Question

我制作了一个用于Android设备的工作注册脚本。唯一剩下的就是向欢迎他的用户发送电子邮件，即“欢迎使用该应用......”等等。但是，由于没有发送电子邮件，我对此感到困惑。

这是我的剧本。

<?php
   require "init.php";
   header('Content-type: application/json');

  $email = $_POST['email'];
  $user_name = $_POST['user_name'];
  $user_pass = $_POST['user_pass'];
  $msg = "Congratulations. You are now registered to the most amazing app    
  ever!";

  $passwordEncrypted = sha1($user_pass);         

    if($email && $user_name && $user_pass){

        $sql_query = "select * from user_info WHERE email ='".mysqli_real_escape_string($con, $email)."' and user_name ='".mysqli_real_escape_string($con, $user_name)."'";

        $result = mysqli_query($con, $sql_query);   

        $results = mysqli_num_rows($result);

        if ($results){
            $don = array('result' =>"fail","message"=>"Email or username exists.");
        }else{

            $sql_query = "insert into user_info values('$email','$user_name','$passwordEncrypted');";

            if(mysqli_query($con,$sql_query)){

                $don = array('result' =>"success","message"=>"Successfully registered!Well done");
                mail("$email","Well done",$msg);
            }
        }
    } if(!$email || !filter_var($_POST['email'], FILTER_VALIDATE_EMAIL)){

        $don = array('result' =>"fail","message"=>"Please enter a valid email");

    }else if(!$user_name){
        $don = array('result' =>"fail","message"=>"Please enter your username");
    }else if(!$user_pass || (strlen($user_pass)<8 || (!preg_match('`[A-Za-z0-9]`', $user_pass)))){
        $don = array('result' =>"fail","message"=>"Password must contain 8 letters or more with at least one capital letter ");

    }


  echo json_encode($don);

?>

关于那个的任何想法？

感谢。

Answer 1

您的$email变量不应该在引号("$email")中，它应该是这样的 -

改变这个 -

mail("$email","Well done",$msg);

到此 -

mail($email,"Well done",$msg);

从php发送电子邮件不起作用

1 个答案: