What's the best way to get the current URL in Spring MVC? - url

Question

I'd like to create URLs based on the URL used by the client for the active request. Is there anything smarter than taking the current HttpServletRequest object and it's getParameter...() methods to rebuilt the complete URL including (and only) it's GET parameters.
Clarification: If possible I want to resign from using a HttpServletRequest object.

Answer 1

Well there are two methods to access this data easier, but the interface doesn't offer the possibility to get the whole URL with one call. You have to build it manually:
public static String makeUrl(HttpServletRequest request)
{
return request.getRequestURL().toString() + "?" + request.getQueryString();
}
I don't know about a way to do this with any Spring MVC facilities.
If you want to access the current Request without passing it everywhere you will have to add a listener in the web.xml:
<listener>
<listener-class>org.springframework.web.context.request.RequestContextListener</listener-class>
</listener>
And then use this to get the request bound to the current Thread:
((ServletRequestAttributes) RequestContextHolder.currentRequestAttributes()).getRequest()

Answer 2

Instead of using RequestContextHolder directly, you can also use ServletUriComponentsBuilder and its static methods:
ServletUriComponentsBuilder.fromCurrentContextPath()
ServletUriComponentsBuilder.fromCurrentServletMapping()
ServletUriComponentsBuilder.fromCurrentRequestUri()
ServletUriComponentsBuilder.fromCurrentRequest()
They use RequestContextHolder under the hood, but provide additional flexibility to build new URLs using the capabilities of UriComponentsBuilder.
Example:
ServletUriComponentsBuilder builder = ServletUriComponentsBuilder.fromCurrentRequestUri();
builder.scheme("https");
builder.replaceQueryParam("someBoolean", false);
URI newUri = builder.build().toUri();

Answer 3

Java's URI Class can help you out of this:
public static String getCurrentUrl(HttpServletRequest request){
URL url = new URL(request.getRequestURL().toString());
String host = url.getHost();
String userInfo = url.getUserInfo();
String scheme = url.getProtocol();
String port = url.getPort();
String path = request.getAttribute("javax.servlet.forward.request_uri");
String query = request.getAttribute("javax.servlet.forward.query_string");
URI uri = new URI(scheme,userInfo,host,port,path,query,null)
return uri.toString();
}

Answer 4

in jsp file:
request.getAttribute("javax.servlet.forward.request_uri")

Answer 5

You can also add a UriComponentsBuilder to the method signature of your controller method. Spring will inject an instance of the builder created from the current request.
#GetMapping
public ResponseEntity<MyResponse> doSomething(UriComponentsBuilder uriComponentsBuilder) {
URI someNewUriBasedOnCurrentRequest = uriComponentsBuilder
.replacePath(null)
.replaceQuery(null)
.pathSegment("some", "new", "path")
.build().toUri();
//...
}
Using the builder you can directly start creating URIs based on the current request e.g. modify path segments.
See also UriComponentsBuilderMethodArgumentResolver

Answer 6

If you need the URL till hostname and not the path use Apache's Common Lib StringUtil, and from URL extract the substring till third indexOf /.
public static String getURL(HttpServletRequest request){
String fullURL = request.getRequestURL().toString();
return fullURL.substring(0,StringUtils.ordinalIndexOf(fullURL, "/", 3));
}
Example: If fullURL is https://example.com/path/after/url/ then
Output will be https://example.com

Answer 7

System.out.println(((HttpServletRequest)request).getRequestURI());
I used it. hope it's useful.