Question

我有一张选票，每位选民获得3票，从10名不同候选人中选出。投票1分配3分，投票2分2分，投票3分1分。

我有以下SQL查询总计从每个投票中获得的积分数（因此投票1,2和3的单独结果）。

我需要做的是将所有这些结果放在一个表中，但我不太清楚从哪里开始。

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

以下候选名称表：

cand_number     cand_name
cand_1          Dave
cand_2          Sarah
cand_3          Nigel

所以我正在寻找的结果看起来像是：

Candidate       Votes
Dave            6
Sarah           7
Nigel           5

Answer 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选票

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

编辑：行正在乘以JOIN，这就是为什么它对于cand2和cand3来说是错误的

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;

Answer 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

Answer 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

总计投票结果

3 个答案: