java - Dispatcher servlet spring 和 url 模式

Question

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，如果我按如下方式在 tomcat 服务器中部署 Web 应用程序，“/”符号实际上适用于什么:host:port/或 host:port/myWeb/

Answer 1

模式 / 将使您的 servlet 成为应用程序的默认 servlet，这意味着它将选择没有另一个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:

1. 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.
2. 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.
3. 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.
4. 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.