Question

对不起，标题太烂了。我想将这两个查询基本上都转换为一个查询，结果有两列：

select count(columnA) as prev from myTable where set_id = 1530880217;

select count(columnA) as curr from myTable where set_id = 1530901756;

输出：

prev | curr

1000 | 1500

Answer 1

使用条件聚合：

select sum(case when set_id = 1530880217 then 1 else 0 end) as prev,
       sum(case when set_id = 1530901756 then 1 else 0 end) as curr
from myTable
where set_id in (1530880217, 1530901756);

这假设columnA从未为NULL。如果您确实要NULL进行检查：

select sum(case when set_id = 1530880217 and ColumnA is not null then 1 else 0 end) as prev,
       sum(case when set_id = 1530901756 and ColumnA is not null then 1 else 0 end) as curr
from myTable
where set_id in (1530880217, 1530901756);

SQL / Hive如何将两个不同的查询合并到一个具有不同列的结果中

1 个答案: