您好现在已经从事了一段时间，但似乎无法弄清为什么它不会更新。 我已经写了该语句，但是当它运行时它根本不会更新记录，但是我看不出有什么问题。
$query = ("INSERT INTO tbldvd(Title, Certificate, Director, Genre, Synopsis, RentalAmount, BeingRented) VALUES('$Title','$Certification','$Director','$Genre','$Synopsis','$Price','$Rented') WHERE DVDID= ".$ID."");
是我的查询，运行时出现错误
ERROR 404: Could not able to execute INSERT INTO tbldvd(Title, Certificate, Director, Genre, Synopsis, RentalAmount, BeingRented) VALUES ('Secret Life Of Walter Mitty','18','Ben Stiller','Family','Day Dreamer','5.00','No') WHERE DVDID= 3 You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'WHERE DVDID= 3' at line 1
已经尝试使用update命令来获得相同的问题
$query = mysqli_query($con, "UPDATE tbldvd SET Title=[$Title],Certificate=[$Certification],Director=[$Director],Genre=[$Genre],Synopsis=[$Synopsis],RentalAmount=[$Price],BeingRented=[$Rented] WHERE DVDID = $ID");
我哪里出错了
ERROR 404: Could not able to execute You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '[Secret Life Of Walter Mitty],Certificate=[18],Director=[Ben Stiller],`Genre' at line 1