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.">&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

我有一个用例来生成cURL命令(我可以在终端使用)从httpServletRequest实例。我创建了一个这样的方法。您可以直接在终端中复制粘贴此方法的输出

private StringBuilder generateCURL(final HttpServletRequest httpServletRequest) {
    final StringBuilder curlCommand = new StringBuilder();
    curlCommand.append("curl ");

    // iterating over headers.
    for (Enumeration<?> e = httpServletRequest.getHeaderNames(); e.hasMoreElements();) {
        String headerName = (String) e.nextElement();
        String headerValue = httpServletRequest.getHeader(headerName);
        // skipping cookies, as we're appending cookies separately.
        if (Objects.equals(headerName, "cookie")) {
            continue;
        }
        if (headerName != null && headerValue != null) {
            curlCommand.append(String.format(" -H \"%s:%s\" ", headerName, headerValue));
        }
    }

    // iterating over cookies.
    final Cookie[] cookieArray = httpServletRequest.getCookies();
    final StringBuilder cookies = new StringBuilder();
    for (Cookie cookie : cookieArray) {
        if (cookie.getName() != null && cookie.getValue() != null) {
            cookies.append(cookie.getName());
            cookies.append('=');
            cookies.append(cookie.getValue());
            cookies.append("; ");
        }
    }
    curlCommand.append(" --cookie \"" + cookies.toString() + "\"");

    // appending request url.
    curlCommand.append(" \"" + httpServletRequest.getRequestURL().toString() + "\"");
    return curlCommand;
}

// 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;
}

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();

在Spring项目中您可以使用

UriComponentsBuilder.fromHttpRequest(new ServletServerHttpRequest(request)).build().toUriString()