javax.servlet.ServletException: Circular view path []: would dispatch back to the current....

解决：javax.servlet.ServletException: Circular view path []: would dispatch back to the current…

原因：
当没有声明ViewResolver时，spring会注册一个默认的ViewResolver，就是JstlView的实例，该对象继承自InternalResoureView。
JstlView用来封装JSP或者同一Web应用中的其他资源，它将model对象作为request请求的属性值暴露出来, 并将该请求通过javax.servlet.RequestDispatcher转发到指定的URL.
Spring认为，这个view的URL是可以用来指定同一web应用中特定资源的，是可以被RequestDispatcher转发的。
也就是说，在页面渲染(render)之前，Spring会试图使用RequestDispatcher来继续转发该请求。

错误代码：
示例1：

@RequestMapping("/index")
public void query(){//代码内容
}

示例2：

@RequestMapping("/index")
public String query(){//代码内容return null;
}

解决方案：
通过原因分析，造成问题有两个因素：

缺省转发
view和path同名
所以，解决方案如下
1.消除缺省转发
2.修改view和path，让他们不同名

例：

@RequestMapping("/index")
public String query(){//代码内容return "succeed"; //这里需要返回一个和"/index"不同名的view
}

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