我如何从两个不同的表(叫他们tab1和tab2)选择计数(*)有作为结果:

Count_1   Count_2
123       456

我试过了:

select count(*) Count_1 from schema.tab1 union all select count(*) Count_2 from schema.tab2

但我所拥有的只有:

Count_1
123
456

当前回答

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。

其他回答

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

SELECT  (
        SELECT COUNT(*)
        FROM   tab1
        ) AS count1,
        (
        SELECT COUNT(*)
        FROM   tab2
        ) AS count2
FROM    dual
SELECT  (
        SELECT COUNT(*)
        FROM   tbl1
        )
        +
        (
        SELECT COUNT(*)
        FROM   tbl2
        ) 
    as TotalCount

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。

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。