我知道之前发过了类似的问题,我已经通过了它,这里是链接Login code doesn't work, stay stuck on log in page,但我们的代码完全不同,我已经完成了3个小时但是没有解。这是我的代码在我设计之前最初看起来的样子。
<body>
<?php
require('db.php');
session_start();
// If form submitted, insert values into the database.
if (isset($_POST['username'])){
$username = stripslashes($_REQUEST['username']); // removes backslashes
$username = mysqli_real_escape_string($con,$username); //escapes special characters in a string
$password = stripslashes($_REQUEST['password']);
$password = mysqli_real_escape_string($con,$password);
//Checking is user existing in the database or not
$query = "SELECT * FROM `users` WHERE username='$username' and password='".md5($password)."'";
$result = mysqli_query($con,$query) or die(mysql_error());
$rows = mysqli_num_rows($result);
if($rows==1){
$_SESSION['username'] = $username;
header("Location:../user/page/emergency.php"); // Redirect user to index.php
}else{
echo "<div class='form'><h3>Username/password is incorrect.</h3><br/>Click here to <a href='login.php'>Login</a></div>";
}
}else{
?>
<div class="form">
<h1>Log In</h1>
<form action="" method="post" name="login">
<input type="text" name="username" placeholder="Username" required />
<input type="password" name="password" placeholder="Password" required />
<input name="submit" type="submit" value="Login" />
</form>
<br /><br />
</div>
<?php } ?>
</body>
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以下是我尝试使用css和javascript设置样式后的样子。
<body>
<h2> Selamat Datang, Sila Log Masuk Untuk Kemaskini</h2>
<?php
require('db.php');
session_start();
// If form submitted, insert values into the database.
if (isset($_POST['username'])){
$username = stripslashes($_REQUEST['username']); // removes backslashes
$username = mysqli_real_escape_string($con,$username); //escapes special characters in a string
$password = stripslashes($_REQUEST['password']);
$password = mysqli_real_escape_string($con,$password);
//Checking is user existing in the database or not
$query = "SELECT * FROM `users` WHERE username='$username' and password='".md5($password)."'";
$result = mysqli_query($con,$query) or die(mysql_error());
$rows = mysqli_num_rows($result);
if($rows==1){
$_SESSION['username'] = $username;
header("Location:../user/page/emergency.php"); // direct user to update page
}else{
echo "<div class='form'><h3>Username/password is incorrect.</h3><br/>Click here to <a href='login.php'>Login</a></div>";
}
}else{
?>
<div class="container">
<section id="content">
<form action="" method="post" name="login">
<h1>Daftar Masuk</h1>
<div>
<input type="text" placeholder= "Username" required ="" id ="username"/>
</div>
<div>
<input type="password" placeholder= "Password" required ="" id ="password"/>
</div>
<input name="submit" type="submit" value="Login" />
</form>
</section>
</div>
<?php } ?>
<script src="js/index.js"></script>
</body>
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一旦我设计了它,表单就会停留在登录页面上。帮帮我,我该怎么办?非常感谢任何建议和指导。
更新
好吧,伙计们,我已经找到了问题的根源,我只需要在第二张表格上删除。答案 0 :(得分:0)
事实证明,在我做样式的表单上,我只需要删除标记 p>
所以来自
<div class="container">
<section id="content">
<form action="" method="post" name="login">
<h1>Daftar Masuk</h1>
<div>
<input type="text" placeholder= "Username" required ="" id ="username"/>
</div>
<div>
<input type="password" placeholder= "Password" required ="" id ="password"/>
</div>
<input name="submit" type="submit" value="Login" />
</form>
</section>
</div>
&#13;
到
<div class="container">
<section id="content">
<form action="" method="post" name="login">
<h1>Daftar Masuk</h1>
<input type="text" placeholder= "Username" required ="" id ="username"/>
<input type="password" placeholder= "Password" required ="" id ="password"/>
<input name="submit" type="submit" value="Login" />
</form>
</section>
</div>
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