3 回答

TA貢獻(xiàn)1744條經(jīng)驗(yàn) 獲得超4個(gè)贊
SELECT (
SELECT COUNT(*)
FROM tab1
) AS count1,
(
SELECT COUNT(*)
FROM tab2
) AS count2
FROM dual

TA貢獻(xiàn)1790條經(jīng)驗(yàn) 獲得超9個(gè)贊
只是因?yàn)樗杂胁煌?/p>
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
它給出了換位的答案(每張表一行而不是一列),否則我認(rèn)為這沒(méi)有太大不同。我認(rèn)為在性能方面,它們應(yīng)該是等效的。

TA貢獻(xiàn)1829條經(jīng)驗(yàn) 獲得超6個(gè)贊
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
/
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