Answers:
SELECT (
SELECT COUNT(*)
FROM user_table
) AS tot_user,
(
SELECT COUNT(*)
FROM cat_table
) AS tot_cat,
(
SELECT COUNT(*)
FROM course_table
) AS tot_course
如果使用MyISAM表,最快的方法是直接查询统计信息:
select table_name, table_rows
from information_schema.tables
where
table_schema='databasename' and
table_name in ('user_table','cat_table','course_table')
如果您拥有InnoDB,则必须使用count()进行查询,因为information_schema.tables中的报告值是错误的。
您当然可以使用由Ben James假定的Select Agregation语句,但是这将导致视图具有与表一样多的列。一种替代方法可能如下:
SELECT COUNT(user_table.id) AS TableCount,'user_table' AS TableSource FROM user_table
UNION SELECT COUNT(cat_table.id) AS TableCount,'cat_table' AS TableSource FROM cat_table
UNION SELECT COUNT(course_table.id) AS TableCount, 'course_table' AS TableSource From course_table;
这样的方法的好处是,您可以显式编写Union语句并生成视图或创建临时表,以保存使用变量代替表名从Proc cal中连续添加的值。我倾向于选择后者,但是这实际上取决于个人喜好和应用。如果您确定表永远不会改变,则您希望数据采用单行格式,并且不会添加表。坚持本·詹姆斯的解决方案。否则,我建议您保持灵活性,您可以随时修改交叉表结构。
select RTRIM(A.FIELD) from SCHEMA.TABLE A where RTRIM(A.FIELD) = ('10544175A')
UNION
select RTRIM(A.FIELD) from SCHEMA.TABLE A where RTRIM(A.FIELD) = ('10328189B')
UNION
select RTRIM(A.FIELD) from SCHEMA.TABLE A where RTRIM(A.FIELD) = ('103498732H')
我知道这是一个旧堆栈,但是我将发布此Multi-SQL select案例
SELECT bp.bizid, bp.usrid, bp.website,
ROUND((SELECT SUM(rating) FROM ratings WHERE bizid=bp.bizid)/(SELECT COUNT(*) FROM ratings WHERE bizid=bp.bizid), 1) AS 'ratings',
(SELECT COUNT(*) FROM bzreviews WHERE bizid=bp.bizid) AS 'ttlreviews',
bp.phoneno, als.bizname,
(SELECT COUNT(*) FROM endorsment WHERE bizid=bp.bizid) AS 'endorses'
, als.imgname, bp.`location`, bp.`ownership`,
(SELECT COUNT(*) FROM follows WHERE bizid=bp.bizid) AS 'followers',
bp.categories, bp.openhours, bp.bizdecri FROM bizprofile AS bp
INNER JOIN alluser AS als ON bp.usrid=als.userid
WHERE als.usertype='Business'