What are “Resource#'s”?

问题:

HI

I am getting Resource#6 and Resource#7 when I print the following variables:

$salty_password = sha1($row['salt'], $_POST['password']);

if(isset($_POST['subSignIn']) && !empty($_POST['email']) && !empty($_POST['password'])) {


    $query =  "SELECT `salt` FROM `cysticUsers` WHERE `Email` = '" . $_POST['email'] . "'";
    $request = mysql_query($query,$connection) or die(mysql_error());
    $result = mysql_fetch_array($request);



    $query2 = "SELECT * FROM `cysticUsers` WHERE `Email` = '". $_POST['email']."' AND `Password` = '$salty_password'";
    $request2 = mysql_query($query2,$connection) or die(mysql_error());
    $result = mysql_fetch_array($request2);

    print_r($request);
    print_r($request2);

if(@mysql_num_rows($request,$request2)) {


        $_SESSION['CLIFE']['AUTH'] = true;
        $_SESSION['CLIFE']['ID'] = $result['id'];

        // UPDATE LAST ACTIVITY FOR USER
        $query = "UPDATE `cysticUsers` SET `LastActivity` = '" . date("Y-m-d") . " " . date("g:i:s") . "' WHERE `id` = '" . mysql_real_escape_string($_SESSION['CLIFE']['ID']) . "' LIMIT 1";
        mysql_query($query,$connection);


        if(!empty($_POST['return'])) {
            header("Location: " . $_POST['return']);

        }else{
            header("Location: CysticLife-Dashboard.php?id=" . $_SESSION['CLIFE']['ID']);
            }
        }

    }else{

        $_SESSION['CLIFE']['AUTH'] = false;
        $_SESSION['CLIFE']['ID'] = false;

    }


?>

Trying to troubleshoot this code chunk and not sure what that means. I am trying to sign back in with the clear text password I signed up with after its been hashed and salted. I feel like I'm very close but something is slightly wrong. Help on why that is not working would be greatly appreciated as well.

Thanks in advance

回答1:

mysql_query() returns result sets as objects of type resource (they're not objects in terms of PHP OOP code but I can't think of a better word). These contain binary data that can only be read by certain functions, for example the mysql_fetch_*() functions.

To debug your MySQL queries you should check for errors using mysql_error() and mysql_errno() and/or save your SQL statements in variables and print those.

From what I see, you're performing two queries but overwriting the same $result variable, without doing anything about the first. Also, mysql_num_rows() can only count one result set at a time, so you can't pass two result sets into the same call.

回答2:

Those are PHP's internal data types called resource.

They cannot be serialized (i.e. there's no "toString()") and are hence displayed as Resource#X.

回答3:

SQL queries through PHP are done using a variable known as a resource. This variable, in your code, is completely useless other than to pass to each function you want to perform (i.e. change a database, execute a query, grab the last error, etc.).

That being said, executing a query doesn't return any information from the database, just a reference to that record set (where in PHP it's storing the information). You would then use that variable in a call such as mysql_fetch_array to retrieve the actual row information.