Question

var warp1_data = {
  "id": "warp1",
  "x": 371,
  "y": 95,
  "images": 'map/town/balamb.png',
  "framerate": 32,
  "animations":
    {
      "down": [0, 3, "down", 0.25],
      "left": [4, 7, "left", 0.25],
      "right": [8, 11, "right", 0.25],
      "up": [12, 15, "up", 0.25],
      "downLeft": [16, 19, "downLeft", 0.25],
      "downRight": [20, 23, "downRight", 0.25],
      "upLeft": [24, 27, "upLeft", 0.25],
      "upRight": [28, 31, "upRight", 0.25]
    },
  "frames":
    {
      "height": 32,
      "width":32,
      "regX": 0,
      "regY": 0,
      "count": 32
    },
  "movement":
    {
      "type" : '8dir',
      "distanceOnPress" : 4,
      "keycodeUp" : -1,
      "keycodeDown" : -1,
      "keycodeLeft" : -1,
      "keycodeRight" : -1
    },
  "collisionObject": true,
  "collisionType": "pixel",
  "collisionAlphaThreshold": 10,
  "onCollision": function() {
    //only allow the collision text once every 5 seconds
    if (!window.playerCollisionTxt) {
      window.playerCollisionTxt = true;
      setTimeout(function(){ window.playerCollisionTxt = false; }, 5000);


        $.ajax({
            url: "update_loc.php",
            type: "GET",
            data: { 'username': user, 'liked': '1' },                   
            success: function()
                        {
                            alert("ok");                                    
                        }
        });

      location.reload();
    }
  }
};

这段代码有问题吗？我已将此添加到我的HTML中：

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.12.0/jquery.min.js"></script>

我想要的只是将JavaScript AJAX连接到SQL数据库。

无论如何，我正在使用深渊网络服务器最新版本＆amp;这是我的PHP文件：

<?php

include "config.php";
$get_move_loc_1 = mysql_query("SELECT * FROM ".$CONFIG['database'].".login WHERE userid='".$_SESSION['ff_account']."' LIMIT 1");
$get_move_loc_2 = mysql_fetch_array($get_move_loc_1);

$get_move_loc_3 = mysql_query("SELECT * FROM ".$CONFIG['database'].".char WHERE account_id='".$get_move_loc_2['account_id']."' LIMIT 1");
$get_move_loc_4 = mysql_fetch_array($get_move_loc_3);

$char_id = $get_move_loc_4['char_id'];

mysql_query("UPDATE ".$CONFIG['database'].".char SET last_map='Balamb Garden' WHERE char_id='$char_id'");
mysql_query("UPDATE ".$CONFIG['database'].".char SET map_x='371' WHERE char_id='$char_id'");
mysql_query("UPDATE ".$CONFIG['database'].".char SET map_y='115' WHERE char_id='$char_id'");

＆GT;

Answer 1

您的jQuery版本来自Stone Age（版本1.2真的??）。请更新到最新版本。
你的ajax请求没问题。如果达到update_loc.php则没问题。
但是在ajax调用之后，你有location.reload();重新加载页面。我认为这会让你感到困惑。删除它就是全部。

ajax不起作用，无法加载update_loc.php

1 个答案: