servlets之映射 servlet 来满足我的请求_编程设计

我想映射一个 servlet 来服务包含“app”并以 *.html 结尾的请求，方法如下

<url-pattern>/app/*.html</url-pattern>

但是在运行应用程序时它给了我一个错误

java.lang.IllegalArgumentException: Invalid <url-pattern>  
/app/*.html in servlet mapping

请帮我绘制一下。请提供我可以了解这些 url 映射规则和约定的链接。

请您参考如下方法:

这确实是无效的。通配符必须是第一个或最后一个字符，分别表示后缀或前缀模式。

<url-pattern>*.html</url-pattern>

或

<url-pattern>/app/*</url-pattern>

这在Servlet API specification的第12.2节中都有明确规定。。以下是相关性摘录:

12.2 Specification of Mappings

In the Web application deployment descriptor, the following syntax is used to define mappings:

A string beginning with a ‘/’ character and ending with a ‘/*’ suffix is used for path mapping.

A string beginning with a ‘*.’ prefix is used as an extension mapping.

The empty string ("") is a special URL pattern that exactly maps to the application's context root, i.e., requests of the form http://host:port/<contextroot>/. In this case the path info is ’/’ and the servlet path and context path is empty string (““).

A string containing only the ’/’ character indicates the "default" servlet of the application. In this case the servlet path is the request URI minus the context path and the path info is null.

All other strings are used for exact matches only.

要解决此问题，您有 2 个选择:

使用 /app/*模式，并且不要将非 HTML 文件放入 /app 中。将它们放在其他地方。

使用不同的前缀模式，例如 /controller/*并创建一个 Filter映射到 /app/*并在 doFilter() 中执行以下操作方法:

String uri = ((HttpServletRequest) request).getRequestURI(); 
if (uri.endsWith(".html")) { 
    request.getRequestDispatcher("/controller" + uri).forward(request, response); 
} else { 
    chain.doFilter(request, response); 
}

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servlets之映射 servlet 来满足我的请求

12.2 Specification of Mappings

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