Question

假设我有两个名为tbl1和tbl2的表（不同的列和ID），并且使用以下查询获得每个表的每月总数：

query1：

SELECT MONTHNAME(tbl1_date), SUM(col1+col2+col3+col4) 
FROM myTbl1 
GROUP BY YEAR(tbl1_date), MONTH(tbl1_date);

query2：

SELECT MONTHNAME(tbl2_date), SUM(col1+col2+col3+col4+col5+col6) 
FROM myTbl2 
GROUP BY YEAR(tbl2_date), MONTH(tbl2_date);

如何从query1中减去query2并获得SUM的差值？

Answer 1

加入两个查询，然后减去。

SELECT t1.month, t1.year, t1.total - t2.total AS diff
FROM (
    SELECT MONTHNAME(tbl1_date) AS month, YEAR(tbl1_date) AS year, SUM(col1+col2+col3+col4) AS total
    FROM myTbl1 
    GROUP BY YEAR(tbl1_date), MONTH(tbl1_date)
) AS t1
JOIN (
    SELECT MONTHNAME(tbl2_date) AS month, YEAR(tb2_date) AS year, SUM(col1+col2+col3+col4+col5+col6) AS total
    FROM myTbl2 
    GROUP BY YEAR(tbl2_date), MONTH(tbl2_date)
) AS t2 ON t1.month = t2.month AND t1.year = t2.year

Answer 2

如果表可以具有不同的月份和，则您需要所有行，那么您确实需要full join。 MySQL不支持该功能，但是在这种情况下，您可以使用union all和group by：

SELECT yyyy, mon, SUM(s) - SUM(s2) as diff
FROM ((SELECT YEAR(tbl1_date) as yyyy,
              MONTHNAME(tbl1_date) as mon, 
              SUM(col1+col2+col3+col4) as s1, 0 as s2
       FROM myTbl1 
       GROUP BY YEAR(tbl1_date), MONTH(tbl1_date)
      ) UNION ALL
      (SELECT YEAR(tbl2_date) as yyyy,
              MONTHNAME(tbl2_date) as mon, 
              0, SUM(col1+col2+col3+col4) as s2
       FROM myTbl12
       GROUP BY YEAR(tbl2_date), MONTH(tbl2_date)
      )
     ) ym
GROUP BY yyyy, mon;

SQL：从另一个查询中减去一个查询

2 个答案: