Actually, that is exactly what you are doing. $result is the handle to the
dataset, not the dataset itself. Try this:
//log to the server
$db=mysql_connect("localhost","root","");
if ($db) {
//test if the connexion works
$sel=mysql_select_db("test");
if ($sel) {
$sql_query="SELECT * FROM mytable WHERE
login='$login'";
$result=mysql_query($sql_query);
if ($result) {
//check if there is a result
$rows = mysql_fetch_array($result)
foreach ($rows as $key => $value)
{
echo $key . " :: ". $value;
}
}
//the query didn't work
else echo mysql_error();
}
else echo mysql_error();
}
Bastien
From: Marc Henri <marchenri1@xxxxxxxx>
To: php-db@xxxxxxxxxxxxx
Subject: Error: "Resource id #3"
Date: Sat, 14 May 2005 12:16:58 +0200 (CEST)
Hello,
I'm starting to learn how to manage databases with
MySQL/PHP. The program is very basic but I have a
strange error: "Resource id #3".
I read many things on Internet and understood that
others have this error because they are trying to echo
the pointer of the query and not the result itself.
It's not my case so I don't understand.
Thank you very much for your help.
<?
/*This program will log on to a users DB to check if
the user is allowed to log in or not.
The program checks the login and password of the user.
*/
$login="";
if (isset($_POST['login'])) {
$login=$_POST['login'];
}
if (isset($_POST['password'])) {
$password=$_POST['password'];
}
//log to the server
$db=mysql_connect("localhost","root","");
if ($db) {
//test if the connexion works
$sel=mysql_select_db("test");
if ($sel) {
$sql_query="SELECT * FROM mytable WHERE
login='$login'";
$result=mysql_query($sql_query);
if ($result) {
//check if there is a result
echo $result;
}
//the query didn't work
else echo mysql_error();
}
else echo mysql_error();
}
?>
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