PHP/MySQL查询返回不在数据库中的结果

我正在尝试创建一个页面，检查您的登录信息是否正确。出于某种原因，当我回显从数据库中选择数据的结果查询时，我没有从数据库中获得我应该得到的数据，我不确定它是什么数据。以下是页面上显示的内容：

Connected successfully
Resource id #3

以下是我的php代码：

<?php
//start session
session_start();

//connect to MySQL Database
$con = mysql_connect('localhost', 'root', ''); 
if (!$con) { 
    die('Could not connect: ' . mysql_error()); 
} 
echo 'Connected successfully<br />'; 

// make members the current db
$db_selected = mysql_select_db('members_db', $con);
if (!$db_selected) {
    die ('Can\'t use members database : ' . mysql_error());
}


$email = $_SESSION['email'];

$result = mysql_query("SELECT email,password FROM `members` WHERE 'email' = '$email'");
if (!$result) {
    echo 'Could not run query: ' . mysql_error();
    exit;
}
$row = mysql_fetch_row($result);

if ($email == $row[0] && $_SESSION['password'] == $row[1])
{
    $_SESSION['loggedin']="YES";
    $url= "Location: welcome.php";
    header($url);
    exit;
}

echo $result;

/*
//Problems:
$problem = "";
if ($email == $row[0] && $_SESSION['password'] != $row[1])
{
    $problem = "invalidPassword";
}
if ($email != $row[0])
{
    $problem = "InvalidUser";
}
$url = "Location: index.php?problem=$problem";
header($url);
exit;*/
?>

---

如果这有助于检查您是否获得了正确的数据，那么看起来您甚至没有进入正确的用户验证区域

 $row = mysql_fetch_row($result);
 if ($email == $row[0] && $_SESSION['password'] == $row[1])
 {

 $_SESSION['loggedin']="YES";
 $url= "Location: welcome.php";
 header($url);
 exit;
}
else //Validation failed
{

echo $row[0] . ' SESSION EMAIL ' . $email; //Which would echo the email that was returned and the email stored in the session, if they do not match it means the password was wrong.
}

mysql_查询返回一个资源，所以这就是您在回显它时得到的资源。您应该查看
$row
中的内容，它将告诉您为什么您的if FailedBligatory
mysql
被折旧使用
mysqli
尝试
echo$row[0]；echo$row[1]
在
$row=mysql\u fetch\u row（$result）之后。还要确保您的查询返回一行。您可以使用查找返回的行数。