所以我第一次做了一個單獨的登錄頁面，也是第一次使用 PHP 代碼。如您所見，下面是我的表單代碼。<form method="post" action="validate_login.php"><div class="login-inhoud"> <h1>Login</h1> <div class="login-input"> <input type="email" id="users_email" name="users_email" size="30" maxlength="30" placeholder="Email"> <br><br> <input type="password" id="users_pass" name="users_pass" size="30" maxlength="30" placeholder="Wachtwoord"> </div> <br> <div class="login-submit"> <button type="submit"><b>submit</b></button> </div></div>但是，當我添加此 Php 代碼時，它不起作用。另外 mysql_connect 和 mysqli_connect 之間有什么區別？ <?php// Grab User submitted information$email = $_POST["users_email"];$pass = $_POST["users_pass"];// Connect to the database$con = mysqli_connect("localhost", "root", "");// Make sure we connected successfullyif ($con->connect_errno) { echo "Failed to connect to MySQL: " . $con->connect_error; exit(); }// Select the database to use mysqli_select_db("my_dbname", $con); $result = mysqli_query("SELECT users_email, users_pass FROM users WHERE users_email = $email"); $row = mysqli_fetch_array($result); if ($row["users_email"] == $email && $row["users_pass"] == $pass) echo "You are a validated user."; else echo "Sorry, your credentials are not valid, Please try again."; ?>
1 回答
慕雪6442864
TA貢獻1812條經驗 獲得超5個贊
你的代碼中有很多錯誤......
試試這個，希望它有效：
<?php
// Grab User submitted information
$email = $_POST["users_email"];
$pass = $_POST["users_pass"];
// Connect to the database
$con = mysqli_connect("localhost", "root", "", "my_dbname"); //Why complicate life?!
// Make sure we connected successfully
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
exit;
}
$result = mysqli_query($con, "SELECT users_email, users_pass FROM users WHERE users_email = $email");
$row = mysqli_fetch_array($result);
if ($row["users_email"] == $email && $row["users_pass"] == $pass)
echo "You are a validated user.";
else
echo "Sorry, your credentials are not valid, Please try again.";
?>
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