Mysql 合并sql中的联合结果

我正试图把来自英国的工会的结果结合起来

SELECT MONTHNAME(terms) AS month, COUNT(DISTINCT project_num) as total 
FROM projects
WHERE terms >= '2017/01/01' AND Building_designer='SOMEPERSON'
GROUP BY MONTH(terms)
UNION
SELECT MONTHNAME(terms) AS month, COUNT(DISTINCT project_num) as total 
FROM archive
WHERE terms >= '2017/01/01' AND Building_designer='SOMEPERSON'
GROUP BY MONTH(terms)

我得到以下结果：来自SQL语句的结果

我正在努力使它的总数将是一个月的多个实例的组合。 sql表完全相同

这就是我希望它看起来的样子：

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一个快速的想法就是做这样的事情。您基本上希望对每个表的计数求和

select month, sum(total) from 
(
SELECT MONTHNAME(terms) AS month, COUNT(DISTINCT project_num) as total FROM projects WHERE terms >= '2017/01/01' AND Building_designer='SOMEPERSON' GROUP BY MONTH(terms)
UNION
SELECT MONTHNAME(terms) AS month, COUNT(DISTINCT project_num) as total FROM archive WHERE terms >= '2017/01/01' AND Building_designer='SOMEPERSON' GROUP BY MONTH(terms)
) group by month;

---

转换为派生表，放置别名，然后聚合

select 
x.month,
sum(x.total) [Total]
from (
SELECT
  MONTHNAME(terms) AS month,
  COUNT(DISTINCT project_num) AS total
FROM projects
WHERE terms >= '2017/01/01'
AND Building_designer = 'SOMEPERSON'
GROUP BY MONTH(terms)
UNION
SELECT
  MONTHNAME(terms) AS month,
  COUNT(DISTINCT project_num) AS total
FROM archive
WHERE terms >= '2017/01/01'
AND Building_designer = 'SOMEPERSON'
GROUP BY MONTH(terms)
) x

group by x.month

一个完整的外部连接将是理想的。但在您的情况下，让我们进行两个级别的聚合：

SELECT month, MAX(total_projects) as total_projects, MAX(total_archive) as total_archive
FROM ((SELECT MONTHNAME(terms) AS month, COUNT(DISTINCT project_num) as total_projects, 0 as total_archive
       FROM projects
       WHERE terms >= '2017/01/01' AND Building_designer = 'SOMEPERSON'
       GROUP BY MONTH(terms)
      ) UNION ALL
      (SELECT MONTHNAME(terms) AS month, 0, COUNT(DISTINCT project_num
       FROM archive
       WHERE terms >= '2017/01/01' AND Building_designer = 'SOMEPERSON'
       GROUP BY MONTH(terms)
      )
     ) pa
GROUP BY month
ORDER BY month;

编辑：

哎呀。您只需要一列。如果要计算每个月不同项目的数量，请执行union all，然后在下一个更高级别合并结果：

SELECT month, COUNT(DISTINCT project_num) as total
FROM ((SELECT MONTHNAME(terms) AS month, project_num
       FROM projects
       WHERE terms >= '2017/01/01' AND Building_designer = 'SOMEPERSON'
      ) UNION ALL
      (SELECT MONTHNAME(terms) AS month, project_num
       FROM archive
       WHERE terms >= '2017/01/01' AND Building_designer = 'SOMEPERSON'
      )
     ) pa
GROUP BY month
ORDER BY month;

---

可以尝试在整个查询上创建求和表达式

SELECT month, SUM (total) FROM
(SELECT MONTHNAME(terms) AS month, COUNT(DISTINCT project_num) as total FROM projects WHERE terms >= '2017/01/01' AND Building_designer='SOMEPERSON' GROUP BY MONTH(terms)
UNION
SELECT MONTHNAME(terms) AS month, COUNT(DISTINCT project_num) as total FROM archive WHERE terms >= '2017/01/01' AND Building_designer='SOMEPERSON' GROUP BY MONTH(terms))
GROUP BY month

---

问题的所有要点都需要出现在问题本身中，而不是出现在外部站点上。不清楚您是希望两个表之间存在差异，还是希望两个表中的差异值的总和存在差异。假设使用UNION All手头的数据质量是安全的？