0">Select Fund Name</option>';

 while($row = mssql_fetch_array($result)) // SET RESULTS
 {
     $select .= '
                 <option value="' . $row['fund_ID'] . '">' . $row['fund_name'] . '</option>';  
 } // END WHILE 
$select .= '</select>';
?>

[obrienkev]

Thanks Guys!!