--============= FIRST WAY (Shows as Multiple Row) ===============
SELECT 'tblProducts' [TableName], COUNT(P.Id) [RowCount] FROM tblProducts P
UNION ALL
SELECT 'tblProductSales' [TableName], COUNT(S.Id) [RowCount] FROM tblProductSales S


--============== SECOND WAY (Shows in a Single Row) =============
SELECT  
(SELECT COUNT(Id) FROM   tblProducts) AS ProductCount,
(SELECT COUNT(Id) FROM   tblProductSales) AS SalesCount

SELECT (SELECT COUNT(*) FROM table1) + (SELECT COUNT(*) FROM table2) FROM dual;

因为我找不到其他答案了。

如果你不喜欢子查询并且在每个表中都有主键，你可以这样做:

select count(distinct tab1.id) as count_t1,
       count(distinct tab2.id) as count_t2
    from tab1, tab2

但是就性能而言，我认为Quassnoi的解决方案更好，也是我会使用的解决方案。

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

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
/