PHP：获取SQL结果并插入Cookie值

Question

I am currently trying to run a SQL statement and pull the result out into its own variable and set it as a cookie value but am unsure of how to tackle this and being stuck on it for quite some time Id love some insight. I am using phpmyadmin,what am I doing wrong?

    <?php
    include("db_connect.php");
    session_start();

    $tbl_name="users_table"; // Table name 

    // username and password sent from form
    $myusername=$_GET['inputEmail'];
    $mypassword=$_GET['inputPassword'];

    $sql="SELECT * FROM $tbl_name WHERE Email='$myusername' and Password='$mypassword'";

    if ($result = $conn->query($sql)) {

        /* determine number of rows result set */
        $row_cnt = $result->num_rows;
        if($row_cnt==1){
    $_SESSION['username']=($myusername);
    $_SESSION['password']=($mypassword);
    $cookie_name1 = "user";
    $cookie_name2 = "content";

    setcookie($cookie_name1, $_SESSION['username'], time() + (86400 * 30), "/"); // 86400 = 1 day
$sql="SELECT Pref_Game FROM $tbl_name WHERE Email='$myusername'";
$result = $conn->query($sql);
$row = $result->fetch_object();
setcookie($cookie_name2,$row->Pref_Game, time() + (86400 * 30), "/"); // 86400 = 1 day
    }
    else {
    echo "<script>
    alert('Wrong email or password entered!');
    </script>";
    }

     $result->close();
    }
    ?>

Answer 1

I am assuming that you are using the Mysqli class to query the database. You first need to query the database as you are only setting the $result variable to the query.

$result = $conn->query($sql);

$row = $result->fetch_object();

So if the $sql contains the last query shown above ("SELECT Pref_Game FROM $tbl_name WHERE Email='$myusername'") you can get the value from the $row object by $row->Pref_Game

You need to put the $row = $result->fetch_object(); in a while loop if more than one row is being queried.