Question

我在PHP 5.4中创建了一个在SQL Server 2008 r2中创建记录的代码。 PHP代码创建一个日期：

$date = date("Y-m-d");

其中：

'2015-06-23'

该日期用于对SQL服务器的查询：

$insert = 'INSERT INTO AdAttend (AdEnrollSchedID, AdEnrollID, Date, StartTime, Attend, Absent, unitType, Status, Comment, Excused, ModFlag, UserID, DateAdded, DateLstMod, AdClassAttendID, Type)
        VALUES ('.$AdEnrollSchedID.','.$AdEnrollID.','.$date.',\'1899-12-30 07:00:00.000\','.$present.','.$absent.',\'M\',\'A\',\''.$comment.'\',0,\'A\','.$instructorID.
        ','.$date.','.$date.','.$AdClassAttendID.',\'A\')';

查询似乎工作正常。记录按原样插入。但是当我检查表中的记录时，日期是：

 '1905-06-23 00:00:00.000'

所以问题是，为什么将2015年改为1905年？

Answer 1

插入转义引号或在单引号中使用双引号

$insert = 'INSERT INTO AdAttend (AdEnrollSchedID, AdEnrollID, Date, StartTime, Attend, Absent, unitType, Status, Comment, Excused, ModFlag, UserID, DateAdded, DateLstMod, AdClassAttendID, Type)
        VALUES ('.$AdEnrollSchedID.','.$AdEnrollID.',\''.$date.'\',\'1899-12-30 07:00:00.000\','.$present.','.$absent.',\'M\',\'A\',\''.$comment.'\',0,\'A\','.$instructorID.
        ','.$date.','.$date.','.$AdClassAttendID.',\'A\')';

OR

$insert = 'INSERT INTO AdAttend (AdEnrollSchedID, AdEnrollID, Date, StartTime, Attend, Absent, unitType, Status, Comment, Excused, ModFlag, UserID, DateAdded, DateLstMod, AdClassAttendID, Type)
        VALUES ('.$AdEnrollSchedID.','.$AdEnrollID.',"'.$date.'",\'1899-12-30 07:00:00.000\','.$present.','.$absent.',\'M\',\'A\',\''.$comment.'\',0,\'A\','.$instructorID.
        ','.$date.','.$date.','.$AdClassAttendID.',\'A\')';

从php写日期到SQL，日期改变了

1 个答案: