如何添加我自己的评分系统? [英] How can I add my own rating system?

问题描述

如何添加评分系统?我的博客从数据库中获取帖子.

如果不存在则创建表`posts`(`id` int(11) NOT NULL AUTO_INCREMENT,`video` varchar(255) 非空，`title` 长文本非空，`text` 长文本非空，`posted_by` 长文本非空，主键(`id`)) ENGINE=InnoDB AUTO_INCREMENT=4 DEFAULT CHARSET=latin1 ROW_FORMAT=COMPACT;

在 index.php 中，我使用 echo 调用它们，一切正常.

示例:

 
<div class="row"><div class="col-md-8"><h1 style="颜色:白色;"class="my-4">我的博客<div class="card mb-4"><div class="embed-responsive embed-responsive-16by9"><iframe style="border:0;"class="embed-responsive-item" width="450" height="240" src=""allowfullscreen></iframe>
<div class="card-body"><h2 class="card-title">Title</h2><p>文本</p>

<div class="card-footer text-muted">页脚</div>

解决方案

For Rating System 创建表例如post_rating 将列作为id 为 int(自动增量)，post_id 为 int，user_id 为 int(如果适用于您的项目)评级为 int

然后当用户想要给 posts_rating 表提供评分时插入值.

要显示帖子的当前评分，请执行 sql 语句:

选择 round(AVG(rating),1) 作为 currentRating FROM post_rating WHERE post_id=".$post_id;

How can I add a rating system? My blog gets its posts from database.

CREATE TABLE IF NOT EXISTS `posts` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `video` varchar(255) NOT NULL,
  `title` longtext NOT NULL,
  `text` longtext NOT NULL,
  `posted_by` longtext NOT NULL,
  PRIMARY KEY (`id`)
) ENGINE=InnoDB AUTO_INCREMENT=4 DEFAULT CHARSET=latin1 ROW_FORMAT=COMPACT;

In index.php I call them using echo, everything works fine.

Example:

    <div class="container">
      <div class="row">
        <div class="col-md-8">
          <h1 style="color:white;" class="my-4"> My Blog
          </h1>
            <div class="card mb-4">
           <div class="embed-responsive embed-responsive-16by9">
           <iframe style="border:0;" class="embed-responsive-item" width="450" height="240" src="" allowfullscreen></iframe>
           </div>
            <div class="card-body">
              <h2 class="card-title">Title</h2>
              <p>Text</p>
            </div>
            <div class="card-footer text-muted">Footer</div>
          </div>

解决方案

For Rating System create a table e.g. posts_rating having columns as id as int(auto incrememt), post_id as int, user_id as int(if applicable in your project) rating as int

Then when a user wants to give rating insert value to the posts_rating table.

To display current rating for the post execute the sql statement:

select round(AVG(rating),1) as currentRating FROM post_rating WHERE post_id=".$post_id;

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