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simjay
02-08-2006, 12:26 AM
Im not that good at PHP, have a look at this and i think ul find it funny but i aint got a clue. How would i make it list all the usernames?


<?php

$sql = "SELECT * FROM ALS_signup";
$result = mysql_query($sql, $conn) or die(mysql_error());
while ($newArray = mysql_fetch_array($result)) {

$username = $newArray['username'];

echo" <select name='test'>
<option>$username</option>
</select>
";

}

?>

SocoNaTromba
02-08-2006, 12:32 AM
You're missing the mysql_connect and database selection procedures.

simjay
02-08-2006, 12:42 AM
I have that,

I just need to know how to get the list menu working

firepages
02-08-2006, 01:39 AM
assumin you have a field in your database called 'username' then your code should work as-is , whats the problem you are having ?

simjay
02-08-2006, 01:41 AM
Warning: mysql_query(): supplied argument is not a valid MySQL-Link resource in /home/****/public_html/members/messages.php on line 508

firepages
02-08-2006, 03:13 AM
so $conn is not valid , do other queries fail or just this one ?

how do you connect to the database ? try just using
$result = mysql_query($sql) ;//without the $conn
if that fails show us your mysql_connect(); statement

Rich Pedley
02-08-2006, 12:23 PM
after you have sorted out the mysql problem you may want to redo things slightly:


<?php
$sql = "SELECT * FROM ALS_signup";
$result = mysql_query($sql, $conn) or die(mysql_error());
echo "<select name='test'>";
while ($newArray = mysql_fetch_array($result)) {
$username = $newArray['username'];
echo" <option>$username</option>";
}
echo "</select>";
?>
otherwise you will be creating a select box for each username.

degsy
02-08-2006, 03:38 PM
http://www.zend.com/php/beginners/php101-8.php#Heading5



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