java 如何在 Servlet 2.4 版本的 init() 方法中获取 ContextPath
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How to get ContextPath in init() method of Servlet version 2.4
提问by Lama
I'm using version 2.4 of Servletand I need to get the ContextPaththrough the init()method which is called on server start, so I don't have any Request objectthat could call getContextPath()and because the Servlet version I do not have getContextPath()method in the ServletContexteither.
我使用的是 2.4 版Servlet，我需要ContextPath通过init()在服务器启动时调用的方法，所以我没有任何可以调用的Request 对象，getContextPath()因为 Servlet 版本我也没有getContextPath()方法ServletContext。
Is there a way to get this ContextPath()somehow ?
有没有办法以ContextPath()某种方式得到这个？
回答by Ian Roberts
One web application can be published at several different context paths, so thecontext path (singular) is only meaningful in the context of a particular request. Servlet 2.5 added getContextPath()to ServletContext, specified to return the "primary" context path for this web application, but there's no container-independent way to access this information in earlier spec versions.
一个Web应用程序可以在多个不同的上下文路径被公开，所以在上下文路径（单数）是仅在特定请求的上下文中有意义。Servlet 2.5 添加getContextPath()到ServletContext，指定返回此 Web 应用程序的“主要”上下文路径，但在早期规范版本中没有独立于容器的方式来访问此信息。
There may be tricks that work for certain containers, for example on Tomcat the ServletContext.getResource()method returns URLs with a custom scheme, of the form jndi://hostname/context/.... Thus you may be able to use
可能有一些技巧适用于某些容器，例如在 Tomcat 上，该ServletContext.getResource()方法返回带有自定义方案的 URL，形式为jndi://hostname/context/...。因此，您可以使用
ctx.getResource("/").getPath()
to get the context path on Tomcat (or possibly getResource("/WEB-INF/web.xml")and trim off the tail, as getResource()is specified to return nullif you ask it for a file that does not exist). You will have to experiment with different containers to find similar tricks that work on those.
获取 Tomcat 上的上下文路径（或者可能getResource("/WEB-INF/web.xml")并修剪掉尾部，如果您要求它提供一个不存在的文件，getResource()则指定返回null）。您将不得不尝试不同的容器以找到适用于这些容器的类似技巧。
回答by krampstudio
It seems to be only possible form servlet 2.5 as explained in this post: ServletContext getContextPath()
正如这篇文章中所解释的，似乎只有 servlet 2.5 才有可能：ServletContext getContextPath()
回答by Ernesto Campohermoso
You are right in Servlet 2.4 the object ServeltContext does not have the method getContextPath.
您在 Servlet 2.4 中是对的，对象 ServeltContext 没有 getContextPath 方法。
I can suggest two options:
我可以建议两种选择：
Set the context path as parameter of the servlet:
<servlet><servlet-name>initServlet</servlet-name> <servlet-class>net.cirrus-it.InitServlet`</servlet-class> <init-param> <param-name>contextPath</param-name> <param-value>/myApp</param-value> </init-param></servlet>Try to determine the context path from the method getRealPath()
将上下文路径设置为 servlet 的参数：
<servlet><servlet-name>initServlet</servlet-name> <servlet-class>net.cirrus-it.InitServlet`</servlet-class> <init-param> <param-name>contextPath</param-name> <param-value>/myApp</param-value> </init-param></servlet>尝试从方法 getRealPath() 确定上下文路径
According to the documentation:
根据文档：
Returns a String containing the real path for a given virtual path. For example, the path "/index.html" returns the absolute file path on the server's filesystem would be served by a request for "http://host/contextPath/index.html", where contextPathis the context pathof this ServletContext.
返回一个包含给定虚拟路径的真实路径的字符串。例如，路径“/index.html”返回服务器文件系统上的绝对文件路径，该路径将由对“http://host/contextPath/index.html”的请求提供服务，其中contextPath是此 ServletContext的上下文路径.
回答by Pradeep Simha
Try this code:
试试这个代码：
class demo extends HttpServlet {
public void init(ServletConfig config) {
String path = config.getServletContext().getRealPath("/");
}
}
It should work
它应该工作
