为什么我的用户注册表不能正常工作？

Question

我想用php和mysql创建一个用户注册表单。当我尝试点击提交按钮时，没有新记录添加到我的数据库中。该数据库是功能性的，并已与其他形式。

HTML/FORM

<?php 
include 'header.php'; 
?> 
    <section> 
     <div class="form"> 
      <form action="signup.php" method="post"> 
       <h1> Sign Up!</h1> 

       <p>First name: 
        <input type="text" name="fName" maxlength="15" required pattern="^[a-zA-Z]{3,20}$" placeholder="Enter Name" /> 
       </p> 

       <p>Last name: 
        <input type="text" name="lName" maxlength="15" pattern="^[a-zA-Z]{3,20}$" required placeholder="Enter Last Name" /> 
       </p> 

       <p>Email: 
        <input type="email" name="email" maxlength="40" required placeholder="Enter Email" /> 
       </p> 

       <p>Username: 
        <input type="text" name="username" maxlength="20" ^[A-Za-z0-9_]{1,15}$ required placeholder="Enter Username" /> 
       </p> 

       <p>Password: 
        <input type="password" name="password" maxlength="20" pattern="(?=^.{8,}$)((?=.*\d)|(?=.*\W+))(?![.\n])(?=.*[A-Z])(?=.*[a-z]).*$" required placeholder="Enter Password" /> 
       </p> 

       <p>Re-type Password: 
        <input type="password" name="password2" maxlength="20" pattern="^(?=.*\d)(?=.*[a-z])(?=.*[A-Z])(?!.*\s).*$" required placeholder="Re-type Password" /> 
       </p> 

       <p> 
        <button type="submit" name="signupbutton"> Sign up </button> 
       </p> 
      </form> 
     </div> 
    </section> 

<div class="footerspecial"> 
    <?php 
include 'footer.php'; 
?> 
</div> 

PHP/SQL

<?php 

//have they submitted at least once? 
if(isset($POST['$password2'])){ 
    $fName = $_POST['fName']; 
    $lName = $_POST['lName']; 
    $username = $_POST['username']; 
    $email = $_POST['email']; 
    $password = $_POST['password']; 
    $password2 = $_POST['password2']; 
    //do the passwords NOT match? 
     if ($password !== $password2) {//do string comparison here 
       echo'<h2>Error: passwrods don\'t match!</h2>'; 
       require ('registerform.php'); 
     } 
    else { 
      //does the username already exist? 
      $sql = mysql_query("SELECT * FROM users WHERE username=='$username'"); 
     if ($results=$con->query($sql)){ 
       echo'<h2>Error: username is already taken</h2>'; 
       require ('registerform.php'); 
     } 
     else { 

      $sql = mysql_query("SELECT * FROM users WHERE email=='$email'"); 
     if ($results=$con->query($sql)){ 
      echo'<h2>Error: email already used</h2>'; 
      require ('registerform.php'); 
      } 
      else { 
    // If the values are posted, insert them into the database. 
        $sql= "INSERT INTO users (fName, lName, email, username, password, password2) VALUES ('$fName', '$lName', '$email', '$username', '$password', $password2)"; 
        if (!$con->query($sql)){ 
         echo 'Error: coulndt do suff'; 
        } 
        else { 
         echo 'Account made'; 


        }//ENDS SUCCESSFUL INSURT 

      }//ENDS EMAIL VALIDATION 

     }//ENDS THE USERNAME VALIDATION 
    }//END PASSWORD VALIDATION 
} 
?> 

图片形式的真的不知道，如果它的帮助，但ya'know

https://gyazo.com/418b86ecb5090604a1f229e1e94fe3bf

Answer 1

我这里猜测你的数据库没有password2列（似乎有点毫无意义），所以试图插入它会给出一个错误。

你应该阅读有关MySQLi error reporting

在你的PHP文件的开始同时添加error_reporting(-1);显示PHP错误。

P.S.你的代码很容易被SQL注入，你应该使用prepared statements来保证安全。

Answer 2

首先，您可能会遇到多个问题，而您在$ password2周围没有单引号。这可能导致插入失败。

VALUES（ '$ FNAME'， '$ L-NAME'， '$电子邮件'， '$用户名'， '$密码'，$密码2）“;

另外我想呼应的SQL错误出如你不这样做，你可以很容易地做这个测试的if语句为真不假

   if ($con->query($sql)){ 
        //if true then runs your code; 
       } 
       else { 
        echo "Error: " . $sql . "<br>" . $con->error; // This will echo out any sql errors you may have 
       }