我真的很新,所以我赞成如果你能保持答案真的很简单和基本所以我能理解。 所以我试图建立一个简单的登录平台,但它不起作用.. 我认为我对sql代码有一些问题,因为$ rows标记并不算作适合包含在数据库中的用户的一行。 thx in advanced。
<?php
session_start();
?>
include "config.php";
$wrong_dt="";
if(isset($_POST['email']) && $_POST['email']!="" && isset($_POST['password']) && $_POST['password']!=""){
$pw = mysqli_prepare($conn , "SELECT `email` FROM `Users` WHERE `email` = ? AND `password` = ? LIMIT 1");
if($pw){
echo "good";
echo $_POST['email'];
echo $_POST['password'];
mysqli_stmt_bind_param($pw, "ss" , $_POST['email'] , $_POST['Password']);
mysqli_stmt_execute($pw);
$result= mysqli_stmt_get_result($pw);
echo mysqli_num_rows($result);
if(mysqli_num_rows($result) > `0){
$_SESSION['email']=$_POST['email'];
header("Location:Find.php");
}
else{
$wrong_dt="wrong details";
}
}
}
?>
MusicRun
</head>
<header>
<h1> MusicRun </h1>
<h5>Same Hobby.Same Music.</h5>
<p>MusicRun is a platform for people who likes running in a group with other people who likes the same music as they do</p>
</header>
<body background= "http://www.spyderonlines.com/images/nike_sports_athletics_run_running_81188.jpg">
<main>
<form method="post" action="login.php">
<div class="form-group">
<label for="exampleInputEmail1">Email address</label>
<input type="email" class="form-control" name="email" id="exampleInputEmail1" aria-describedby="emailHelp" placeholder="Enter email">
<small id="emailHelp" class="form-text text-muted">We'll never share your email with anyone else.</small>
</div>
<div class="form-group">
<label for="exampleInputPassword1">Password</label>
<input type="password" class="form-control" name="password" id="exampleInputPassword1" placeholder="Password">
<button type="submit" class="btn btn-primary" id="loginBtn">Login</button>
</div>
<p><?php echo $wrong_dt ?></p>
<div class="newAccount">
<p>Not a User yet?</p>
<button type="button" class="btn btn-primary" id="login_button" onclick="window.location.href='http://matanhm.myweb.jce.ac.il/Final%20Project/register.php'">Sign up</button>
</div>
</form>
</main>
<footer>
</footer>
</body>