在MySQL中在两个表之间合并一列

这是我上周问的一个问题的继续。它涉及来自两个不同表的表列的组合

请考虑这个例子。

我有一个表（投票顺序a），其中包含以下值：

+----------------+---------------------+----------------+
| VOTE_CANDIDATE | RANKED_CHOICE_VOTES | ORIGINAL_VOTES |
+----------------+---------------------+----------------+
|              2 |                   4 |              0 |
|              3 |                   1 |              0 |
|              4 |                   2 |              0 |
|              5 |                   1 |              0 |
+----------------+---------------------+----------------+

SELECT vote_candidate, COUNT(*) ranked_choice_votes, 0 original_votes
FROM vote_orders a
INNER JOIN
(
  SELECT vote_id, MIN(vote_order) AS min_vote_order
  FROM vote_orders
  WHERE vote_candidate NOT IN (1,6)
  GROUP BY vote_id
) b
ON a.vote_id = b.vote_id
AND a.vote_order = b.min_vote_order
INNER JOIN
(
  SELECT vote_id
  FROM vote_orders
  WHERE vote_candidate = 1
  AND vote_order = 1
) c
ON a.vote_id = c.vote_id
GROUP BY vote_candidate;

+----------------+
| ORIGINAL_VOTES |
+----------------+
|              1 |
|              2 |
|              4 |
|              2 |
+----------------+

SELECT COUNT(*) original_votes
FROM vote_orders
WHERE vote_order = 1
AND vote_candidate NOT IN (1,6)
GROUP BY vote_candidate;

我有另一个表（投票顺序）包含以下值：

+----------------+---------------------+----------------+
| VOTE_CANDIDATE | RANKED_CHOICE_VOTES | ORIGINAL_VOTES |
+----------------+---------------------+----------------+
|              2 |                   4 |              0 |
|              3 |                   1 |              0 |
|              4 |                   2 |              0 |
|              5 |                   1 |              0 |
+----------------+---------------------+----------------+

SELECT vote_candidate, COUNT(*) ranked_choice_votes, 0 original_votes
FROM vote_orders a
INNER JOIN
(
  SELECT vote_id, MIN(vote_order) AS min_vote_order
  FROM vote_orders
  WHERE vote_candidate NOT IN (1,6)
  GROUP BY vote_id
) b
ON a.vote_id = b.vote_id
AND a.vote_order = b.min_vote_order
INNER JOIN
(
  SELECT vote_id
  FROM vote_orders
  WHERE vote_candidate = 1
  AND vote_order = 1
) c
ON a.vote_id = c.vote_id
GROUP BY vote_candidate;

+----------------+
| ORIGINAL_VOTES |
+----------------+
|              1 |
|              2 |
|              4 |
|              2 |
+----------------+

SELECT COUNT(*) original_votes
FROM vote_orders
WHERE vote_order = 1
AND vote_candidate NOT IN (1,6)
GROUP BY vote_candidate;

现在，我想合并这两个表，使生成的表如下所示（注意，

原始投票

现在包含

投票顺序。原始投票

[second queries]内容）：

我想它需要一个高级的

UNION

和

JOIN

组合，但我就是不知道如何组合它们才能得到这个结果。你可以用它来做这个实验。

---

如果有任何关于如何解决这个问题的想法，我将不胜感激。

啊，我想我现在知道你想要什么了

可能只需添加另一个子查询即可获得每个候选人的主要投票数。然后在投票候选人字段中，将此结果与其他结果合并

SELECT a.vote_candidate, COUNT(*) ranked_choice_votes, d.original_votes
FROM vote_orders a
INNER JOIN
(
  SELECT vote_id, MIN(vote_order) AS min_vote_order
  FROM vote_orders
  WHERE vote_candidate NOT IN (1,6)
  GROUP BY vote_id
) b
ON a.vote_id = b.vote_id
AND a.vote_order = b.min_vote_order
INNER JOIN
(
  SELECT vote_id
  FROM vote_orders
  WHERE vote_candidate = 1
  AND vote_order = 1
) c
ON a.vote_id = c.vote_id
LEFT OUTER JOIN
(
    SELECT vote_candidate, COUNT(*) AS original_votes
    FROM vote_orders 
    WHERE vote_order = 1
    GROUP BY vote_candidate
) d
ON a.vote_candidate = d.vote_candidate
GROUP BY vote_candidate;

---

啊-你是救世主。我已经在互联网上搜索过这个解决方案。我将深入研究这方面的基本原理，以便从中学习一两件事。：）