PHP表单提交到MySQL数据库

Question

I have a registration form. In the database, the username and email are unique index. When the form submits and username or email are already present in the database, the values are not inserted. I want to notify the user that the values were not inserted. How can i do this?

HTML

<form action="register.php" method="post" id="reg" onsubmit='return validate();'>
    Company Name: 
    <input type="text"  class="inputs" name="name" id="name" /><br />
    Email:
    <input type="text" class="inputs" name="email" id="txtEmail" /><br />
    User name:
    <input type="text"  class="inputs"  name="uname" id="uname"/><br />
    Password:
    <input type="password" class="inputs" name="pass" id="pass1"/><br />
    Conferm Password:
    <input type="password" class="inputs" name="cpass"  id="pass2"/><br /><br /> 
    <input type="submit" value="Register" class="button" />
</form>

register.php:

include ("db.php"); 
if (isset($_POST['register'])) { 
    echo $name = ($_POST["name"]); 
    echo $email = ($_POST["email"]); 
    echo $uname = ($_POST["uname"]); 
    echo $password = ($_POST["pass"]); 
    mysqli_query($con,"INSERT INTO company_profile(user_name, password, company_name, email, phone, country, activation_string) VALUES ('$uname','$password','$name','$email','','','')"); 
} 

Answer 1

* Sweet And Short *

First check that username or email is exist or not using select query if resulting is 0 (it means not exists), Insert query will run ahead

<?php 
   if($_POST['register']){   
      $uname = $_POST['uname'];
      $email = $_POST['email'];
      $name= $_POST['name'];
      $pass= $_POST['pass'];
      $result =  mysqli_query($con, 'SELECT * from TABLE_NAME where email_id = "'.$email.'" or username = "'.$uname.'" ');
       if(mysqli_num_rows($result) > 0){
          echo "Username or email already exists.";
       }else{
         $query = mysqli_query($con , 'INSERT INTO TABLE_NAME (`email_id`, `username`,`name`,`pass`) VALUES("'.$email.'", "'.$email.'", "'.$uname.'","'.$name.'", "'.$pass.'")');

         if($query){
            echo "data are inserted successfully.";
         }else{
          echo "failed to insert data.";
         }
     } 
}
  ?>

Answer 2

查询方法将返回true或false，这取决于是否已插入行。

Answer 3

Try the following Code

    include ("db.php"); 
    if (isset($_POST['register'])) 
    { 
    echo $name = ($_POST["name"]); 
    echo $email = ($_POST["email"]); 
    echo $uname = ($_POST["uname"]);
    echo $password = ($_POST["pass"]);
   $var = mysqli_query('SELECT * from company_profile where email_id = "'.$email.'" or username = "'.$uname.'" ');
$num = mysqli_num_rows($var);
if($num==0)
{
    $result = INSERT INTO company_profile(user_name, password, company_name, email, phone, country, activation_string) VALUES ('$uname','$password','$name','$email','','','');
    $res = mysqli_query($result);
        if($res)
        {
        echo "Records Inserted Successfully!!";
        }
        else
        {
        echo "Records Inserted Failed!!";
        }
}
else
{
echo "User with the Details Already exists!!"
}
     }