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。

我的经验是使用SQL Server，但是你能做到:

select (select count(*) from table1) as count1,
  (select count(*) from table2) as count2

在SQL Server我得到的结果，你是后。

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
/

作为附加信息，要在SQL Server中完成同样的事情，您只需要删除查询的“FROM dual”部分。

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
)

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。