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。

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。

选择 (select count() from tab1 where field like 'value') + (select count() from tab2 where field like 'value') 数

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

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。

select (select count(*) from tab1) count_1, (select count(*) from tab2) count_2 from dual;