如果表(或者至少是一个键列)是相同类型的，那么就先做联合，然后计数。

select count(*) 
  from (select tab1key as key from schema.tab1 
        union all 
        select tab2key as key from schema.tab2
       )

或者把你的语句加上另一个和()。

select sum(amount) from
(
select count(*) amount from schema.tab1 union all select count(*) amount from schema.tab2
)

这是我的分享

选项1 -计数从相同的域从不同的表

select distinct(select count(*) from domain1.table1) "count1", (select count(*) from domain1.table2) "count2" 
from domain1.table1, domain1.table2;

选项2 -同一表从不同的域计数

select distinct(select count(*) from domain1.table1) "count1", (select count(*) from domain2.table1) "count2" 
from domain1.table1, domain2.table1;

选项3 -计数从不同的领域为同一表与“联合所有”有行计数

select 'domain 1'"domain", count(*) 
from domain1.table1 
union all 
select 'domain 2', count(*) 
from domain2.table1;

享受SQL，我总是这样做:)

只是因为它略有不同:

SELECT 'table_1' AS table_name, COUNT(*) FROM table_1
UNION
SELECT 'table_2' AS table_name, COUNT(*) FROM table_2
UNION
SELECT 'table_3' AS table_name, COUNT(*) FROM table_3

它给出了转置的答案(每个表一行而不是一列)，否则我不认为它有多大不同。我认为在性能方面，它们应该是相等的。

为了完整起见，这个查询将创建一个查询，为您提供给定所有者的所有表的计数。

select 
  DECODE(rownum, 1, '', ' UNION ALL ') || 
  'SELECT ''' || table_name || ''' AS TABLE_NAME, COUNT(*) ' ||
  ' FROM ' || table_name  as query_string 
 from all_tables 
where owner = :owner;

输出是这样的

SELECT 'TAB1' AS TABLE_NAME, COUNT(*) FROM TAB1
 UNION ALL SELECT 'TAB2' AS TABLE_NAME, COUNT(*) FROM TAB2
 UNION ALL SELECT 'TAB3' AS TABLE_NAME, COUNT(*) FROM TAB3
 UNION ALL SELECT 'TAB4' AS TABLE_NAME, COUNT(*) FROM TAB4

然后你可以运行它来得到你的计数。有时它只是一个方便的脚本。

其他略有不同的方法:

with t1_count as (select count(*) c1 from t1),
     t2_count as (select count(*) c2 from t2)
select c1,
       c2
from   t1_count,
       t2_count
/

select c1,
       c2
from   (select count(*) c1 from t1) t1_count,
       (select count(*) c2 from t2) t2_count
/

SELECT  (
        SELECT COUNT(*)
        FROM   tab1
        ) AS count1,
        (
        SELECT COUNT(*)
        FROM   tab2
        ) AS count2
FROM    dual