在 php 中創建一個非常簡單的 1 用戶名/密碼登錄
[英]Creating a very simple 1 username/password login in php
問題描述
我想只為 1 個用戶進行一次登錄而不存儲在數據庫中，但我似乎無法讓它工作。
我的代碼：login.php
<html>
<head>
<title>Login</title>
</head>
<h3>Add entry</h3>
<p> Add another Article</p>
<form action="trylog.php" method = "post">
<label for="username">Username</label> <input type="username" id="usename" name="username"><br /><br />
<label for="password">Password:</label> <input type="text" id="password" name="password"><br /><br />
<button type = "submit">Login</button>
</form>
</html>
嘗試日志文件
<html>
<title>Login</title>
<body>
<?php
$usr = "admin";
$psw = "password";
$username = '$_POST[username]';
$password = '$_POST[password]';
//$usr == $username && $psw == $password
session_start();
if ($_SESSION['login']==true || ($_POST['username']=="admin" && $_POST['password']=="password")) {
echo "password accepted";
$_SESSION['login']=true;
}else {
echo "incorrect login";
}
?>
<form name="input" action="adminportal.php" method="get">
<input type="submit" value="Home">
</form>
</body>
</html>
5 個解決方案
解決方案1
28 已采納 2013-10-23 01:34:33
您的代碼可能看起來更像：
<?php
session_start();
$errorMsg = "";
$validUser = $_SESSION["login"] === true;
if(isset($_POST["sub"])) {
$validUser = $_POST["username"] == "admin" && $_POST["password"] == "password";
if(!$validUser) $errorMsg = "Invalid username or password.";
else $_SESSION["login"] = true;
}
if($validUser) {
header("Location: /login-success.php"); die();
}
?>
<!DOCTYPE html>
<html>
<head>
<meta http-equiv="content-type" content="text/html;charset=utf-8" />
<title>Login</title>
</head>
<body>
<form name="input" action="" method="post">
<label for="username">Username:</label><input type="text" value="<?= $_POST["username"] ?>" id="username" name="username" />
<label for="password">Password:</label><input type="password" value="" id="password" name="password" />
<div class="error"><?= $errorMsg ?></div>
<input type="submit" value="Home" name="sub" />
</form>
</body>
</html>
現在，當基於header('LOCATION:wherever.php)重定向頁面時，將session_start()放在頁面頂部並測試以確保$_SESSION['login'] === true 。 請記住， ==將是真實的，如果$_SESSION['login'] == 1為好。 當然，出於安全原因，這是一個壞主意，但我的示例可能會教您另一種使用 PHP 的方式。
解決方案2
13 2017-04-13 16:20:32
這是一個用於登錄的簡單 php 腳本和一個只能由登錄用戶訪問的頁面。
登錄.php
<?php
session_start();
echo isset($_SESSION['login']);
if(isset($_SESSION['login'])) {
header('LOCATION:admin.php'); die();
}
?>
<!DOCTYPE html>
<html>
<head>
<meta http-equiv='content-type' content='text/html;charset=utf-8' />
<title>Login</title>
<meta charset="utf-8">
<meta name="viewport" content="width=device-width, initial-scale=1">
<link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/css/bootstrap.min.css">
</head>
<body>
<div class="container">
<h3 class="text-center">Login</h3>
<?php
if(isset($_POST['submit'])){
$username = $_POST['username']; $password = $_POST['password'];
if($username === 'admin' && $password === 'password'){
$_SESSION['login'] = true; header('LOCATION:admin.php'); die();
} {
echo "<div class='alert alert-danger'>Username and Password do not match.</div>";
}
}
?>
<form action="" method="post">
<div class="form-group">
<label for="username">Username:</label>
<input type="text" class="form-control" id="username" name="username" required>
</div>
<div class="form-group">
<label for="pwd">Password:</label>
<input type="password" class="form-control" id="pwd" name="password" required>
</div>
<button type="submit" name="submit" class="btn btn-default">Login</button>
</form>
</div>
</body>
</html>
admin.php （只有登錄用戶才能訪問）
<?php
session_start();
if(!isset($_SESSION['login'])) {
header('LOCATION:login.php'); die();
}
?>
<html>
<head>
<title>Admin Page</title>
</head>
<body>
This is admin page view able only by logged in users.
</body>
</html>
解決方案3
1 2013-10-23 01:17:36
首先，你需要把session_start(); 在任何輸出到瀏覽器之前，通常在頁面頂部。 看看說明書。
其次，這不會影響您的結果，但這些行沒有在任何地方使用，應該刪除：
$usr = "admin";
$psw = "password";
$username = '$_POST[username]';
$password = '$_POST[password]';
...最后兩行不起作用，您需要將引號放在方括號內：
$username = $_POST['username'];
如果您將session_start()放在頁面頂部（即在<html>標記之前等），這應該可以正常工作。
解決方案4
0 2015-09-04 11:34:35
<?php
session_start();
mysql_connect('localhost','root','');
mysql_select_db('database name goes here');
$error_msg=NULL;
//log out code
if(isset($_REQUEST['logout'])){
unset($_SESSION['user']);
unset($_SESSION['username']);
unset($_SESSION['id']);
unset($_SESSION['role']);
session_destroy();
}
//
if(!empty($_POST['submit'])){
if(empty($_POST['username']))
$error_msg='please enter username';
if(empty($_POST['password']))
$error_msg='please enter password';
if(empty($error_msg)){
$sql="SELECT*FROM users WHERE username='%s' AND password='%s'";
$sql=sprintf($sql,$_POST['username'],md5($_POST['password']));
$records=mysql_query($sql) or die(mysql_error());
if($record_new=mysql_fetch_array($records)){
$_SESSION['user']=$record_new;
$_SESSION['id']=$record_new['id'];
$_SESSION['username']=$record_new['username'];
$_SESSION['role']=$record_new['role'];
header('location:index.php');
$error_msg='welcome';
exit();
}else{
$error_msg='invalid details';
}
}
}
?>
// replace the location with whatever page u want the user to visit when he/she log in
解決方案5
0 2017-01-06 14:44:38
您的代碼可能看起來更像：
<?php
session_start(); $username = $password = $userError = $passError = '';
if(isset($_POST['sub'])){
$username = $_POST['username']; $password = $_POST['password'];
if($username === 'admin' && $password === 'password'){
$_SESSION['login'] = true; header('LOCATION:wherever.php'); die();
}
if($username !== 'admin')$userError = 'Invalid Username';
if($password !== 'password')$passError = 'Invalid Password';
}
?>
<!DOCTYPE html>
<html xmlns='http://www.w3.org/1999/xhtml' xml:lang='en' lang='en'>
<head>
<meta http-equiv='content-type' content='text/html;charset=utf-8' />
<title>Login</title>
<style type='text.css'>
@import common.css;
</style>
</head>
<body>
<form name='input' action='<?php echo $_SERVER['PHP_SELF'];?>' method='post'>
<label for='username'></label><input type='text' value='<?php echo $username;?>' id='username' name='username' />
<div class='error'><?php echo $userError;?></div>
<label for='password'></label><input type='password' value='<?php echo $password;?>' id='password' name='password' />
<div class='error'><?php echo $passError;?></div>
<input type='submit' value='Home' name='sub' />
</form>
<script type='text/javascript' src='common.js'></script>
</body>
</html>
創建沒有密碼硬編碼的簡單密碼登錄[PHP，Javascript，MySQL]
[英]Creating a simple Password login without the password hardcoded [PHP,Javascript,MySQL]
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