数据库更改中的自动通知：类似于Facebook的朋友请求 [英] automatic notification in database change :similar with facebook friend request

问题描述

我想开发一个基于php的mysql社交网站。注册用户可以选择将其他用户添加为朋友，就像在Facebook中一样。

如果用户A点击用户B的个人资料上的添加好友链接朋友请求记录将在A和B的数据库中进行。当B访问等待的朋友请求 - 显示页面（这样的个人资料页面）时，将显示该请求查询B的数据库。这很简单，我认为。

但是当B在线时，C可以向B发送一个朋友请求B.要通知B，C已经提出了这样的请求，即使B没有刷新他/她的个人资料页面（或任何页面）可以选择显示等待的朋友请求）。对于通知类型，它可以是一个显示朋友请求等待总数的框。单击该框将显示详细信息。或者它可以是任何其他形式。

我的兴趣是如何让B知道一个新的朋友请求，而B在线，而不使他/她刷新包含朋友请求的页面

解决方案

为了完成答案，我假设你理解数据库/通知DATA逻辑，现在您只关心如何将其传送到浏览器。我将在这里列出所有要点。

1. HTTP是一个 PULL 协议。


2. 所以，您可以做的是不断轮询服务器的新通知。

您可以进行两种类型的投票： -

1. 短轮询 - 您向服务器发送一个Ajax请求以获取新的通知。
   
   
   
   • 这是一个简短的请求，但是您会在一段时间内持续执行（例如10秒）
   
   
   • 服务器会立即处理PHP
   
   
   • 您处理返回的数据（例如显示通知）
   
   


2. 长轮询 - 您发送与上述相同的请求。 BUT ...
   
   
   
   • 在这里，请求的PHP文件进入无限循环，连续检查数据库。
   
   
   • 在那里，PHP文件回显它并结束循环。现在，请求完成，并在浏览器中收到数据。
   
   
   • 您处理数据。
   
   
   • 启动另一个长时间轮询的Ajax请求，步骤重复。
   
   
   • （额外：如果PHP没有在特定的超时时间内完成，则中止当前请求并启动新的请求。）
   
   

接下来，您可以通过两种不同的方式实现以下任一方法： -

1. 编写自己的Javascript代码 - 需要一些专业知识，但您可以了解它！


2. 使用一些图书馆 - 方便，省时。您可以使用 PubNet ， BeaconPush 等。

我希望这能给您一个清晰的想法！

i wish to develeop a php mysql based social networking site. Registered users will have the option to add another user as a friend just as is done in Facebook.

If user A clicks on the 'add friend' link on user B's profile, friend-request records will be made in A's and B's databases accordingly. When B visits the waiting friend-requests- showing-page ( as such the profile page), the request will be shown querying B's db.This much is pretty simple to do i think.

But while B is online, C can make a friend-request to B. I want to make a notification to B that C has made such a request even if B does not refresh his/her profile page(or any page with the option of showing the waiting friend-requests ). As to the type of notification, it can be a box showing the total number of friend-requests waiting. Clicking on the box will show the details. Or it could be in any other form.

My point of interest is how to make B aware of a new friend request while B is online without making him/her refresh the page containing the friend-requests?

解决方案

To finalize the answers, I assume that you understand the database/notification DATA logic well, and now you are concerned only about HOW TO deliver it to the browser. I will list all the points here.

1. HTTP is a PULL protocol. You cannot push data from Server to Client.
2. So, what you can do is - continuously poll the server for new notifications.

You can do two types of polling :-

1. Short-polling - You send an Ajax request to the server for new notifications.
   • This is a short request, but you do it continuously in an interval (say 10s)
   • The server process the PHP and immediately returns.
   • You process the returned data (eg. Show notifications)
2. Long-polling - You send the same request as above. BUT...
   • In this, the requested PHP file goes into an infinite loop, checking DB continuously.
   • When a DB change is there, the PHP file "echo" it and ends the loop. Now the request is completed, and data is received at Browser.
   • You process the data.
   • Start another long-poll Ajax request, and the steps repeat again.
   • (Extra: If the PHP is not completed in a particular timeout, then abort the current request and start new one.)

Next, you can implement any of these in two different ways:-

1. Write your own Javascript code - Needs some expertise, but you can learn it!
2. Use some libraries - Easy, and time-saving. You can use PubNet, BeaconPush etc.

I hope that this makes a clear idea to you!

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