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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