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

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

Declare @all int
SET @all = (select COUNT(*) from tab1) + (select count(*) from tab2)
Print @all

or

SELECT (select COUNT(*) from tab1) + (select count(*) from tab2)

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。

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

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