Question

所以我将用户的排名保存为SQL数据库作为枚举器。然后我想尝试检查他们的等级并将其与字符串进行比较。如果排名等于字符串，那么我想允许他们访问。这是我到目前为止的代码（哪些不起作用）：

    $email = stripslashes($email);
    $password = stripslashes($password);
    $email = mysql_real_escape_string($email);
    $password = mysql_real_escape_string($password);
    $sql = "SELECT RANK FROM $table WHERE EMAIL = '$email' and PASSWORD = '$password'";
    $result = mysql_query($sql);

    $count = mysql_num_rows($result);

    if($count == 1){
        if($result == "OWNER") {
            header("location: panel.php");
        } else {
            $message = "Permission denied.";
        }
    } else {
        $message = "Incorrect email or password.";
    }

Answer 1

重要通知！

自PHP 5.5.0起，不推荐使用

mysql_*个函数。我们强烈建议使用MySQLi或PDO。

回答您的问题，您必须获取查询结果或使用mysql_result()

<强>查询

$sql = "SELECT RANK
        FROM $table
        WHERE EMAIL = '$email' and PASSWORD = '$password'
        LIMIT 1";
$result = mysql_query($sql);

获取结果：

$rank = mysql_fetch_array($result);
$rank = $rank['RANK'];

或者改为使用mysql_result()：

$rank = mysql_result($result);

Answer 2

你需要mysql_fetch_assoc $ result var才能读取表格行。我也建议你使用mysqli函数。在这里，尝试这个而不是你的（没有mysqli，也没有自己测试）：

$email = mysql_real_escape_string(stripslashes($email));
$password = mysql_real_escape_string(stripslashes($password));
$sql = "SELECT `RANK` FROM ".$table." WHERE `EMAIL` = '".$email."' and `PASSWORD` = '".$password."'";
$result = mysql_query($sql);

$count = mysql_num_rows($result);

$result = mysql_fetch_assoc($result);

if($count == 1){
    if($result == "OWNER") {
        header("location: panel.php");
    } else {
        $message = "Permission denied.";
    }
} else {
    $message = "Incorrect email or password.";
}

如何在PHP中获取SQL数据库的枚举值

2 个答案: