Question

我有一份登记表。在数据库中，用户名和电子邮件是唯一索引。当表单提交并且数据库中已存在用户名或电子邮件时，不会插入值。我想通知用户没有插入值。我怎么能这样做？

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

* 甜蜜和短暂*

首先使用 select 查询检查用户名或电子邮件是否存在，如果结果为0（表示不存在）， Insert 查询将提前运行

<?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，具体取决于是否已插入行。

Answer 3

尝试以下代码

    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!!"
}
     }

php表单提交到mysql数据库

3 个答案: