Problems with getting data via ajax from mysql db

Question

Im opening a inappbrowser window for a payment service. After the Payment is done and sucessfull entries in my db will be updated.

user table - rows like

ID - USERNAME - VIPSTATUS - CALLSTATUS

The vipstatus and callstatus gets changed to "1" after payment.When the inappbrowser will be closed i want to get the new data from the vipstatus and callstatus from the user and overwrite the existing localstorage items callstatus and vip status. Username is saved in localstorage as username. Im guessing my ajaxcode is wrong, because it dont work, it dont even load the script.

My Ajax Code

$(document).ready( function() {
                  $("#paybutton").click(function() {
                                        var params = "projectpaymentoption=1195&id=",
                                        usernamepay = window.localStorage.getItem("username"),
                                        paymenturl = params + usernamepay;

                                        $.ajax({
                                               type: 'POST',
                                               url: 'http://www..de/phone/encode.php',
                                               data: $.param({"paymenturl": paymenturl}),
                                               success: function(result) {
                                               var paybrowser = window.open(result,'_blank','location=no','closebuttoncaption=Zurück');

                                               paybrowser.addEventListener('exit',function(event) {


                                               $.ajax({
                                               type: 'POST',
                                               url: 'http://www..de/update.php',
                                               data: $.param({"username": username}),
                                               success: function(data) {
                                               window.localStorage.setItem("vipstatus", data[2]);
                                               window.localStorage.setItem("callstatus",data[3]);
                                      }
                         });
                  }

My UPDATE.PHP Code

<?php 


  $dbhost = "blabla";
  $dbuser = "blabla";
  $dbpass = "blabla";
  $dbname = "blabla";
  $tableName = "user";


  print_r($_POST);
  $user = $_POST['data']['username'];
  print PHP_EOL . $user . PHP_EOL;

  $con = mysqli_connect($dbhost,$dbuser,$dbpass);
  $dbs = mysqli_select_db($dbname, $con);
  $result = mysqli_query("SELECT user$ FROM $tableName");         
  $array = mysqli_fetch_row($result);           
  $conn->close();
  ?>

Why don't you tag your posting with JQuery?

Erwin Moller
– Erwin Moller

2015-07-07 13:27:23 +00:00
Commented Jul 7, 2015 at 13:27 — Erwin Moller
– Erwin Moller, Commented Jul 7, 2015 at 13:27

Lukas Boersma · Accepted Answer · 2015-07-07 13:30:35Z

Your Javascript code is not valid Javascript. It misses some closing brackets. Try this:

$(document).ready(function() {
    $("#paybutton").click(function() {
        var params = "projectpaymentoption=1195&id=",
            usernamepay = window.localStorage.getItem("username"),
            paymenturl = params + usernamepay;

        $.ajax({
            type: 'POST',
            url: 'http://www..de/phone/encode.php',
            data: $.param({
                "paymenturl": paymenturl
            }),
            success: function(result) {
                var paybrowser = window.open(result, '_blank', 'location=no', 'closebuttoncaption=Zurück');

                paybrowser.addEventListener('exit', function(event) {

                    $.ajax({
                        type: 'POST',
                        url: 'http://www..de/update.php',
                        data: $.param({
                            "username": username
                        }),
                        success: function(data) {
                            window.localStorage.setItem("vipstatus", data[2]);
                            window.localStorage.setItem("callstatus", data[3]);
                        }
                    });
                });
            }
        });
    });
});

Jamie Harding · Accepted Answer · 2015-07-07 14:30:43Z

1

<?php 

  $dbhost = "blabla";
  $dbuser = "blabla";
  $dbpass = "blabla";
  $dbname = "blabla";
  $tableName = "user";

  $con = mysqli_connect($dbhost,$dbuser,$dbpass);
  $dbs = mysqli_select_db($con, $dbname);

  print_r($_POST);
  $user = mysqli_real_escape_string($con, $_POST['data']['username']);
  print PHP_EOL . $user . PHP_EOL;

  $result = mysqli_query($con, "SELECT $user FROM $tableName");         
  while ($array = mysqli_fetch_row($result)) {
  echo $array['vipstatus'];

  echo $array['callstatus'];
  }           
  $conn->close();
  ?>

Fixed your mysqli_* query. On $result and $dbs.

edited Jul 7, 2015 at 14:30

answered Jul 7, 2015 at 13:27

Jamie Harding

3833 silver badges22 bronze badges

7 Comments

Jamie Harding Over a year ago

Just tidied up the code and sanitized the username, I don't like variables that are not prepared/sanitized.

Alvaro Montoro Over a year ago

Technically, that's just a sanitized variable and may not be enough to prevent SQL injection. It would be better to use parameterized queries

mav Over a year ago

When i use that the the variables in local storage for vipstatus will be "a" and for vipstatus "r". @JamieSterling

Jamie Harding Over a year ago

@mav It will input anything you want it to input into vipstatus.

mav Over a year ago

yeah but i want to get the callstatus and vipstatus from the row for the user and not "a" and "r" :P

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