i'm trying to have php go into a database for me, get something out of a table
and then display it for me, but it's not working and i'm really having a hard
time figuring out what exactly the problem is. the code i'm using is practicly
straight out of the book, although with some minor changes.
the error i'm getting is:
"Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result
resource in f:\inetpub\wwwroot\andrew\admin\login_pr
band"
and the whole code for the page is as follows
----------------------------------------
<html>
<head>
<title>Untitled Document</title>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
</head>
<body>
<?php
$loginname = $HTTP_POST_VARS["login"];
$password = $HTTP_POST_VARS["pass"];
$db = mysql_connect("localhost", "$loginname", "$password");
mysql_select_db("databasename",$db);
$result = mysql_query("SELECT bandname FROM bands");
echo "<table>";
echo "<tr><td>band</td></tr>";
while ($row = mysql_fetch_array($result))
{
echo "<tr><td>";
echo $row["bandname"];
echo "</td></tr>";
}
echo "</table>";
?>
</body>
</html>
----------------------------------------
the error is complaining about this being incorrect:
while ($row = mysql_fetch_array($result))
but that's an unchanged line from the book so it should be correct, shouldn't
it?
if anybody could tell me what the problem is, i'd be most grateful.
[edit] nevermind, i figured it out. it was simply a matter of having "or
die(mysql_error())" at the end of the mysql functions [/edit]
no, wait, that wasn't exactly true.
the _real_ error was that i had put in a wrong name for the database. i changed it at the same time as i added the "or die(mysql_error())" bits and didn't think any more of it. heheheheh, no wonder it didn't want to work