在php中想用sql语句创建一个表，但总是报错啊！

代码是（连接数据库的省略了）：
$create_owndiarytable="CREATE TABLE (owndiary_".$userid[id]." (id INT IDENTITY(6,6) PRIMARY KEY ,email VARCHAR(50),name VARCHAR(30),owndiarydirpath VARCHAR(100)))";
mysql_query($create_owndiarytable) or die('false to create diarytable!'.mysql_error());
报错是：
false to create diarytable!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 '(owndiary_1 (id INT IDENTITY(6,6) PRIMARY KEY ,email VARCHAR(50),name VARCHAR(30' at line 1
大家帮帮忙啊，很急的啊！
我把表名后面那个括号去掉了，但还是不行啊！报错是：
false to create diarytable!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 'IDENTITY(6,6) PRIMARY KEY ,email VARCHAR(50),name VARCHAR(30),owndiarydirpath VA' at line 1
还有我不太明白，那个“IDENTITY()”是用来干什么的，如果没有这个，可以设ID为主键吗？

$create_owndiarytable="CREATE TABLE owndiary_".$userid[id]." (id INT IDENTITY(6,6) PRIMARY KEY ,email VARCHAR(50),name VARCHAR(30),owndiarydirpath VARCHAR(100))";

应该是你多了一对括号， CREATE TABLE 后直接加表名，不用再加括号了

温馨提示：内容为网友见解，仅供参考

当前网址：https://22.t2y.org/zz/fc22st000.html