select * from user where name like "%1%%" escape 1; select *,RIGHT(name,LENGTH(name) - POSITioN("@"
select * from user where name like "%1%%" escape 1;
select *,RIGHT(name,LENGTH(name) - POSITioN("@" in name)) from user;
#对name列中b不为 null的计数 select COUNT(name = "b" or null) as b , COUNT(name = "c" or null) as c ,(select COUNT(id) from user where name is null) as nan from user ;
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