在php和mysql中計算前5名

Question

嘿，我已經在php和mysql中創建了評分系統，現在我想選擇前5名，我做了一些事情，但是如果用戶添加了評分（最大）它是nr 1排名。

你們是如何做到這一點在PHP和MySQL？

表看起來是這樣的：

- ID

- 中期

- UID

- 遊客

該評級是從1到數5

在此先感謝！

Answer 1

由於@chaos指出，這是想法：

SELECT `mid`, SUM(`rating`) AS `total` 
FROM `rating` 
GROUP BY `mid` 
ORDER BY `total` DESC 
LIMIT 5 

然而，以確保不會與極少數評級物品進入前5，你可以添加一個門檻，只允許文章擁有超過X收視率將在結果顯示，也許能給前5名的更準確的圖片：

SELECT `mid`, SUM(`rating`) AS `total`, COUNT(1) AS `nrRatings` 
FROM `rating` 
GROUP BY `mid` 
HAVING nrRatings > 5 // This is the threshold. Only articles with more than 
        // 5 ratings will be considered 
ORDER BY `total` DESC 
LIMIT 5 

Answer 2

SELECT `mid`, SUM(`rating`) AS `total` 
FROM `rating` 
GROUP BY `mid` 
ORDER BY `total` DESC 
LIMIT 5 

Answer 3

如果我理解你的寫法，看起來像一個等級1的解釋第1條ST？ ORDER BY應該是ASC。我認爲AVG（）比SUM（）更合適。與閾值的想法完全一致，所以這樣的事情（無恥的C/P重寫; THX @Machine：P）：

 
SELECT `mid`, AVG(`rating`) AS `avg_rating`, SUM(1) AS `nrRatings` 
FROM `rating` 
GROUP BY `mid` 
HAVING nrRatings > 5 
ORDER BY `avg_rating` ASC 
LIMIT 5