如何从两个不同的表中选择count(*)
(称之为tab1
和tab2
),结果如下:
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
但我所拥有的只是:
Count_1
123
456
答案 0 :(得分:281)
SELECT (
SELECT COUNT(*)
FROM tab1
) AS count1,
(
SELECT COUNT(*)
FROM tab2
) AS count2
FROM dual
答案 1 :(得分:73)
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。
答案 2 :(得分:30)
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。
答案 3 :(得分:23)
我的经验是使用SQL Server,但您可以这样做:
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。
答案 4 :(得分:8)
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
/
答案 5 :(得分:6)
因为我看不到任何其他答案。
如果您不喜欢子查询并且在每个表中都有主键,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
但是表现明智我认为Quassnoi的解决方案更好,而且我会使用它。
答案 6 :(得分:5)
select (select count(*) from tab1) count_1, (select count(*) from tab2) count_2 from dual;
答案 7 :(得分:5)
这是我的分享
选项1 - 从不同表中的相同域计算
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算
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算“union all”以计算行数
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:)
答案 8 :(得分:5)
SELECT (SELECT COUNT(*) FROM table1) + (SELECT COUNT(*) FROM table2) FROM dual;
答案 9 :(得分:4)
为了一点完整性 - 此查询将创建一个查询,以便为您计算给定所有者的所有表。
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;
输出类似于
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
然后您可以运行以获取您的计数。这只是一个方便的脚本。
答案 10 :(得分:3)
快速刺了一下:
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,因此不需要From Dual
(因此存在差异)。
答案 11 :(得分:3)
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
答案 12 :(得分:1)
如果表格(或至少一个关键列)属于同一类型,则首先进行联合,然后计算。
select count(*)
from (select tab1key as key from schema.tab1
union all
select tab2key as key from schema.tab2
)
或者拿走你的声明并在它周围加上另一笔钱。
select sum(amount) from
(
select count(*) amount from schema.tab1 union all select count(*) amount from schema.tab2
)
答案 13 :(得分:1)
Declare @all int
SET @all = (select COUNT(*) from tab1) + (select count(*) from tab2)
Print @all
或
SELECT (select COUNT(*) from tab1) + (select count(*) from tab2)
答案 14 :(得分:1)
--============= 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
答案 15 :(得分:0)
选择
(从tab1中选择计数(),其中field
喜欢' value')+
(从tab2中选择计数(),其中field
喜欢'值')
计数
答案 16 :(得分:0)
计数两个表中的行
SELECT (SELECT COUNT(*) FROM table1 ) +
(SELECT COUNT(*) FROM table2 )
AS total FROM dual;
答案 17 :(得分:-1)
select @count = sum(data) from
(
select count(*) as data from #tempregion
union
select count(*) as data from #tempmetro
union
select count(*) as data from #tempcity
union
select count(*) as data from #tempzips
) a