Errors in code (linking in with drop box question)

[brockie99]

Select Member by Surname:
<select name = "memselect">
<?php
if ($memselect)
{
$i;
$select = mysql_query("SELECT surname FROM members;",$db);
while ($myrow = mysql_fetch_array($select))
{
echo "<option value = "$i">%s";
$i++;
}
}
?>
</select>

Can anyone tell me what is wrong with this piece of code. I am trying to pull the surnames from the table 'members', and put them in a drop down box.
I keep getting the error: Parse error: parse error, expecting ','' or';''
Thanks

[rachel2004]

You put the semicolon inside of your quotes for the query.

Try this:

$query = "SELECT surname FROM members";
$select = mysql_query($query, $db);

BTW, what are you doing about escaping the nested double quotes here:

echo "<option value = "$i">%s";

Maybe try this:

echo '<option value = "$i">%s';

[TheDefender]

Really quick glance, make the following changes and see if that doesn't work better for you:

<select name = "memselect"> 
<?php 
if ($memselect) 
{ 
$select = mysql_query("SELECT surname,id FROM members",$db); 
while($myrow = mysql_fetch_array($select)) 
{ 
echo "<option value=\"".$myrow['id']."\">".$myrow['surname']."</option>"; 
} 
} 
?> 
</select>

Select your Unique Identifying ID and use that for the value of your options... Your life will be much easier... You are using a unique identifying ID right?