Question

我有以下注册php脚本。我希望得到一个json回复 {＆＃34;导致＆＃34;：＆＃34;成功＆＃34;＆＃34;消息＆＃34;：＆＃34; 123＆＃34;}

其中123是注册用户的id。我想要这个id，以后用户可以将数据发布到其他表。

但是我明白了。

{"result":"fail","message":null}

这是我的剧本。

<?php
  session_start();
  require "init.php";
  header('Content-type: application/json');
  $id = $_POST['id'];
  $email = $_POST['email'];
  $user_name = $_POST['user_name'];

  $user_pass = $_POST['user_pass'];
  $passwordEncrypted = sha1($user_pass);  

  $confirmPass = $_POST['confirm_pass'];
  $confPasswordEncrypted = sha1($confirmPass);  

  $msg = "Congratulations. You are now registered to the most amazing app   
  ever!";            

        if(!filter_var($email, FILTER_VALIDATE_EMAIL)){

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

        }    

if($email && $user_name && $user_pass && $confirmPass && filter_var($email, FILTER_VALIDATE_EMAIL)){


    $sql_query = "select * from user_info WHERE email  ='".mysqli_real_escape_string($con, $email)."' or user_name 
    ='".mysqli_real_escape_string($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{
        //This is where I am trying to get the id
        while($row = mysqli_fetch_array($result)) {             
            $posts['id'] = $row['id'];


        }   

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

        if(mysqli_query($con,$sql_query)){
            $_SESSION['id'] = mysqli_insert_id($con);
            //And this is the json response I was talking about
            $don = array('result' =>"success","message"=>$posts['id']);
            mail($email,"Well done. You are registered to my sample app!",$msg);

        }
    }
}else if(!$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){

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

    }else if(!confirmPass){

        $don = array('result' =>"fail","message"=>"Please confirm your    
        password");

    }     

  echo json_encode($don);

 ?>

Answer 1

更改

$don = array('result' =>"success","message"=>$posts['id']);

到

$don = array('result' =>"success","message"=>$_SESSION['id']);

$ posts [＆＃39; id＆＃39;]始终为null，因为该行未插入数据库。删除该代码。

Answer 2

更改

$don = array('result' =>"success","message"=>$posts['id']);

要：

$don = array('result' =>"success","message"=>mysqli_insert_id($con));

问题在于您指的是始终为$posts['id']的{{1}}您尝试在此设置：

null

请注意，如果$results = mysqli_num_rows($result); if ($results){ $don = array('result' =>"fail","message"=>"Email or username exists."); }else{ while($row = mysqli_fetch_array($result)) { $posts['id'] = $row['id']; } ...不包含任何行，我们只会到达while。因此，$result为false，并且此循环从不执行。事实上，这个循环在这个脚本中没用，应该删除。

这超出了您的问题的范围，但您应该考虑以下事项：

mysqli_fetch_array($result)的重点是什么？你永远不会检查密码和确认通行证匹配。
鉴于两者应该相同，为什么要存储密码和确认传递？
在不使用预准备语句的情况下，不应在SQL查询中使用用户提供的值而不使用它们，甚至更好。您的$confirmPass查询至少会转义值，但您的SELECT查询却没有。这让你对SQL注入攻击持开放态度。
INSERT不是散列pwd进行存储的好方法。请改用PHP的sha1($user_pass)和password_hash函数。 See the guide

Answer 3

您不需要以下行，因此请在选择查询之前删除它们。

//This is where I am trying to get the id
while($row = mysqli_fetch_array($result)) {             
   $posts['id'] = $row['id'];

}

您可以替换

 $don = array('result' =>"success","message" => $posts['id']);

带

$don = array('result' =>"success","message"=> mysqli_insert_id($con));

也不需要会话。

在我的json响应中获取null值。

3 个答案: