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:
- 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 is null .
- All other strings are used for exact matches only.
路径映射规则:
- 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.
(编辑:李大同)
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