如何从ExternalContext获取webapp的绝对URL？

Question

我正在尝试从ExternalContext检索Web应用程序的根URL，但无法理解使用哪种方法...

Answer 1

更简洁的方法是：

HttpServletRequest request = (HttpServletRequest) externalContext.getRequest();
String url = request.getRequestURL().toString();
String baseURL = url.substring(0, url.length() - request.getRequestURI().length()) + request.getContextPath() + "/";

然后，当方案为http且端口为80时，您不需要忽略端口，依此类推。

Answer 2

您可以从FacesContext获取ExternalContext ，然后从外部上下文中提取request

String file = request.getRequestURI();
if (request.getQueryString() != null) {
   file += '?' + request.getQueryString();
}
URL reconstructedURL = new URL(request.getScheme(),
                               request.getServerName(),
                               request.getServerPort(),
                               file);
reconstructedURL.toString();

资源

Answer 3

这是我发现的最简单的方法，它不涉及URL的各个部分的神秘字符串操作。 它似乎适用于所有情况，包括不同的协议和端口。

String getAbsoluteApplicationUrl() throws URISyntaxException {
        ExternalContext externalContext = FacesContext.getCurrentInstance().getExternalContext();
        HttpServletRequest request = (HttpServletRequest) externalContext.getRequest();
        URI uri = new URI(request.getRequestURL().toString());
        newUri = new URI(uri.getScheme(), null,
                uri.getHost(),
                uri.getPort(),
                request.getContextPath().toString(),null, null);
        return newUri.toString();
 }

Answer 4

我有一个类似于BalusC的：

FacesContext context = FacesContext.getCurrentInstance();
HttpServletRequest request = (HttpServletRequest)context.getExternalContext().getRequest();
String requestURL = request.getRequestURL().toString();
String url = requestURL.substring(0, requestURL.lastIndexOf("/"));

Answer 5

让我重新说一下Jigar的答案：

final ExternalContext ectx = context.getExternalContext();
String url = ectx.getRequestScheme()
  + "://" + ectx.getRequestServerName()
  + ":" + ectx.getRequestServerPort()
  + "/" + ectx.getRequestContextPath();