Ajax calls PHP file but some datas doesn't return

sedric alonso 0
Viewed 40

"undone" OR "error" RETURNS. BUT "done" DOESN'T RETURN. When mail adress exists 'undone' returns, inserting new user to database works but 'done' doesn't return. I wanna see "undone" from $data["status"] return when user is inserted. How can i fix?

js

$(document).ready(function(){
   $("#form").submit(function(){
     $.ajax({
       url: "file.php",
       type: "POST",
       dataType: "json",
       data: $('#form').serialize(),
       success: function(data){
         alert(data.status);
       }
     })
   })
 })

file.php

<?php
   session_start();
   if (isset($_POST['submit'])){

     include 'config.php';
     $data = array();
     $name = $conn -> real_escape_string($_POST['username']);
     $mail = $conn -> real_escape_string($_POST['mail']);
     $psw = $conn -> real_escape_string($_POST['password']);
     $code = $conn -> real_escape_string(uniqid());
 
 
 
     $qry = $conn->query("SELECT * FROM member WHERE mail='{$mail}'");
     if (mysqli_num_rows($qry)>0) {
       $data["status"] = "undone";
     }else {
       $qry2 = $conn->prepare("INSERT INTO member (name, mail, psw, code) VALUES ('{$name}','{$mail}','{$psw}','{$code}')");
       if($qry2){
         $qry2->execute();
         $_SESSION['SESSION_MAILADDRESS'] = $mail;
         $data["status"] = "done";
 
         mysqli_free_result($qry2);
       }else {
         $data["status"] = "error";
       }
     }
     echo json_encode($data);
     mysqli_free_result($qry);
     $conn->close();
   }
  ?>
mysqli ajax php
1 Answers

You need to remove

mysqli_free_result($qry2);

from your code.

As per https://php.net/manual/en/mysqli-result.free.php you're supposed to pass in a mysqli result object to that function, but now I look closely, you passed it a mysqli statement object. So that should cause a crash. I expect your AJAX call returned a 500 (internal server error) status and an error message which ought to be visible from the browser's Network tool as I described.

In any case using the mysqli_free_result command makes no sense in this context, because you don't have a result to free - a INSERT query does not generate a result set.

ADyson0
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