从多个表中选择count(*)

Question

如何选择count(*),从两个不同的表(叫他们tab1和tab2),其结果为:

Count_1   Count_2
123       456

我试过这个:

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

但我只有:

Count_1
123
456

Answer 1

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

Answer 2

作为附加信息,要在SQL Server中完成相同的操作,您只需要删除查询的"FROM dual"部分.

Answer 3

仅仅因为它略有不同:

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

它给出了转换的答案(每个表一行而不是一列),否则我认为它没有太大的不同.我认为在性能方面他们应该是等同的.

Answer 4

我的经验是SQL Server,但你能做到:

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

在SQL Server中,我得到你想要的结果.

Answer 5

    select 
    t1.Count_1,t2.Count_2
    from 
(SELECT count(1) as Count_1 FROM tab1) as t1, 
(SELECT count(1) as Count_2 FROM tab2) as t2

Answer 6

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
/

Answer 7

一个快速的尝试提出了：

Select (select count(*) from Table1) as Count1, (select count(*) from Table2) as Count2

注意：我在 SQL Server 中对此进行了测试，因此From Dual没有必要（因此存在差异）。

Answer 8

因为我看不出任何其他答案.

如果您不喜欢子查询并且每个表中都有主键,则可以执行以下操作:

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

但是表现明智我相信Quassnoi的解决方案更好,而且我会使用它.

Answer 9

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

Answer 10

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

Answer 11

这是来自我的分享

选项1 - 从不同表中的相同域计数

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

选项2 - 从同一个表的不同域计数

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

选项3 - 从同一个表的不同域计数"union all"以具有计数行

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:)

Answer 12

如果表（或至少一个键列）是相同类型的，只需先进行联合然后计数。

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

或者接受你的陈述并在它周围加上另一个 sum() 。

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

Answer 13

出于完整性考虑，该查询将创建一个查询，以向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

然后可以运行以获取计数。有时只是一个方便的脚本。

Answer 14

--============= 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