我是
spring框架的新手今天我遇到了web.xml文件中的调度程序servlet配置,我想出了一个关于url模式的问题,比如这个语法/.那么,如果我在tomcat服务器中部署web应用程序,实际上“/”符号适用的是：host：port /或host：port / myWeb /
解决方法
模式/将使您的servlet成为应用程序的默认servlet,这意味着它将获取没有其他完全匹配的每个模式.
URL模式映射：
@H_404_29@
- 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.- 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 isnull.- All other strings are used for exact matches only.
路径映射规则：
@H_404_29@
- The container will try to find an exact match of the path of the request to the path of the servlet. A successful match selects the servlet.
- The container will recursively try to match the longest path-prefix. This is done by stepping down the path tree a directory at a time,using the
/character as a path separator. The longest match determines the servlet selected.- If the last segment in the URL path contains an extension (e.g.
.jsp),the servlet container will try to match a servlet that handles requests for the extension. An extension is defined as the part of the last segment after the last.character.- If neither of the prevIoUs three rules result in a servlet match,the container will attempt to serve content appropriate for the resource requested. If a default servlet is defined for the application,it will be used.
