PHP+Ajax：添加好友系统

发布于 2024-09-29 10:28:12 字数 1075 浏览 6 评论 0原文

我正在尝试制作一个添加朋友的小项目。当您单击“添加好友”按钮时，您将发送带有好友 ID 和信息的 Ajax 调用。 username（它们是每个添加按钮的 id 和 name 属性）添加到 add.php 文件中。（mysql结构是：1个users表+X个以用户名命名的表（列：friendid，ispending））在 PHP 文件中只有 2 个 MySQLi 查询：这是添加文件的代码：

session_start();
$friendid = $_POST['id'];
$myname = $_SESSION['username'];
$friendname = $_POST['name'];
$myid = $_SESSION['id'];
$add = new Mysqlconnect();
$add->db->query("INSERT INTO $myname VALUES($friendid, 'yes')");
$add->db->query("INSERT INTO $friendname VALUES ($myid, 'yes')");
$add->db_Close();

需要 Mysqlconnect 类我只是不希望这里的代码太长。这是 Ajax 调用：

$('.add').click(function(){
 var name = $(this).attr("name");
 var id = $(this).attr("id");
   $.ajax({
      type: "POST",
      data: "&name="+name+"&id="+id,
      url: 'add.php',
      success: function(){
         alert("success");
      }
   });
});

问题： 当我单击“添加朋友”时，它确实会提醒“成功”，但每次只有一张表得到更新，甚至根本没有表。虽然有一次我点击了它并且它确实起作用了（那次我没有更改代码，我尝试每 20 秒点击一次）。

我该如何解决这个问题？

原文

I'm trying to make a little project for adding friends.
When you click the add friend button you are sending an Ajax call with the friend id & username (they are the id and name attributes of each add button) to the add.php file.
(The mysql structure is: 1 users table + X tables named the user's name(columns: friendid, ispending))
In the PHP file there are only 2 MySQLi queries: Here is the code for the add file:

session_start();
$friendid = $_POST['id'];
$myname = $_SESSION['username'];
$friendname = $_POST['name'];
$myid = $_SESSION['id'];
$add = new Mysqlconnect();
$add->db->query("INSERT INTO $myname VALUES($friendid, 'yes')");
$add->db->query("INSERT INTO $friendname VALUES ($myid, 'yes')");
$add->db_Close();

The Mysqlconnect class is required i just didn't want the code here to be too long.
Here's the Ajax call:

$('.add').click(function(){
 var name = $(this).attr("name");
 var id = $(this).attr("id");
   $.ajax({
      type: "POST",
      data: "&name="+name+"&id="+id,
      url: 'add.php',
      success: function(){
         alert("success");
      }
   });
});

THE PROBLEM :
When I click "add friend" it does alert "success" but only one table gets updated everytime or even no table at all.
Though one time I clicked it and it did work (I did not change the code that time, I tried clicking like every 20 seconds).

How can I solve this?

分享到QQ

分享到微博

如果你对这篇内容有疑问，欢迎到本站社区发帖提问参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。

发布评论

需要登录才能够评论，你可以免费注册一个本站的账号。

游魂 2024-10-06 10:28:12

如果我理解正确的话，你似乎正在为每个用户的朋友创建一个新表（1个用户表+ X个名为用户名的表），这不是一个好方法，你最好只使用2个表：users和user_friends如下：

drop table if exists users;
create table users
(
user_id int unsigned not null auto_increment primary key,
username varbinary(32) unique not null
)
engine=innodb;

drop table if exists user_friends;
create table user_friends
(
user_id int unsigned not null,
friend_user_id int unsigned not null,
created_date datetime not null,
primary key (user_id, friend_user_id) -- note clustered composite PK (innodb only)
)
engine=innodb;

可以在此处找到完整的示例脚本：http://pastie.org/1242699

希望这会有所帮助。

if i understood you correctly you seem to be creating a new table for each user's friends (1 users table + X tables named the user's name) this isnt a good approach and you'd be better off with just 2 tables: users and user_friends as follows:

drop table if exists users;
create table users
(
user_id int unsigned not null auto_increment primary key,
username varbinary(32) unique not null
)
engine=innodb;

drop table if exists user_friends;
create table user_friends
(
user_id int unsigned not null,
friend_user_id int unsigned not null,
created_date datetime not null,
primary key (user_id, friend_user_id) -- note clustered composite PK (innodb only)
)
engine=innodb;

A full example script can be found here : http://pastie.org/1242699

Hope this helps.

回复收藏 0 原文

~没有更多了~