The message that you receive is not an error, it
is a description of the object $result after the
query is done.
So, you will not get anywhere just by echo $result,
you have to process that information using one
of the mysql_fetch_array(),mysql_fetch_assoc(),
mysql_result().
In your case try :
...
if ($result) {
//check if there is a result
//echo $result;
$row=mysql_fetch_assoc($result);
var_dump($row); // to see the exact structure
// and you can you the array like
echo $row["password"]; //or what the column name is
}
...
Hope it helps,
capi
On Sat, 2005-05-14 at 12:16 +0200, Marc Henri wrote:
> 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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