平均评级与评级存储在2列

Question

我已经搜索类似的网站和所有的答案是指一个单独的评级表，而不是像这样的实现。

我已经转换一个WordPress网站在PHP/MySQL的一个自建的网站，想象实际的岗位是一个审查和评论是与原来的岗位实际的评论。

我遇到的问题是我的职位表和评价领域的分级字段为每个评论输入。

我试图使用别名AvgRating作为一个，但它只返回评论评级（c.rating），如果该行被删除它从posts表中带回评级，有没有加入方式别名作为平均评分？显然只有这部分是错误的，但不会抛出错误。

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; 

我在尝试获取模式聚合计数标记的平均值。有一种方法可以在一个查询中实现此目的吗？或者是我唯一的选择使用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; 

上测试数据库恢复运行此在phpMyAdmin：AvgRating：当前选择不包含唯一的列。电网编辑，复选框，编辑，拷贝和删除功能不可用

（这是可以预料的，因为我错误地认为这将令两者的平均值为别名列avgRating的一个实例）

结果：

post_id title  AvgRating AvgRating 
    2  site1.com 5.0000  2.2000 

所以这是正确的在一定意义上，而是如何在原岗位等级的5.000到的评论的评价2.2000因素？ （5 + 5 + 3 + 1 + 0 + 2）/6=2.66667，期望的结果应该是2.6667而不是2.2000。 2.2000是唯一的评论评分的平均值，并在原岗位等级的5.000

Answer 1

一种方法是不保理与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

我想通了使用PHP & SQL的混合物：（我本来想找一个不太混乱的方式做到这一点的SQL虽然）

<?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; 
} 
?> 

或作为一个例子的功能：

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); 
?>