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simjay
04-08-2007, 09:17 PM
example: $letter = a

<?
$sql = "SELECT * FROM name WHERE *****first letter = $letter *****";
$result = mysql_query($sql, $conn) or die(mysql_error());
while ($newArray = mysql_fetch_array($result)) {
$name = $newArray['name'];

echo "$name *<br>";
}
?>

I am trying to make a script so i can display a list of name beging with the letter *.

Any help would be great.

iLLin
04-08-2007, 09:20 PM
$sql = "SELECT * FROM name WHERE first letter = '%".$letter/"';


That will return all results with the name that begins with "a";

simjay
04-08-2007, 09:25 PM
sorry but its not liking that?

simjay
04-08-2007, 09:27 PM
this is what i have


$sql = "SELECT * FROM user ORDER BY user_name WHERE user_name = '%".$letter/"'";

iLLin
04-08-2007, 09:28 PM
im sorry


$sql = "SELECT * FROM name WHERE first letter LIKE '%".$letter."';


Notice the LIKE

iLLin
04-08-2007, 09:28 PM
and the slash was a typo, should have been a period

simjay
04-08-2007, 09:31 PM
still not working sorry, are the speach marks right?

iLLin
04-08-2007, 09:33 PM
sql = "SELECT * FROM user WHERE user_name LIKE '%".$letter."' ORDER BY user_name";

simjay
04-08-2007, 09:38 PM
no work, lol im getting ready to give up with this, lol

iLLin
04-08-2007, 09:44 PM
Im sucking today


sql = "SELECT * FROM user WHERE user_name LIKE '".$letter."%' ORDER BY user_name";


The percent should be at the end.

simjay
04-09-2007, 04:52 AM
yay, lol
thank you so much