创建一个类的函数PHP MySQL的AJAX [英] Creating a Like function php mysql ajax

问题描述

好吧，我一直在寻找和发现非常关注这个问题上， 但没有什么令人满意的。

Alright, I have been searching and finding very much about this subject, but nothing satisfying.

我想跟踪谁已经喜欢什么，不只是增加一个+1到表。

I want to keep track of who has liked what, not just add a +1 to a table.

我有三个表：文章，评论和喜欢

I have three tables: posts, comments and likes.

表的设计看起来像这样的时刻

The table design looks like this for the moment

CREATE TABLE IF NOT EXISTS `likes` (
  `like_id` int(11) NOT NULL PRIMARY KEY AUTO_INCREMENT,
  `post_id_fk` INT(11),
  `comment_id_fk` INT(11),
  `uid_fk` int(11) NOT NULL,
  `date` datetime NOT NULL DEFAULT '0000-00-00 00:00:00',
  `ip` varchar(39) COLLATE utf8_unicode_ci NOT NULL DEFAULT '0.0.0.0',
FOREIGN KEY (uid_fk) REFERENCES users(uid),
FOREIGN KEY (post_id_fk) REFERENCES post(post_id),
FOREIGN KEY (comment_id_fk) REFERENCES comments(comments_id),
) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci AUTO_INCREMENT=1;

正如你可以看到我有三个不同的外键，uid_fk用户的uid这样我就可以知道是谁喜欢什么。现在来出现的问题，一个外键的post_id，一个用于comments_id。

And as you can see i have three different foreign keys, uid_fk for the user uid so i can know who has liked what. And now comes there problem, one foreign key to post_id and one for comments_id.

MySQL不会接受一个外键，如果它不存在。如果我想像的注释，它不会让因为post_id_fk的外键。

Mysql won't accept a foreign key if it doesn't exist. if i want to "like" a comment, it won't let be because of the foreign key of post_id_fk.

如何解决这个数据库烂摊子？

How to solve this DB mess?

和AJAX的喜欢/ delike问题：

And the AJAX like/delike problem:

我发现这个<一href="http://stackoverflow.com/questions/13469947/jquery-changing-class-of-button-with-ajax-call">jQuery ：更改类的按钮使用AJAX调用当我正在寻找，它看起来simpel，非常漂亮。 ，我也遵循这一 http://pluscss.com/tutorials / AJAX样脚本使用-PHP-MySQL的，jQuery的教程。但我有问题，将它们结合起来。

I found this jQuery : Changing class of button with AJAX call when i was searching, and it looks simpel and very nice. And i have also follow this http://pluscss.com/tutorials/ajax-like-script-using-php-mysql-jquery tutorial. But I'm having problems combine them.

这是我想要做的：

1. 算喜欢目前的金额
2. 检查，如果用户之前都喜欢它
3. 给用户喜欢的选项（或delike如果pviously喜欢$ P $）

使用Ajax和like.php

with ajax and like.php

可能有人能帮我解决这个我会非常感激！

Could someone help me with this i would be very thankful!

推荐答案

这将是更好的分离表。 创建post_likes表和comments_likes表。 这样，不仅你摆脱你的存在的问题，但结构更加解耦和可重用性。

It would be better to separate the tables. Create a post_likes table and comments_likes table. That way not only you're getting rid of your existing problem but the structure is more decoupled and more reusable.

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