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
// 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;
}
其他回答
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Spring项目中您可以使用
UriComponentsBuilder.fromHttpRequest(new ServletServerHttpRequest(request)).build().toUriString()
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
结合getRequestURL()和getQueryString()的结果应该会得到您想要的结果。
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;
}