我思故我在

select top 10 * from test where day(dateandtime)=day(getdate())select * from test  where datediff(d,day(dateandtime),day(getdate()))=0select   *   from   Attendance   where   convert(char(10),time,111)=convert(char(10),getdate(),111)

<br /> <br />delete from 'tableName:<br />DELETE 语句每次删除一行，并在事务日志中为所删除的每行记录一项。（相当于把房子里家具全丢了，但为了纪念原先的家具，所以原先放家具的那块地以后不再放新家具了）<br /> <br />drop TABLE 'tableName:<br />删除表定义及其数据（相当于除了把家具丢了外，还把房子拆了，回到解放前）<br /> <br /> <br />truncate TABLE 'tableName':<br />TRUN

今天：select * from 表名 where to_days(时间字段名) = to_days(now()); //MYSQL语法 String mysql = "select * from Table where date(时间字段) = curdate(

String[] strs = { "2", "4" }; String strTemp = ""; for (int k = 0; k < strs.length; k++) { strTemp += strs[k] + ","; } strTemp = strTemp.substring(0, strTemp.length() - 1); //