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
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。