2

我对 SQL 有点陌生，而且我有一个选举申请。我已经完成了 80% 的工作，现在坚持计算来自 2 个或更多列的选票。

示例表：

|**Senator_1** | **Senator_2** | **Senator_3**|
----------------------------------------------
   George      |    Carl       |   Anthony
   Carl        |    Tony       |   Stark
   Anthony     |    George     |   Tony
   Anthony     |    George     |   Stark

我想要这样的结果。

|**Candidate_Name** | **Vote_Count**|
-------------------------------------
   George           |   3
   Anthony          |   3
   Carl             |   2
   Stark            |   2
   Tony             |   2

我真的不知道我要使用什么查询。有解决这个问题的想法吗？

顺便说一句，对于从这里开始的混乱和所有争论，我将解释：

我想直接解决我的问题，这就是为什么我刚刚发布了一个示例表。我有一张供选民、候选人和选票使用的表格。所有表都有它的 ID 等，所以我很确定它是规范化的。

4

2 回答 2

11

您遇到的主要问题是您的表格未标准化。我强烈建议您修复当前的表结构。一个可能的新表结构是：

/* Table for unique voters */
CREATE TABLE voters (
    id INT UNSIGNED NOT NULL PRIMARY KEY,
    name VARCHAR(255) NOT NULL 
) ENGINE=InnoDB;

/* Table for unique candidates */
CREATE TABLE candidates (
    id INT UNSIGNED NOT NULL PRIMARY KEY,
    name VARCHAR(255) NOT NULL
) ENGINE=InnoDB;

/* Many-to-many table storing votes by voters for candidates */
CREATE TABLE votes (
    voter_id INT UNSIGNED NOT NULL,
    candidate_id INT UNSIGNED NOT NULL,
    PRIMARY KEY (voter_id, candidate_id),
    CONSTRAINT FOREIGN KEY (voter_id) REFERENCES voters (id),
    CONSTRAINT FOREIGN KEY (candidate_id) REFERENCES candidates (id)
) ENGINE=InnoDB;

/* Populate data */
INSERT INTO voters (name)
VALUES ('Voter 1'), ('Voter 2'), ('Voter 3'), ('Voter 4');

INSERT INTO candidates (name)
VALUES ('George'), ('Carl'), ('Anthony'), ('Tony'), ('Stark');

INSERT INTO votes (voter_id, candidate_id)
VALUES (1,1), (1,2), (1,3),
       (2,2), (2,4), (2,5),
       (3,3), (3,1), (3,4),
       (4,3), (4,1), (4,5);

然后，您可以通过连接两个表轻松获得结果：

/* query showing all voters and the candidates they voted for */
SELECT voters.name, candidates.name
FROM votes
    INNER JOIN voters on votes.voter_id = voters.id
    INNER JOIN candidates ON votes.candidate_id = candidates.id;

/* Finally, a query showing votes per candidate */
SELECT candidates.name, COUNT(*) AS votes
FROM votes
    INNER JOIN candidates ON votes.candidate_id = candidates.id
GROUP BY candidates.id;

请参阅带有演示的 SQL Fiddle

但是，如果您无法更改表的设计，则可以通过取消透视多列中的数据来获得结果。您可以使用 UNION ALL 将多列取消透视为行以获取计数：

select name, count(*) TotalVotes
from
(
  select senator_1 name
  from yt
  union all
  select senator_2 name
  from yt
  union all
  select senator_3 name
  from yt
) d
group by name
order by totalVotes desc;

请参阅带有演示的 SQL Fiddle

于 2013-06-11T13:01:05.617 回答
0

我认为您正在寻找总数。每个名称在不同列中的出现次数。基于此，我认为以下内容可能会有所帮助-

select senator, sum(cnt) as 'count' from (
select senator_1 as 'senator', count(1) 'cnt' from election_table group by senator_1
union all
select senator_2 as 'senator', count(1) 'cnt' from election_table group by senator_2
union all
select senator_3 as 'senator', count(1) 'cnt' from election_table group by senator_3
) x group by x.senator
于 2013-06-11T13:01:28.390 回答