通过ajax从mysqldb获取数据的问题


Problems with getting data via ajax from mysql db

我正在为支付服务打开一个inapp浏览器窗口。付款完成并成功后,我数据库中的完整条目将被更新。

用户表-类似的行

ID - USERNAME - VIPSTATUS - CALLSTATUS

付款后,vipstatus和callstatus更改为"1"。当inapp浏览器关闭时,我想从用户的vipstatus和callstatus中获取新数据,并覆盖现有的本地存储项callstatus和vip status。用户名作为用户名保存在本地存储中。我猜我的ajaxcode是错误的,因为它不起作用,甚至没有加载脚本。

我的Ajax代码

$(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]);
                                      }
                         });
                  }

我的更新.PHP代码

<?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();
  ?>

您的Javascript代码不是有效的Javascript。它漏掉了一些括号。试试这个:

$(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]);
                        }
                    });
                });
            }
        });
    });
});
<?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();
  ?>

修复了您的mysqli_*查询。关于CCD_ 1和CCD_。