You will need to put the option tag in a loop to get all of the records in
the table.
<select name="canidate"> (this can go anywhere but in while loop)
<?
$db = mysql_connect($dbserver, $dbuser, $dbpass);
mysql_select_db($dbname,$db);
$sortby = "name ASC";
$sql="SELECT * FROM webl_players ORDER BY $sortby";
$result=mysql_query($sql,$db);
$num=mysql_numrows($result);
$i=0;
while ($i < $num) {
$row = mysql_fetch_array($result);
$FirstName = $row["FirstName"];
$LastName = $row["LastName"];
echo "<option value='$LastName, $FirstName'><?php echo '$LastName,
$FirstName'?></option>";
$i++;
}
?>
Also, without the value in the option tag, you will not be passing anything
to the form.
HTH
Maureen Biorn
Todd Williamsen <[EMAIL PROTECTED]> said:
> I am trying to get data from two columns, FirstName and Last name and
> displaying all the records LastName, FirstName in a drop down menu.
>
> The weird thing is that it only displays one record. I thought the table
> was hosed, but its not. I tried it through another database and still
> doesn't work.
>
> Here is the code:
>
> <?
> $db = mysql_connect($dbserver, $dbuser, $dbpass);
> mysql_select_db($dbname,$db);
> $sortby = "name ASC";
> $sql="SELECT * FROM webl_players ORDER BY $sortby";
> $result=mysql_query($sql,$db);
> $row = mysql_fetch_array($result);
> $FirstName = $row["FirstName"];
> $LastName = $row["LastName"];
> ?>
>
> <select name="canidate">
> <option selected>
> <?php echo "$LastName, $FirstName"; ?>
> </option>
>
> I would list the whole page but its long! There is another SQL statement at
> the top of the page:
>
> <?php
>
> $db = mysql_connect($dbserver, $dbuser, $$dbpass);
> mysql_select_db($dbname, $db);
> $sql = "SELECT * FROM Canidate SORT BY 'LastName'";
> $result = mysql_query($sql);
> ?>
>
>
>
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