先看下面几个sql语句
1SETQUOTED_IDENTIFIER ON
2SELECT*FROM"USER" WHEREa='netasp'
3
4SETQUOTED_IDENTIFIER ON
5SELECT*FROM[USER]WHEREa='netasp'
6
7SETQUOTED_IDENTIFIER OFF
8SELECT*FROM[USER]WHEREa="netasp"
9
10SETQUOTED_IDENTIFIER OFF
11SELECT*FROM[USER]WHEREa='netasp'
2SELECT*FROM"USER" WHEREa='netasp'
3
4SETQUOTED_IDENTIFIER ON
5SELECT*FROM[USER]WHEREa='netasp'
6
7SETQUOTED_IDENTIFIER OFF
8SELECT*FROM[USER]WHEREa="netasp"
9
10SETQUOTED_IDENTIFIER OFF
11SELECT*FROM[USER]WHEREa='netasp'
1 SETQUOTED_IDENTIFIER ON
2 SELECT*FROM"USER" WHEREa='netasp'
3
4 SETQUOTED_IDENTIFIER ON
5 SELECT*FROM[USER]WHEREa='netasp'
6
7 SETQUOTED_IDENTIFIER OFF
8 SELECT*FROM[USER]WHEREa="netasp"
9
10 SETQUOTED_IDENTIFIER OFF
11 SELECT*FROM[USER]WHEREa='netasp'
2 SELECT*FROM"USER" WHEREa='netasp'
3
4 SETQUOTED_IDENTIFIER ON
5 SELECT*FROM[USER]WHEREa='netasp'
6
7 SETQUOTED_IDENTIFIER OFF
8 SELECT*FROM[USER]WHEREa="netasp"
9
10 SETQUOTED_IDENTIFIER OFF
11 SELECT*FROM[USER]WHEREa='netasp'
当在数据库中新建一个名字是USER的表的时候，常常会带来一些麻烦，因为USER是SQL中的关键字，但是上面的几个语句不会报错。再说一个概念：标示符是SQL中的中括号[]。
当SET QUOTED_IDENTIFIER值为ON时，双引号内的字符被当作是数据库对象。就是说双引号" "和标识符[]效果是一样样的，他们都表示引用的字符是数据库对象。单引号'表示字符串的边界。
当SET QUOTDE_IDENTIFIER OFF时，双引号被解释为字符串的边界，和单引号的作用是类似的。就是说双引号"不能当做标识符使用，但是可以当做字符边界，和单引号'的效果是一样样的。
可以做一个总结：当SET QUOTED_IDENTIFIER ON " "等同于[ ] 表示数据库对象；当SET QUOTED_IDENTIFIER OFF" "等同于' '表示字符串边界；还有这里的双引号" 并不是两个单引号'合起来的，是shift+”打出来的，初学者可能会犯这样的错误。
