摘要:
结果为: 表A和表B是1对多的关系 A.ID => B.AID SELECT ID,NAME FROM A WHERE EXIST (SELECT * FROM B WHERE A.ID=B.AID) 执行结果为 1 A1 2 A2 原因可以按照如下分析 SELECT ID,NAME FROM A 阅读全文
摘要:
MySQL: select * from tableName where name like '%helloworld%'; Oracle:select * from tableName where instr(name,'helloworld')>0; select BLACK_VALUE, COUNT(*)as total from EC_COUPONS_BLACK ... 阅读全文
摘要:
select sum(t.paid_fee) from order_payment_log t where to_char(to_date(t.edit_time, 'yyyy-MM-dd HH24:mi:ss'), 'yyyy-MM-dd') = to_char(sysdate, 'yyyy-MM 阅读全文