汇总投票结果答案

【问题标题】：Totalling up ballot results汇总投票结果
【发布时间】：2011-09-30 07:16:48
【问题描述】：

我有一张选票，每个选民有 3 张选票，从 10 位不同的候选人中进行选择。第1票得3分，第2票得2分，第3票得1分。

我有以下 SQL 查询来计算从每个投票中获得的点数（因此将投票 1、2 和 3 的结果分开）。

我需要做的是将所有这些结果放在一个表中，但我不太确定从哪里开始。

SELECT cn.cand_name, (count(vote_1) * 3) as vote_1 FROM candidate_votes cv Inner Join candidate_names cn ON cv.vote_1 = cn.cand_number GROUP BY cand_name;

SELECT cn.cand_name, (count(vote_2) * 2) as vote_2 FROM candidate_votes cv Inner Join candidate_names cn ON cv.vote_2 = cn.cand_number GROUP BY cand_name;

SELECT cn.cand_name, (count(vote_3) * 1) as vote_3 FROM candidate_votes cv Inner Join candidate_names cn ON cv.vote_3 = cn.cand_number GROUP BY cand_name;

我有以下结果表：

Voter_number    Vote_1     Vote2      Vote3
123             cand_1     cand_3     cand_2
456             cand_2     cand_1     cand_3
789             cand_2     cand_3     cand_1

还有如下候选名表：

cand_number     cand_name
cand_1          Dave
cand_2          Sarah
cand_3          Nigel

所以我正在寻找的结果将类似于：

Candidate       Votes
Dave            6
Sarah           7
Nigel           5

【问题讨论】：

标签： sql-server count

【解决方案1】：

SELECT
    cn.cand_name, 
    count(cv1.vote_1) * 3 as vote_1, 
    count(cv2.vote_2) * 2 as vote_2, 
    count(cv3.vote_3) as vote_3
FROM
    candidate_names cn
    LEFT JOIN
    candidate_votes cv1 ON cv1.vote_1 = cn.cand_number
    LEFT JOIN
    candidate_votes cv2 ON cv2.vote_2 = cn.cand_number
    LEFT JOIN
    candidate_votes cv3 ON cv3.vote_3 = cn.cand_number
GROUP BY cn.cand_name;

这也允许您添加所有投票

(count(cv1.vote_1) * 3) +
    (count(cv2.vote_2) * 2) +
    count(cv3.vote_3) as totalvotes

编辑：行乘以 JOIN，这就是 cand2 和 cand3 错误的原因

SELECT
    cn.cand_name, 
    SUM(CASE WHEN cv.vote_1 = cn.cand_number THEN 3 ELSE 0 END) as vote_1, 
    SUM(CASE WHEN cv.vote_2 = cn.cand_number THEN 2 ELSE 0 END) as vote_2, 
    SUM(CASE WHEN cv.vote_3 = cn.cand_number THEN 1 ELSE 0 END) as vote_3
FROM
    candidate_names cn
    JOIN
    candidate_votes cv ON cn.cand_number IN (cv.vote_1, cv.vote_2, cv.vote_3)
GROUP BY cn.cand_name;

【讨论】：

@Tom：我的加权乘法有误。请立即尝试
我已经注意到并且已经更正了它......它仍然没有给出正确的结果。
@Tom：好吧，“结果错误”没有帮助。没有样本数据，也没有迹象表明出了什么问题（值错误？候选人太多？）。
你的方法给出了 Dave=6, Sarah=8, Nigel=6
@gbn 让我们continue this discussion in chat

【解决方案2】：

SELECT cn.cand_name
     , COALESCE(cv1.cnt_1,0)
     , COALESCE(cv2.cnt_2,0)
     , COALESCE(cv3.cnt_3,0)
     , 3*COALESCE(cv1.cnt_1,0) + 2*COALESCE(cv2.cnt_2,0)
       + 1*COALESCE(cv3.cnt_3,0) AS total
FROM candidate_names AS cn
  LEFT JOIN 
    ( SELECT vote_1 AS vote
           , COUNT(*) AS cnt_1
      FROM candidate_votes cv
      GROUP BY vote_1
    ) AS cv1
    ON cv1.vote = cn.cand_number
  LEFT JOIN 
    ( SELECT vote_2 AS vote
           , COUNT(*) AS cnt_2
      FROM candidate_votes cv
      GROUP BY vote_2
    ) AS cv2
    ON cv2.vote = cn.cand_number
  LEFT JOIN 
    ( SELECT vote_3 AS vote
           , COUNT(*) AS cnt_2
      FROM candidate_votes cv
      GROUP BY vote_3
    ) AS cv3
    ON cv3.vote = cn.cand_number

【讨论】：

ypercube 最有效的查询

【解决方案3】：

SELECT
  Candidate = n.cand_name,
  Votes = SUM(s.vote_weight)
FROM (
  SELECT
    cand_number = CASE x.weight
      WHEN 1 THEN Vote3
      WHEN 2 THEN Vote2
      WHEN 3 THEN Vote1
    END,
    vote_weight = x.weight
  FROM candidate_votes v
    CROSS JOIN (SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3) x (weight)
) s
  INNER JOIN candidate_names n ON s.cand_number = n.cand_number
GROUP BY n.cand_name

【讨论】：