php - SQL查询中的5分评分公式

Question

       RATING
[  id -  pid = rate]
[  1  -   1  -  5  ]
[  2  -   1  -  5  ]
[  3  -   1  -  4  ]
[  4  -   1  -  1  ]
[  5  -   1  -  1  ]

WHERE pid = 1;
$r5 = 10;
$r4 = 4;
$r3 = 0;
$r2 = 0;
$r1 = 2;
$tt = 16;

从这张表中我试图得出这个数学公式

($r5*5 + $r4*4 + $r3*3 + $r2*2 + $r1) / $tt

不是将所有值调用到数组中并计算它，我可以在查询本身内部做什么？对于网站来说，哪一种实际上更快、最高效？

Answer 1

您可以使用以下解决方案:

SELECT pid,
    (((SUM(CASE WHEN rate = 5 THEN 5 ELSE 0 END) * 5) +
    (SUM(CASE WHEN rate = 4 THEN 4 ELSE 0 END) * 4) +
    (SUM(CASE WHEN rate = 3 THEN 3 ELSE 0 END) * 3) +
    (SUM(CASE WHEN rate = 2 THEN 2 ELSE 0 END) * 2) +
    (SUM(CASE WHEN rate = 1 THEN 1 ELSE 0 END) * 1)) / SUM(rate)) AS c_rate
FROM table_name
WHERE pid = 1
GROUP BY pid

要格式化输出而不舍入结果，您可以使用 TRUNCATE 。您还可以使用ROUND结果是否应四舍五入并截断。

使用TRUNCATE的解决方案:

SELECT pid,
    TRUNCATE((((SUM(CASE WHEN rate = 5 THEN 5 ELSE 0 END) * 5) +
    (SUM(CASE WHEN rate = 4 THEN 4 ELSE 0 END) * 4) +
    (SUM(CASE WHEN rate = 3 THEN 3 ELSE 0 END) * 3) +
    (SUM(CASE WHEN rate = 2 THEN 2 ELSE 0 END) * 2) +
    (SUM(CASE WHEN rate = 1 THEN 1 ELSE 0 END) * 1)) / SUM(rate)), 1) AS c_rate
FROM table_name
WHERE pid = 1
GROUP BY pid

使用ROUND的解决方案:

SELECT pid,
    ROUND((((SUM(CASE WHEN rate = 5 THEN 5 ELSE 0 END) * 5) +
    (SUM(CASE WHEN rate = 4 THEN 4 ELSE 0 END) * 4) +
    (SUM(CASE WHEN rate = 3 THEN 3 ELSE 0 END) * 3) +
    (SUM(CASE WHEN rate = 2 THEN 2 ELSE 0 END) * 2) +
    (SUM(CASE WHEN rate = 1 THEN 1 ELSE 0 END) * 1)) / SUM(rate)), 1) AS c_rate
FROM table_name
WHERE pid = 1
GROUP BY pid