HttpUtil已弃用，这是正确的方法

StringBuffer url = req.getRequestURL();
String queryString = req.getQueryString();
if (queryString != null) {
    url.append('?');
    url.append(queryString);
}
String requestURL = url.toString();

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();
    }
}

在HttpServletRequest对象上使用以下方法

以getRequestURI () 返回该请求的URL部分，从协议名到HTTP请求第一行的查询字符串。

java.lang.StringBuffer getRequestURL () -重建客户端用来发出请求的URL。

以getQueryString () -返回包含在请求URL中路径后面的查询字符串。

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.">&lt;www.ziesemer.com&gt;</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

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();
}

如果你使用了.getRequestURL()中的StringBuffer的构建器模式，你可以用三元写一个简单的一行代码:

private String getUrlWithQueryParms(final HttpServletRequest request) { 
    return request.getQueryString() == null ? request.getRequestURL().toString() :
        request.getRequestURL().append("?").append(request.getQueryString()).toString();
}

但这只是语法糖。