我使用以下代码将用户插入到名为“accounts”的表中
session_start();
include("include/connect.php");
//Posted information from the form put into variables
$username = mysqli_escape_string($conn, $_POST["username"]);
$email = mysqli_escape_string($conn,$_POST["email"]);
$password = mysqli_escape_string($conn,$_POST["password_1"]);
//check if user already exists
$usernameCheck="SELECT * FROM accounts WHERE username='$username'";
//query the db
$usernameResult = mysqli_query($conn, $usernameCheck) or die(mysqli_error($conn));
//if no users exits then add data
if(mysqli_num_rows($usernameResult) == 0){
$AddUser = "INSERT INTO `accounts` (`username`, `email`, `password`) VALUES ('$username', '$email', '$password')";
$newUserSend = mysqli_query($conn, $AddUser) or die(mysqli_error($conn));
mysqli_close($conn);
$_SESSION["user_session"]= $username;
}
然而,每次运行时,它都会插入一行包含正确数据的行,然后在下面插入一个空白行。例如
ID Name email password
1 test [email protected] test
2
这是发布数据的表单:
<form method="POST" class="signup-form" action="RegisterProcess.php">
<h2 class="form-title">Create account</h2>
<div class="form-group">
<input type="text" value="<?php echo $username; ?>" class="form-input" name="username" placeholder="Your Name" required/>
</div>
<div class="form-group">
<input type="email" value="<?php echo $email; ?>" class="form-input" name="email" placeholder="Your Email Address" required/>
</div>
<div class="form-group">
<input type="password" class="form-input" name="password_1" placeholder="Your Password" required/>
<span toggle="#password" class="zmdi zmdi-eye field-icon toggle-password"></span>
</div>
<div class="form-group">
<input type="password" class="form-input" name="password_2" placeholder="Confirm your password" required/>
<span toggle="#password" class="zmdi zmdi-eye field-icon toggle-password"></span>
</div>
<div class="form-group">
<input type="submit" class="form-submit" value="Sign up"/>
</div>
</form>
None
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