查询一天:select * from 表名 where to_days(时间字段名) = to_days(now()); 昨天 SELECT * FROM 表名 WHERE TO_DAYS( NOW( ) ) - TO_DAYS( 时间字段名) <= 1 7天 SELECT * FROM 表名 where DATE_SUB(CURDATE(), INTERVAL 7 DAY) <= date(时间字段名) 查询一周内数据 select * from Tabel名 where 时间字段名 bet
SQL SERVER 查询第20行到30之间的数据 1.先查询前20行的ID,后查询除去20条记录的前10条记录 SELECT TOP 10 * FROM tbBank WHERE BankID NOT IN(SELECT TOP 20 BankID FROM tbBank ORDER BY BankID ASC) 2.先查询前20行记录最大的ID,后查询大于该值的前10条记录 SELECT TOP 10 * FROM tbBank WHERE BankID>(SELECT MAX(BankID
SELECT HOUR(e.time) as Hour,count(*) as Count FROM error_log e WHERE e.date = '2017-09-02' GROUP BY HOUR(e.time) ORDER BY Hour(e.time); 下面是查询结果截图 在另一篇文章里,我总结了查询每半小时统计一次的方法.Mysql 查询一天中每半小时记录的数量
1. Refactor the following code (C#) to use polymorphism instead of if-else. Please provide your answer in C#, Java or C++. Alternatively for partial points, you could describe the type of change using UML to aid your response. public class Calculato
1.先查询前20行的ID,后查询除去20条记录的前10条记录 SELECT TOP * FROM tbBank WHERE BankID NOT IN(SELECT TOP BankID FROM tbBank ORDER BY BankID ASC) 2.先查询前20行记录最大的ID,后查询大于该值的前10条记录 SELECT TOP * FROM tbBank WHERE BankID>(SELECT MAX(BankID) FROM (SELECT TOP BankID FROM tbBa
1.取得当前时间 function getNowFormatDate() { var date = new Date(); var seperator1 = "-"; var seperator2 = ":"; var month = date.getMonth() + 1; var strDate = date.getDate(); var hours = date.getHours(); var minutes = date.getMinutes(); var