Question

我被困在如何修改我的查询以获得我正在寻找的排名结果。我一直在关注SO的问题和疑问，但无法得到结果。我可以得到查询来进行计算并返回数据，但“等级”不正确。任何推动都会很棒！

表1包含学校数据：

+-----+------------+---------------+
| SID | schoolName | schoolCountry |
+-----+------------+---------------+
|   1 | ASD        | UAE           |
|   2 | ASIJ       | Japan         |
|   3 | ASP        | France        |
+-----+------------+---------------+

表2包含评论数据（我的查询有更多列，但这是它的工作原理）。

+-----+----------+--------+----+----+----+----+----+
| RID | schoolID | active | Q1 | Q2 | Q3 | Q4 | Q5 |
+-----+----------+--------+----+----+----+----+----+
|   1 |        1 |      1 |  8 |  9 |  5 |  1 |  9 |
|   2 |        2 |      1 |  7 |  6 |  6 |  7 |  9 |
|   3 |        1 |      0 |  1 |  4 |  7 |  8 |  5 |
|   4 |        3 |      1 |  2 | 10 |  6 |  7 |  5 |
+-----+----------+--------+----+----+----+----+----+

我试图通过平均学校的整体评价得分来创建不同的等级（国家，地区，整体）。我目前正在进行国家排名，目前我的查询是。

SELECT SID, schoolName, rank, average 
FROM (
SELECT (@rank := @rank + 1) AS rank,schools.SID, schools.schoolName,
ROUND(AVG(IF(reviews.active = 1, ((Q1+Q2+Q3+Q4+Q5+Q6+Q7+Q8+Q9+Q10+Q11+Q12+Q13+Q14+Q15+Q16+Q17+Q18+Q19+Q20+Q21+Q22+Q23+Q24+Q25+Q26+Q27+Q28+Q29+Q30+Q31+Q32+Q33+Q34+Q35+Q36+Q37+Q38+Q39+Q40+Q41+Q42+Q43+Q44+Q45+Q46+Q47+Q48+Q49+Q50+Q51+Q52)/(52*10)*10), NULL)) ,1) AS average
FROM schools
RIGHT JOIN reviews ON reviews.schoolID = schools.SID
CROSS JOIN (SELECT @rank := 0) AS vars
WHERE schools.schoolCountry = 'United Arab Emirates'
GROUP BY schools.SID

) as order_ranked  
ORDER BY `order_ranked`.`average`  DESC

我的输出结果如下：

+-----+----------------------------------------+------+---------+--+
| SID |               schoolName               | rank | average |  |
+-----+----------------------------------------+------+---------+--+
| 568 | GEMS Wellington Primary School         |    3 | 8.3     |  |
|   1 | American School of Dubai               |    1 | 8.1     |  |
| 561 | Dubai American Academy                 |    4 | 7.9     |  |
| 560 | Deira International School             |   11 | 7.7     |  |
| 569 | GEMS World Academy Dubai               |   10 | 7.0     |  |
| 570 | Greenfield Community School            |    8 | 6.7     |  |
| 565 | GEMS American Academy Abu Dhabi        |    6 | 6.0     |  |
| 584 | Universal American School              |    5 | 5.9     |  |
| 558 | American Academy In Al Mizhar          |    7 | 5.5     |  |
| 579 | The Cambridge High School Abu Dhabi    |    9 | 4.8     |  |
| 576 | Ras Al Khaimah English Speaking School |    2 | 4.3     |  |
+-----+----------------------------------------+------+---------+--+

你可以看到它排名，但不正确。我只是无法弄清楚原因。

Answer 1

排名应取决于平均排序。但内部查询中没有排序依据。首先计算平均值并计算其后的排名。此外，我将where条件移至join，因为您拥有它的方式相当于内连接。

SELECT SID, schoolName, @rank := @rank + 1 AS rank, average 
FROM (
SELECT schools.SID, schools.schoolName,
ROUND(AVG(IF(reviews.active = 1, ((Q1+Q2+Q3+Q4+Q5+Q6+Q7+Q8+Q9+Q10+Q11+Q12+Q13+Q14+Q15+Q16+Q17+Q18+Q19+Q20+Q21+Q22+Q23+Q24+Q25+Q26+Q27+Q28+Q29+Q30+Q31+Q32+Q33+Q34+Q35+Q36+Q37+Q38+Q39+Q40+Q41+Q42+Q43+Q44+Q45+Q46+Q47+Q48+Q49+Q50+Q51+Q52)/(52*10)*10), NULL)) ,1) AS average
FROM schools
RIGHT JOIN reviews ON reviews.schoolID = schools.SID AND schools.schoolCountry = 'United Arab Emirates'
GROUP BY schools.SID,schools.schoolName
) as order_ranked 
CROSS JOIN (SELECT @rank := 0) AS vars 
ORDER BY `average` DESC

Answer 2

我不确定为什么要使用right join。在MySQL中，在使用变量之前，您经常需要在子查询中进行排序。

SELECT SID, schoolName, (@rank := @rank + 1) AS rank, average 
FROM (SELECT s.SID, s.schoolName,
ROUND(AVG(Q1+Q2+Q3+Q4+Q5+Q6+Q7+Q8+Q9+Q10+Q11+Q12+Q13+Q14+Q15+Q16+Q17+Q18+Q19+Q20+Q21+Q22+Q23+Q24+Q25+Q26+Q27+Q28+Q29+Q30+Q31+Q32+Q33+Q34+Q35+Q36+Q37+Q38+Q39+Q40+Q41+Q42+Q43+Q44+Q45+Q46+Q47+Q48+Q49+Q50+Q51+Q52)/(52*10)*10)) AS average
      FROM schools s JOIN
           reviews r
           ON r.schoolID = s.SID 
      WHERE s.schoolCountry = 'United Arab Emirates' AND r.isactive = 1
      GROUP BY schools.SID
      ORDER BY average DESC
     ) sr CROSS JOIN
     (SELECT @rank := 0) AS vars
ORDER BY average  DESC

MYSQL排名问题

2 个答案: