如何將密碼用戶名與登錄表單中的php mysql匹配

Question

因此，基本上，我試圖匹配已經在數據庫中的用戶名和密碼，但我不明白爲什麼當密碼沒有重定向到另一頁時和用戶名是正確的

這裏是我的HTML表單

<form class="form-inline" method="post" action= "login.php" > 

<input type="email" placeholder="Enter email" class="form-control" name="logemail"> 
<input type="password" placeholder="password" class="form-control" name="logpass"> 

<button type="submit" class="btn btn-default" id="login" name ="submit1">Log in</button> 
</form> 

這是我的login.php文件

<?php 



$servername = "localhost"; 
$username = "root"; 
$password = ""; 
$dbname = "registration_form"; 

if(isset($_POST['submit'])) 
{ 
$username = $_POST['logemail']; 
$password = $_POST['logpass']; 
$con=mysqli_connect("localhost","root","","registration_form"); 
// Check connection 
if (mysqli_connect_errno()) 
    { 
    echo "Failed to connect to MySQL: " . mysqli_connect_error(); 
    } 
$qz = "SELECT * FROM regis where email1='".$username."' and password3='".$password."'" ; 
$qz = str_replace("\'","",$qz); 
$result = mysqli_query($con,$qz); 
$row = mysqli_num_rows($result); 
if($row == 1) 
    { 
    echo "successfully logged in"; 
    } 
mysqli_close($con); 
} 
?> 

Answer 1

編輯

當你應該檢查submit1時，你重新檢查isset中的提交，這就是爲什麼你的代碼沒有在if塊中執行，但是我更新的答案消除了這一點。

末

更新答::

dologin.php

<?php 

// here u would move these vars onto a seperate file outside 
// the webroot that is readable by the server and "require_once" 
// that file. 
$dbhost = "localhost"; 
$dbuser = "mysql_user_name"; 
// using root is bad only use it for local development but better yet, 
//// make a user for the database in question 
$password = "password"; 
$dbname = "registration_form"; 

// Setup a filter array to sanitize/validate user input. 
// look on php.net for more information on filters available 
$filters = [ 
       'logemail' => ["ARRAY, OF, FILTERS"], 
       'logpass' => ["ARRAY, OF, FILTERS"] 
      ]; 

// NEVER NEVER NEVER access $_POST without filtering it. 
$posted = filter_var_array(INPUT_POST, $filters, true); 

// check our posted array is not empty, even if its submitted it 
// could be empty values. 
if (!empty($posted)) { 

    // I have omitted the $dbname in this procedural example 
    // and move it after the error No, this is just to avoid 
    // confusion and limit the number of possible things to go wrong. 
    $con = mysqli_connect($dbhost, $dbuser, $dbpass); 

    $username = $posted['logemail']; 
    $password = $posted['logpass']; 

    // Check no connection error 
    if (!mysqli_connect_errno()) { 
     mysqli_select_db($con, $dbname) or die("Database select error" . mysqli_error()); 
    } else { 
     die("Failed to connect to MySQL: " . mysqli_connect_error()); 
    } 

    $qz = "SELECT * FROM regis WHERE email1 = $username AND password3 = $password"; 


    $result = mysqli_query($con, $qz); 
    if (mysqli_num_rows($result) == 1) { 

     // initialise the session. 
     session_start(); 

     // add an entry "loggedin" and set it to true. 
     $_SESSION['loggedin'] = true; 
     header('location :index.php');   
    } 
    mysqli_close($con); 
} 

現在，你需要做你的索引頁。

的index.php

<?php 

// Start up the session its needed to maintain logins 
session_start(); 

// We don't know that the raw $_SESSION is safe 
$session_unsafe = $_SESSION; 

// Lets play with some more filtering (php.net) 
$session = filter_var($session_unsafe, FILTER_VALIDATE_BOOLEAN); 

// add more filtering as required. 

// remember the boolean in dologin.php here === makes sure it 
// matches the "type" of the variable too because: 
// 1 == yes == true 
// 0 == no == false 
// but === means an exact match of type (integer/string/boolean) as 
// well as its value. True === 1 would return false. 
if ($session === true) { 

    // show logged in stuff 


} else { 

    // do your none logged in actions 
} 

結論：這不是解決這個最好的方法，你可以使用使用庫MySQLi或PDO的OOP方式，實現更清潔更看代碼可管理相同的結果。由於您似乎剛剛從樹上移開，您可能希望查看這兩種方法之一併在繼續之前瞭解這些方法。同時注意保護會話。我給你的例子會讓你的登錄工作，但它不會是安全的，你也不會做很多事情。

所以你的檢查要點：

PHP：面向對象，會話，過濾器這些關鍵點我想看看作爲一個優先事項，然後再繼續。

現在，你會選擇sql查詢，但它不會同時學習它們。所以找到一些用於學習SQL的好資源。

Answer 2

既然你已經使用isset（$ _ POST [ '提交']）修改您的按鈕的HTML作爲

<button type="submit" class="btn btn-default" id="login" name ="submit">Log in</button> 

，另外檢查，你沒有寫任何代碼重定向，而不是

echo "successfully logged in"; 

寫

header("location:new_page.php"); 
exit();