用戶提交的前5名和排序按熱門程度

Question

數據庫設置（MySQL的）

表：top_fives

id, uid, first,  second, third, fourth, fifth, creation_date 
1, 1, cheese, eggs, ham,  bacon, ketchup, 2010-03-17 
2, 2, mayonaise, cheese, ketchup, eggs, bacon, 2010-03-17 

用戶可以提交他們的前5某一主題的。現在我想根據受歡迎程度排序的前五名的總結。

每列都有自己的點值。列「第一」是獎勵5分，「第二」四點，「第三」三個點，等等...

所以，在我的例子應該是這樣的：

1 Cheese  (9 points = 5 + 4 -> 1 time in 'first' column and 1 time in 'second' column) 
2 Eggs  (6 points) 
3 Mayonaise (5 points) 
4 Ketchup (4 points) 
5 Bacon  (3 points) 
6 Ham  (3 points) 

什麼是這種情況下最簡單的解決方案（PHP）？

Answer 1

最好的解決辦法是擺在首位，以已恢復正常數據。唯一可行的解​​決方案是模擬正常規範化數據庫的行爲。當然，解決方案不應涉及任何PHP代碼，並應在數據庫上進行：

SELECT type, SUM(score) 
FROM 
(
(SELECT first as type, COUNT(*)*5 as score 
FROM top_fives 
GROUP BY first 
) UNION 
(SELECT second AS type, COUNT(*)*4 as score 
FROM top_fives 
GROUP BY second 
) UNION 
(SELECT third AS type, COUNT(*)*3 as score 
FROM top_fives 
GROUP BY third 
) UNION 
(SELECT fourth AS type, COUNT(*)*2 as score 
FROM top_fives 
GROUP BY fourth 
) UNION 
(SELECT fifith AS type, COUNT(*) as score 
FROM top_fives 
GROUP BY fifth 
) 
) 
GROUP By type 
ORDER BY SUM(score) DESC; 

C.

Answer 2

解決方案將正常化你的表（見下文）。

如果你沒有，你應該能夠做到可以：

Select name, Sum(points) total_points 
From (
    Select first name, 5 points 
    From top_fives 
    Union 
    Select second name, 4 points 
    From top_fives 
    Union 
    ... 
) 
Group By name 
Order By total_points Desc 

---

標準化的解決方案：

food 

food_id, food_name 
1  cheese 
2  eggs 
3  ham 
... 

food_rating 
------ 
uid, food_id, points 
1 1  5 
1 2  4 
1 3  3 
2 1  4 

。

Select f.food_name, Sum(r.points) total_points 
From food_rating r 
Join food f On (f.food_id = r.food_id) 
Group By food_name 
Order By total_points Desc