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我想使用php和mysql进行注册表单。
我使用xampp和phpmyadmin从这个和我的sql查询表'用户'工作正常。
但是当我尝试插入使用PHP代码时,它显示“用户注册失败”。下面
是我的代码
register.php
我在做什么这个register.php php代码错了?
<?php
require('includes/connect.php');
// If the values are posted, insert them into the database.
if (isset($_POST['username']) && isset($_POST['psw'])){
$username = $_POST['username'];
$email = $_POST['email'];
$password = $_POST['psw'];
$query = "INSERT INTO `user` (username, password, email) VALUES('$username', '$password', '$email')";
$result = mysqli_query($connection, $query);
if($result){
$smsg = "User Created Successfully.";
}else{
$fmsg ="User Registration Failed";
}
}
?>
<html>
<head>
<title>BondOnNet | Register</title>
<meta http-equiv="Content-type" content="text/html; charset=utf-8" />
<!-- JS -->
<script src="js/jquery-1.4.1.min.js" type="text/javascript"></script>
<script src="js/jquery.jcarousel.pack.js" type="text/javascript"></script>
<script src="js/jquery-func.js" type="text/javascript"></script>
<!-- End JS -->
<link rel="stylesheet" type="text/css" href="css/register.css">
</head>
<body>
<div id="Register_header">
<h1 id="logo"><a href="index.html">BondOnNet</a></h1>
</div>
<div id="register_container">
<form method="POST" style="border:1px solid #ccc">
<?php if(isset($smsg)){ ?><div class="alert alert-success" role="alert"> <?php echo $smsg; ?> </div><?php } ?>
<?php if(isset($fmsg)){ ?><div class="alert alert-danger" role="alert"> <?php echo $fmsg; ?> </div><?php } ?>
<div class="imgcontainer">
<img src="images/avatar.png" alt="Avatar" class="avatar">
</div>
<div class="container">
<label><b>Username</b></label>
<input type="text" placeholder="Enter Username" name="username" required>
<label><b>Email</b></label>
<input type="text" placeholder="Enter Email" name="email" required>
<label><b>Password</b></label>
<input type="password" placeholder="Enter Password" name="psw" required>
<label><b>Repeat Password</b></label>
<input type="password" placeholder="Repeat Password" name="psw-repeat" required>
<input type="checkbox" checked="checked"> Remember me
<p>By creating an account you agree to our <a href="terms.html" style="color: #4CAF50;">Terms & Privacy</a></p>
<div class="clearfix">
<button type="submit" class="signupbtn">Sign Up</button>
</div>
</div>
</form>
</div>
</body>
</html>
connect.php
<?php
$connection = mysqli_connect('localhost', 'root', '');
if (!$connection){
die("Database Connection Failed" . mysqli_error($connection));
}
$select_db = mysqli_select_db($connection, 'test');
if (!$select_db){
die("Database Selection Failed" . mysqli_error($connection));
}
?>
我试图代码,所以很多小的变化,但没有它的工作!
PLZ帮我理清什么即时做错了
'echo'你'$ query'变量,复制和直接在phpmyadmin中运行打印的查询。您会对查询失败的原因有所了解。
此外,使用'mysqli_error($连接)'在执行后打印查询的错误 – gaganshera
你的查询变量$ result的var_dump()是什么?将它放在if(){}语句中,以确保它在那里传递逻辑。 您是否尝试将原始查询注入SQL phpmyadmin(如果可以查看查询在SQL环境中是否正常运行? –
尝试使用[prepared statements](http://php.net/manual/en/mysqli.quickstart.prepared-statements.php)而不是引用字符串。这将有助于防止SQL注入并逃脱任何奇怪的字符。 – ethrbunny