我有两个表，即tbl_votes和tbl_candidates：
tbl_votes：包含ff列：
voteID president vicePresident secretary treasurer rep1 rep2 rep3
1 1 3 9 12 15
2 1 4 6 8 12 15
3 2 3 5 7 9 12
而tbl_candidates包含ff列：
idcandidate fName mName lName position
1 Jefferson Garcia Roxas president
2 Carlson Mcvoy Zafiro president
3 Andrew Efron Anderson vice president
4 Jessica Garcia Roxas secretary
...
9 Jayson Sanchez Dela Cruz representative
10 Arille Alisa Villanueva Valdez representative
11 Arnold Castro Baltazar representative
12 Kristina de Leon Guillermo representative
13 Jayson Jones Dela Cruz representative
14 Karla Santos Abrogena representative
15 Jason Chua representative
我想要的查询是：
Name votes
Jefferson Garcia Roxas 2
Carlson Mcvoy Zafiro 1
Andrew Efron Anderson 2
Jessica Garcia Roxas 1
...
Jayson Sanchez Dela Cruz 2
Arnold Castro Baltazar 0
Kristina de Leon Guillermo 3
Jason Chua 2
我被困在哪里数票，这是我的代码：
SELECT `Name`, Votes
FROM (SELECT COUNT(*) as Votes FROM tbl_votes) a
CROSS JOIN (Select (Concat(fName ,' ',mName,' ', lName)) as `Name` from tbl_candidates) b;
此查询的结果是：
Name Votes
Jefferson Garcia Roxas 3
Carlson Mcvoy Zafiro 3
Andrew Efron Anderson 3
Jessica Garcia Roxas 3
Jayson Sanchez Dela Cruz 3
Arnold Castro Baltazar 3
Kristina de Leon Guillermo 3
Jason Chua 3
我将如何计算特定于idcandidate的计数？ 非常感谢您的大力帮助。谢谢！
答案 0 :(得分：2)
虽然 可以<{1}}（但与你提出的不同）并且对结果进行适当的汇总，但这是一种不能很好扩展的糟糕方法。当然，当你背负着糟糕的数据模型时，没有真正好的方法，就像你一样。
有几种方法可以解决这个问题，其中没有一种方法特别好，例如：
CROSS JOIN按位置分解问题，使用每个位置的一个内联视图为每个候选人为该位置生成一行。然后它通过SELECT `Name`, COUNT(*) AS `votes`
FROM
(
SELECT
CONCAT(fName, ' ', mName, ' ', lName) as `Name`
FROM
tbl_candidates c
JOIN tbl_votes v
ON c.idcandidate = v.president
WHERE
c.position = 'president'
UNION ALL
SELECT
CONCAT(fName, ' ', mName, ' ', lName) as `Name`
FROM
tbl_candidates c
JOIN tbl_votes v
ON c.idcandidate = v.vicePresident
WHERE
c.position = 'vice president'
UNION ALL
SELECT
CONCAT(fName, ' ', mName, ' ', lName) as `Name`
FROM
tbl_candidates c
JOIN tbl_votes v
ON c.idcandidate IN (v.rep1, v.rep2, v.rep3)
WHERE
c.position = 'representative'
) vote_agg
GROUP BY `Name`
将它们组合成一个整体列表，并对结果执行聚合查询以计算每个候选人的投票。
如果现有候选人对他们没有竞选的职位有任何投票（通过对指定数据模型的约束很难或无法阻止），那么这些将被忽略。如果任何一张选票对同一代表候选人有多于一票，那么只会计算一个（可能是期望的行为，也许不是）。
答案 1 :(得分：1)
SELECT fName||' '||mName||' '||lName, votes FROM (
SELECT a.idcandidate, COUNT(b.idcandidate) votes
FROM tbl_candidates a
LEFT JOIN (
SELECT president idcandidate from tbl_votes
UNION ALL
SELECT vicePresident idcandidate from tbl_votes
UNION ALL
SELECT secretary idcandidate from tbl_votes
UNION ALL
SELECT treasurer idcandidate from tbl_votes
UNION ALL
SELECT rep1 idcandidate from tbl_votes
UNION ALL
SELECT rep2 idcandidate from tbl_votes
UNION ALL
SELECT rep3 idcandidate from tbl_votes ) b
ON (a.idcandidate = b.idcandidate)
GROUP BY a.idcandidate ) tab
JOIN tbl_candidates b on (b.idcandidate = tab.idcandidate)
上面的答案是针对SQLite的，我在某种程度上误读了这个问题上的标签。
然而它可能有效，除了第一行需要采用mySQL格式：
SELECT CONCAT_WS(" ", fName, mName, lName), votes FROM (
答案 2 :(得分：0)
你的模型有问题......一旦你解决这些问题，创建起来就非常简单。
... Minimaly：
我会创建一个职位表，以便我可以直接与候选人联系。
positons_table
id position
1 president
2 vice-president
3 secretary
4 ....
然后候选人表...
idcandidate fName mName lName position
1 Jefferson Garcia Roxas 1
2 Carlson Mcvoy Zafiro 1
3 Andrew Efron Anderson 2
4 Jessica Garcia Roxas 3
然后，如果你不想改变你的tbl_votes ......
总统投票：
SELECT CONCAT(c.fName, " ", c.mName, " ", c.lName) as Name, count(v.president) as votes
FROM tbl_votes v
INNER_JOIN tbl_candidates c on v.president = c.idcandidate GROUP BY v.president
副总统的投票
SELECT CONCAT(c.fName, " ", c.mName, " ", c.lName) as Name, count(v.vicePresident) as votes
FROM tbl_votes v
INNER_JOIN tbl_candidates c on v.vicePresident = c.idcandidate GROUP BY v.vicePresident
依旧...... 如果你需要总结一下......
SELECT CONCAT(c.fName, " ", c.mName, " ", c.lName) as Name, count(v.president) as votes
FROM tbl_votes v
INNER_JOIN tbl_candidates c on v.president = c.idcandidate GROUP BY v.president
UNION
SELECT CONCAT(c.fName, " ", c.mName, " ", c.lName) as Name, count(v.vicePresident) as votes
FROM tbl_votes v
INNER_JOIN tbl_candidates c on v.vicePresident = c.idcandidate GROUP BY v.vicePresident
UNION ....