Displaying MySQL Results In A Form Field

[Sharif]

I am trying to get mysql results to show up in a list form field. Here is my code:

echo ("Select Date:<BR>");
?>
<select name="article_date">
<?php
$query = mysql_query ("SELECT date_FORMAT(date,'%m/%d/%Y') AS date FROM news WHERE author=$userid");
if (mysql_num_rows($display_title) > 0) // If given ID exists...
{ 
while ($row = mysql_fetch_array($query)) 
{ 
	$date = $row["date"];
	echo $date; 
}
}
}
?>
<option><?php echo $date; ?><option>
</select>

Where could I have gone wrong?

[Hawkee]

You've got a few problems here. First, I noticed right away that you have no value='' tag in your option which will make your option useless to the server. Also, if you have multiple dates that you're pulling from the SQL server, you'll need to include your <option> tag within the loop so that it's displayed multiple times. Right now you have it appearing after your loop which will only show the last item from the database.

• Hawkee

[Moonglobe]

try something like this:

<?php
$query = mysql_query ("SELECT date_FORMAT(date,'%m/%d/%Y') AS date FROM news WHERE author=$userid");
if (mysql_num_rows($display_title) > 0) // If given ID exists...
{ 
while ($row = mysql_fetch_array($query)) 
{ 
    $date = $row["date"];
    echo "<option value=\"".$date."\">".$date."</option>\n"; 
}
}
}
?>
</select>