Question

我搜索过网站，所有类似的答案都是指单独的评级表，而不是像这样的实现。

我已将wordpress网站转换为PHP / MYSQL中的自建网站，想象实际的帖子是评论，评论是与原帖相关的实际评论。

我遇到的问题是我对帖子表有一个评级字段，并为每个评论输入了评分字段。

我正在尝试使用别名AvgRating作为一个，但它只返回评论评级（c.rating），如果该行被删除它会从posts表中带回评论评级，有没有办法加入别名作为平均评分？显然只有这部分是错误的，但不会引发错误。

SELECT p.post_id
     , p.title
     , AVG(p.rating) AvgRating
     , AVG(c.rating) AvgRating -- returns average from comments field, if line is removed will return posts rating value instead..
  FROM posts p
  LEFT 
  JOIN comments c 
    ON p.post_id = c.post_id 
 WHERE p.post_id = $post_id
   AND c.post_id = $post_id
   AND active = 1  
 GROUP 
    BY p.post_id
     , p.title;

我正在尝试获取架构聚合计数标记的平均值。有没有办法在一个查询中实现这一点？或者我唯一的选择是使用2个查询返回2个平均值并使用PHP函数返回两者之间的平均值？

这是一个测试数据库：

CREATE TABLE `posts` (
`post_id` int(11) NOT NULL,
`cat_id` int(11) NOT NULL,
`rating` int(11) NOT NULL,
`title` varchar(255) NOT NULL,
`content` varchar(5000) NOT NULL,
`date_added` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP,
`author` varchar(255) NOT NULL,
`active` int(11) DEFAULT '0',
`slug` varchar(255) NOT NULL,
`user_submitted` int(11) NOT NULL DEFAULT '1'
) ENGINE=InnoDB AUTO_INCREMENT=2157 DEFAULT CHARSET=latin1;

CREATE TABLE `comments` (
`cmt_id` int(11) NOT NULL,
`post_id` int(11) NOT NULL,
`rating` int(11) NOT NULL,
`comment` varchar(5000) NOT NULL,
`cmt_author` varchar(255) NOT NULL,
`cmt_email` varchar(255) NOT NULL,
`ip_address` varchar(100) NOT NULL,
`cmt_date` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP,
`cmt_status` int(11) NOT NULL DEFAULT '0'
) ENGINE=InnoDB AUTO_INCREMENT=397 DEFAULT CHARSET=latin1;

INSERT INTO `posts` (`post_id`, `cat_id`, `rating`, `title`, `content`, `date_added`, `author`, `active`, `slug`, `user_submitted`) VALUES
(2, 9, 5, 'site1.com', 'Original post review body', '2010-05-07 05:00:00', '', 1, 'site1-com', 1);

INSERT INTO `comments` (`cmt_id`, `post_id`, `rating`, `comment`, `cmt_author`, `cmt_email`, `ip_address`, `cmt_date`, `cmt_status`) VALUES
(1, 2, 5, 'it is a good site test review', 'Anonymous', '', '', '2010-01-06 21:51:00', 1),
(2, 2, 3, 'it is an average site test review', 'Anonymous', '', '', '2010-01-06 20:51:00', 1),
(3, 2, 1, 'it is an bad site test review', 'Anonymous', '', '', '2010-01-06 23:51:00', 1),
(4, 2, 0, 'neutral test review', 'Anonymous', '', '', '2010-01-06 22:51:00', 1),
(5, 2, 2, 'below average test review', 'Anonymous', '', '', '2010-01-06 22:51:00', 1);

查询运行：

SELECT p.post_id
 , p.title
 , AVG(p.rating) AvgRating
 , AVG(c.rating) AvgRating -- returns average from comments field, if line is removed will return posts rating value instead..
FROM posts p
LEFT 
JOIN comments c 
ON p.post_id = c.post_id 
WHERE p.post_id = 2
AND active = 1  
GROUP 
BY p.post_id
 , p.title;

在测试数据库上的PHPmyadmin中运行此命令会返回：AvgRating：当前选择不包含唯一列。网格编辑，复选框，编辑，复制和删除功能不可用

（这是预期的，因为我错误地认为它会将两者的平均值放入别名列avgRating的一个实例中）

结果：

post_id  title      AvgRating  AvgRating
   2     site1.com  5.0000     2.2000

所以这在某种意义上是正确的，但是如何将原评级为5.000的评分纳入2.2000的评论中？

期望的结果应该是2.6667而不是2.2000，因为（5 + 5 + 3 + 1 + 0 + 2）/ 6 = 2.6667。 2.2000只是评论评分的平均值，并不考虑原始评分为5.000

Answer 1

一种方式是使用UNION ......

SELECT post_id
     , AVG(rating) avg_rating
  FROM 
     ( SELECT post_id
            , rating FROM posts
        UNION 
          ALL
       SELECT post_id
            , rating 
         FROM comments
     ) x 
 GROUP 
    BY post_id;

http://sqlfiddle.com/#!9/ad65d/14

Answer 2

我用PHP和＆amp;混合物来解决这个问题。 SQL :(我最初在SQL中寻找一种不那么混乱的方法）

<?php
 include_once('includes/dbcon.php');

foreach($db->query("SELECT 
p.post_id,
p.title,
p.rating AS oreview,
SUM(c.rating) AS creview
FROM posts p
LEFT JOIN comments c ON p.post_id = 2 where active=1 
GROUP BY 
p.post_id,
p.title") as $ratr) {
//get count
$stmt = $db->query('SELECT * FROM posts INNER JOIN comments ON comments.post_id=posts.post_id where comments.post_id=2 AND active=1');
$row_count = $stmt->rowCount();
$numreviews = $row_count+1;
//echo $ratr['creview'] ."<br>";
//echo $ratr['oreview'] ."<br>";
echo ($ratr['creview']+$ratr['oreview'])/$numreviews;
 }
 ?>

或作为一个功能，例如：

function aggRating($db,$post_id){
    foreach($db->query("SELECT 
    p.post_id,
    p.title,
    p.rating AS oreview,
    SUM(c.rating) AS creview
    FROM posts p
    LEFT JOIN comments c ON p.post_id = ".$post_id." where active=1 
    GROUP BY 
    p.post_id,
    p.title") as $ratr) {
//get review count
    $stmt = $db->query("SELECT * FROM posts INNER JOIN comments ON comments.post_id=posts.post_id where comments.post_id=".$post_id." AND active=1");
    $row_count = $stmt->rowCount();
    $numreviews = $row_count+1; //add original post review to total
echo round(($ratr['creview']+$ratr['oreview'])/$numreviews,2);
         }
  }

//return result (2.67)
echo aggRating($db,$post_id); 
?>

评级的平均评分存储在2列中

2 个答案: