将两个SQL查询的结果与UNION合并并融合

Question

我正在编写两个单独的SQL查询以获取两个不同日期的数据，如下所示：

SELECT number, sum(sales) as sales, sum(discount) sa discount, sum(margin) as margin
FROM table_a
WHERE day = '2019-08-09'
GROUP BY number

SELECT number, sum(sales) as sales, sum(discount) sa discount, sum(margin) as margin
FROM table_a
WHERE day = '2018-08-10'
GROUP BY number

我试着像这样融合它们，以便从两个不同的日期获得一行中相同数字的结果：

SELECT number, sum(sales) as sales, sum(discount) sa discount, sum(margin) as margin, 0 as sales_n1, 0 as discount_n1, 0 as margin_n1
FROM table_a
WHERE day = '2019-08-09'
GROUP BY number
UNION
SELECT number, 0 as sales, 0 as discount, 0 as margin, sum(sales_n1) as sales_n1, sum(discount_n1) as discount_n1, sum(margin_n1) as margin_n1
FROM table_a
WHERE day = '2018-08-10'
GROUP BY number

但这没有用，因为我以相同的方式获得了第一个查询的行，并且定义为零的列为零，然后是第二个查询的列。

我该如何纠正它以获得所需的输出？

Answer 1

使用条件聚合：

SELECT number,
       sum(case when day = '2019-08-09' then sales end) as sales_20190809,
       sum(case when day = '2019-08-09' then discount end) sa discount, sum(margin) as margin_20190810,
       sum(case when day = '2019-08-10' then sales end) as sales_20190809,
       sum(case when day = '2019-08-10' then discount end) sa discount, sum(margin) as margin_20190810
FROM table_a
WHERE day IN ('2019-08-09', '2019-08-10')
GROUP BY number;

如果您希望数字位于不同的行中（您似乎不需要），请使用聚合：

SELECT day, number, sum(sales) as sales, sum(discount) as discount, sum(margin) as margin
FROM table_a
WHERE day IN ('2019-08-09', '2019-08-10')
GROUP BY day, number