查看下面的查询。。我的系统在本地运行良好。如果有用请告诉我。

select 'Call Detail - Hive T1 to HDFS Staging - Data Compare',
       case 
       when (sum(cnt1) - sum(cnt2)) > 0 
       then 'PASS' 
       else 'FAIL' 
       end as count_records
  from (select count(*) as cnt1, 0 as cnt2 from students1 
        union all
        select 0 as cnt1, count(*) as cnt2 from students1  ) tbl;

使用 cross join 如果没有按列分组。在这种情况下，它将生成一行，其中包含两个计数：

select s.cnt-s1.cnt diff, case when abs(s.cnt-s1.cnt) > 0 then 'Fail' Else 'Pass' end result
from
(select count(*) cnt  from students1 s1) s
cross join
(select count(*) cnt from students2 s2) s1

如果您将添加一些groupby列来比较更详细的粒度，那么使用 FULL JOIN 按列分组：

select s.col1 s_col1, s1.col1 s1_col1, s.cnt-s1.cnt diff, case when abs(s.cnt-s1.cnt) > 0 then 'Fail' Else 'Pass' end result
from
(select count(*) cnt, col1  from students1 s1 group by col1) s
full join
(select count(*) cnt, col1 from students2 s2 group by col1) s1 
on s.col1 = s1.col1

此查询将返回两个表中已计算差异的联接行和未联接行。