我有一个HttpServletRequest对象。
我如何获得导致这个调用到达我的servlet的完整而准确的URL ?
或者至少尽可能准确,因为有些东西是可以重新生成的(也许是参数的顺序)。
我有一个HttpServletRequest对象。
我如何获得导致这个调用到达我的servlet的完整而准确的URL ?
或者至少尽可能准确,因为有些东西是可以重新生成的(也许是参数的顺序)。
当前回答
当请求被转发时,例如从反向代理,HttpServletRequest.getRequestURL()方法将不会返回被转发的url,而是返回本地url。 当设置了x-forwarded-* Headers时,这可以很容易地处理:
public static String getCurrentUrl(HttpServletRequest request) {
String forwardedHost = request.getHeader("x-forwarded-host");
if(forwardedHost == null) {
return request.getRequestURL().toString();
}
String scheme = request.getHeader("x-forwarded-proto");
String prefix = request.getHeader("x-forwarded-prefix");
return scheme + "://" + forwardedHost + prefix + request.getRequestURI();
}
这缺少查询部分,但可以在其他答案中添加。我来这里,是因为我特别需要转发的东西,希望能帮助别人解决这个问题。
其他回答
// http://hostname.com/mywebapp/servlet/MyServlet/a/b;c=123?d=789
public static String getUrl(HttpServletRequest req) {
String reqUrl = req.getRequestURL().toString();
String queryString = req.getQueryString(); // d=789
if (queryString != null) {
reqUrl += "?"+queryString;
}
return reqUrl;
}
HttpServletRequest有以下方法:
getRequestURL() -在查询字符串分隔符之前返回完整URL的部分? getQueryString() -在查询字符串分隔符后返回完整URL的部分?
所以,要获得完整的URL,只需做:
public static String getFullURL(HttpServletRequest request) {
StringBuilder requestURL = new StringBuilder(request.getRequestURL().toString());
String queryString = request.getQueryString();
if (queryString == null) {
return requestURL.toString();
} else {
return requestURL.append('?').append(queryString).toString();
}
}
有点晚了,但我把它包括在我的MarkUtils-Web库中的WebUtils - Checkstyle-approved和junit - tests:
import javax.servlet.http.HttpServletRequest;
public class GetRequestUrl{
/**
* <p>A faster replacement for {@link HttpServletRequest#getRequestURL()}
* (returns a {@link String} instead of a {@link StringBuffer} - and internally uses a {@link StringBuilder})
* that also includes the {@linkplain HttpServletRequest#getQueryString() query string}.</p>
* <p><a href="https://gist.github.com/ziesemer/700376d8da8c60585438"
* >https://gist.github.com/ziesemer/700376d8da8c60585438</a></p>
* @author Mark A. Ziesemer
* <a href="http://www.ziesemer.com."><www.ziesemer.com></a>
*/
public String getRequestUrl(final HttpServletRequest req){
final String scheme = req.getScheme();
final int port = req.getServerPort();
final StringBuilder url = new StringBuilder(256);
url.append(scheme);
url.append("://");
url.append(req.getServerName());
if(!(("http".equals(scheme) && (port == 0 || port == 80))
|| ("https".equals(scheme) && port == 443))){
url.append(':');
url.append(port);
}
url.append(req.getRequestURI());
final String qs = req.getQueryString();
if(qs != null){
url.append('?');
url.append(qs);
}
final String result = url.toString();
return result;
}
}
可能是目前为止最快和最强大的答案,仅次于Mat Banik的答案——但即使是他的答案也没有考虑到HTTP/HTTPS的潜在非标准端口配置。
参见:
http://blogger.ziesemer.com/2017/08/httpservletrequestgetrequesturl.html https://gist.github.com/ziesemer/700376d8da8c60585438
你可以使用滤镜。
@Override
public void doFilter(ServletRequest arg0, ServletResponse arg1, FilterChain arg2) throws IOException, ServletException {
HttpServletRequest test1= (HttpServletRequest) arg0;
test1.getRequestURL()); it gives http://localhost:8081/applicationName/menu/index.action
test1.getRequestURI()); it gives applicationName/menu/index.action
String pathname = test1.getServletPath()); it gives //menu/index.action
if(pathname.equals("//menu/index.action")){
arg2.doFilter(arg0, arg1); // call to urs servlet or frameowrk managed controller method
// in resposne
HttpServletResponse httpResp = (HttpServletResponse) arg1;
RequestDispatcher rd = arg0.getRequestDispatcher("another.jsp");
rd.forward(arg0, arg1);
}
不要忘记在web.xml的filter mapping中放入<dispatcher>FORWARD</dispatcher>
我使用这个方法:
public static String getURL(HttpServletRequest req) {
String scheme = req.getScheme(); // http
String serverName = req.getServerName(); // hostname.com
int serverPort = req.getServerPort(); // 80
String contextPath = req.getContextPath(); // /mywebapp
String servletPath = req.getServletPath(); // /servlet/MyServlet
String pathInfo = req.getPathInfo(); // /a/b;c=123
String queryString = req.getQueryString(); // d=789
// Reconstruct original requesting URL
StringBuilder url = new StringBuilder();
url.append(scheme).append("://").append(serverName);
if (serverPort != 80 && serverPort != 443) {
url.append(":").append(serverPort);
}
url.append(contextPath).append(servletPath);
if (pathInfo != null) {
url.append(pathInfo);
}
if (queryString != null) {
url.append("?").append(queryString);
}
return url.toString();
}