获取URL到webapp上下文（基本URL） [英] getting URL to a webapp context (the base URL)

问题描述

有时您需要在servlet / JSP /基于 HttpServletRequest 的内容中构建Web应用程序上下文的完整URL。
像 http://server.name:8080/context/ 之类的东西。 Servlet API没有一种方法可以实现这一点。

Sometimes you need to construct a full URL to your web app context inside a servlet/JSP/whatever based on HttpServletRequest. Something like http://server.name:8080/context/. Servlet API doesn't have a single method to achieve this.

直接的方法是将所有URL组件附加到 StringBuffer ，例如

The straightforward approach is to append all URL components to a StringBuffer, like

ctxUrl = sb.append(req.getScheme()).append("://")
.append(req.getgetServerName()).append(":")
.append(req.getServerPort()) etc

我想知道这个替代方案是否有任何问题（速度提高了10倍）：

I wonder if there's anything wrong with this alternative (which is 10 times faster):

ctxUrl = req.getRequestURL();
ctxUrl = ctxUrl.substring(0, ctxUrl.lastIndexOf("/")));

以上两种方法总是会产生相同的结果吗？

Will two above methods always produce the same result?

推荐答案

它被称为基本网址（您可以在HTML中使用的网址< base> 标签）。你可以按如下方式获得它：

It's called the "base URL" (the one you could use in HTML <base> tag). You can obtain it as follows:

StringBuffer url = req.getRequestURL();
String uri = req.getRequestURI();
String ctx = req.getContextPath();
String base = url.substring(0, url.length() - uri.length() + ctx.length()) + "/";

您的 ctxUrl.substring（0，ctxUrl.lastIndexOf（/） ））; 方法将在具有多个文件夹的URL上失败，例如 http://server.name:8080/context/folder1/folder2/folder3 。

Your ctxUrl.substring(0, ctxUrl.lastIndexOf("/"))); approach will fail on URLs with multiple folders like http://server.name:8080/context/folder1/folder2/folder3.

• 如何使用相对路径而不包括上下文根名称？（对于JSP）回答）


• 如何获取基本网址？（对于JSF答案）

• How to use relative paths without including the context root name? (for a JSP answer)
• How get the base URL? (for a JSF answer)

这篇关于获取URL到webapp上下文（基本URL）的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！