檢索MySQL數據庫

Question

我已經做了數據庫列電子郵件，名字，姓氏，密碼..等等數據
我的登錄頁面使用電子郵件和密碼用於登錄
我想要檢索的名字列在鍵入的名字同一行作爲電子郵件
我想知道怎麼做
我的代碼如下所示： -

$query=mysql_query("SELECT * FROM Users WHERE email='$email' AND 
`password`='$encrypted_pass' "); 

在此之後，我希望能夠給一個變量分配給姓氏，所以我可以創建一個會話連同登錄

Answer 1

我認爲這涵蓋了你要求的一切。開始一個會話，檢查它是否好，爲我們的發佈數據設置幾個變量，做一些數據驗證，設置查詢，查詢數據庫，檢查結果，在會話中設置var lastname。

<?php 
header('Content-type: text/html'); 

    session_start(); 
    if (session_id() === '') { 
     echo 'session_id is null!'; 
     exit; 
    } 
    $myemail = null; 
    $mypassword = null; 

    if (isset($_POST['Submit']) == true) { 
     //assuming you have a db connection 
     //mysql_connect($host, $user, $password) || die(mysql_error()); 
     //mysql_select_db($database) || die(mysql_error()); 

    if ((isset($_POST["username"]) === false)|| 
      (isset($_POST["password"]) === false)) { 
      echo 'Please fill in all fields'; 
      exit; 
     } else { 
      // get the post data 
      $myemail = ($_POST["username"]); 
      $mypassword = ($_POST["password"]); 
     } 

    // check the form in database 
    $sql = "SELECT * FROM Users WHERE email = '".$myemail."' 
      AND password = '".$mypassword."'"; 
    $result = mysql_query($sql); 
    $count = mysql_num_rows($result); 

    $_SESSION['loggedin'] = false; 
    if ($count === 1) { 
     $userrecord = mysql_fetch_array($result); 
     $_SESSION['username'] = $userrecord['username']; 
     $_SESSION['password'] = $userrecord['password']; 
     $_SESSION['loggedin'] = true; 
     $_SESSION['lastname'] = $userrecord['lastname']; 
     // assign last name to a session var 

     echo 'You have been logged in successfully"; 
    } else { 
     echo 'Wrong username or password'; 
     exit; 
    } 
} else { 
    echo "No form post detected.<br><br>"; 
} 
?> 

這裏有一些基本的東西使用Mysqli，因爲mysql_命令已被棄用。再次，它使用一些常用名稱，以便相應地進行調整

/********************************/ 
    // using mysqli (untested but should work) 
    /********************************/ 

    $mysqli = new mysqli("localhost", "dbuser", "dbpassword", "default_dbname"); 

    /* check connection */ 
    if (mysqli_connect_errno()) { 
     printf("Connect failed: %s\n", mysqli_connect_error()); 
     exit(); 
    } 

    // ecsape the query params 
    $myemail = $mysqli->real_escape_string($myemail); 
    $mypassword = $mysqli->real_escape_string($mypassword); 

    if ($mysqli->query("SELECT * FROM Users WHERE email = '$myemail' 
         AND password = '$mypassword'")) { 
     printf("%d Rows returned.\n", $mysqli->affected_rows); 

    // here is where you can set the $_SESSION vars 
    // and do any other work you want to do on login. 

    } 


    $mysqli->close(); 
    /********************************/ 
    //end mysqli 
    /********************************/ 

最後一件事。獲取GitHub.com存儲庫設置，以便您可以輕鬆回滾到以前的版本。這真的是一個必須有國際海事組織。

Answer 2

然後你在列列表中指定列：

SELECT LastName 
FROM Users 
WHERE email = '$email' 
... 

然後，您需要獲取結果集將其分配給一個變量。

Answer 3

mysql_職能正式廢棄，不再維護/使用安全。

你應該使用PHP的PDO，而不是因爲它是一個更安全，更面向對象的方法。 在DB_NAME，DB_USER_NAME和DB_USER_PASS只需填寫您的特定憑據，此代碼應爲你工作。

$database = new PDO('mysql:host=localhost;dbname=DB_NAME;charset=UTF-8', 'DB_USER_NAME', 'DB_USER_PASS'); 

$query = $database->prepare("SELECT * FROM Users WHERE `email` = ? AND `password` = ? "); 
$query->bindValue(1, $email); 
$query->bindValue(2, $encrypted_pass); 
$query->execute(); 

if($query->rowCount() > 0) { # If rows are found for query 

    $result = $query->fetch(PDO::FETCH_ASSOC); 
    echo $result[ 'LastName' ]; # <-- The LastName field you were looking for! 

} 
else { echo "No Rows Returned!"; } 

對於PHP PDO的更多信息，請參閱http://www.php.net/manual/en/book.pdo.php