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我正在爲登錄寫代碼。我從另一個資源獲取了代碼。一切似乎都沒有問題，但它始終告訴我Alumni_ID（用戶名）不存在 - 即使它已經存在。PHP登錄用戶名錯誤
的登錄PHP代碼：
<?php
error_reporting(E_ALL); ini_set('display_errors', 1);
$servername = getenv('IP');
$username = getenv('C9_USER');
$password = "";
$database = "c9";
$dbport = 3306;
$conn = mysqli_connect($servername,$username,"",$database,$dbport) or die(mysqli_error());
if($_SERVER["REQUEST_METHOD"] == "POST"){
\t
\t
\t $query = mysql_query("SELECT Alumni_ID, Password from 'Alumni' WHERE Alumni_ID='$Alumni_ID'"); //Query the Alumni table if there are matching rows equal to $Alumni_ID
\t
\t $exists = mysql_num_rows($query); //Checks if Alumni_ID exists
\t $table_Alumni = "";
\t $table_Password = "";
\t
\t if($exists = 0) //IF there are no returning rows or no existing Alumni_ID
\t
\t {
\t \t while($row = mysql_fetch_assoc($query)) //display all rows from query
\t \t {
\t \t \t $table_Alumni_ID = $row['Alumni_ID']; // the first Alumni_ID row is passed on to $table_Alumni, and so on until the query is finished
\t \t \t $table_Password = $row['Password']; // the first Password row is passed on to $table_Alumni, and so on until the query is finished
\t \t }
\t \t if(($Alumni_ID == $table_Alumni_ID) && ($Password == $table_Password)) // checks if there are any matching fields
\t \t {
\t \t \t \t if($Password == $table_Password)
\t \t \t \t {
\t \t \t \t \t $_SESSION['user'] = $Alumni_ID; //set the Alumni_ID in a session. This serves as a global variable
\t \t \t \t \t header("location: loggedalumni/index.php"); // redirects the user to the authenticated home page
\t \t \t \t }
\t \t \t \t
\t \t }
\t \t else
\t \t {
\t \t \t Print '<script>alert("Incorrect Password!");</script>'; //Prompts the user
\t \t \t Print '<script>window.location.assign("index.php");</script>'; // redirects to login.php
\t \t }
\t }
\t else
\t {
\t \t Print '<script>alert("Incorrect Username!");</script>'; //Prompts the user
\t \t Print '<script>window.location.assign("index.php");</script>'; // redirects to login.php
\t }
}
?>這是形式：
<div class="col-lg-12">
<div class="nav-link" style= "text-align: center;"><h3><b>Log In</b></h3></div>
<form id="ajax-login-form" action=" " method="post" role="form" autocomplete="off">
<div class="form-group">
<label for="Alumni_ID" style= "text-align: center;">Alumni ID</label>
<input type="text" name="Alumni_ID" tabindex="1" class="form-control" placeholder="Your Alumni ID" value="" autocomplete="off">
</div>
<div class="form-group">
<label for="password" style= "text-align: center;">Password</label>
<input type="password" name="password" id="password" tabindex="2" class="form-control" placeholder="Password" autocomplete="off">
</div>
<div class="form-group">
<div class="row row-centered" style="padding-left:20px">
<div class="col-xs-7">
<input type="checkbox" tabindex="3" name="remember" id="remember">
<label for="remember" style= "align: center;"> Remember Me</label>
</div>
</div>
<div class="row row-centered">
<div class="col-xs-5" style="padding-left:75px">
<input type="submit" name="login-submit" id="login-submit" tabindex="4" class="form-control btn btn-success" style="background-color: #6a0001;" value="Log In" align="middle">
</div>
</div>
</div>
<div class="form-group">
<div class="row">
<div class="col-lg-12">
<hr>
<div style="padding-left:10px">
<a href="recover.php" tabindex="5" class="forgot-password" style = "color: black">Forgot Password?</a>
</div>
<hr>
<div style="padding-left:10px">
<a href="NewAlumni.php" tabindex="5" class="forgot-password" style = "color: black">Register Account</a>
</div>
<hr>
<div style="padding-left:10px">
<a href="adminlogin.php" tabindex="5" class="forgot-password" style = "color: black">Admin Login</a>
</div>
</div>
</div>
</div>
<input type="hidden" class="hide" name="token" id="token" value="a465a2791ae0bae853cf4bf485dbe1b6">
</form>
</div>
那是一個PHP錯誤或SQL錯誤？你能否包含實際的錯誤信息。嘗試'打印「選擇Alumni_ID，來自'Alumni'的密碼WHERE Alumni_ID ='$ Alumni_ID'」;'在你的代碼中，然後直接在SQL中運行輸出語句。 – Manngo
您正在使用mysql函數。他們貶值，並受到黑客攻擊。改用mysqli函數。 –
@SloanThrasher是正確的，但我看到他們正在混入'mysqli_'語句，這更糟糕。請參閱http://stackoverflow.com/questions/17498216/can-i-mix-mysql-apis-in-php瞭解這可能會導致失敗。 – Manngo