Get Root/Base Url In Spring MVC

Question

What is the best way to get the root/base url of a web application in Spring MVC?

Base Url = http://www.example.com or http://www.example.com/VirtualDirectory

Tenstrike

Where do you need this? In a controller or in a JSP page? – Figurative

Everywhere within a website that has access to the contest/request/response to get it. – Tenstrike

ServletUriComponentsBuilder.fromCurrentServletMapping().toUriString() – Yearning

Answer 1

If base url is "http://www.example.com", then use the following to get the "www.example.com" part, without the "http://":

From a Controller:

@RequestMapping(value = "/someURL", method = RequestMethod.GET)
public ModelAndView doSomething(HttpServletRequest request) throws IOException{
    //Try this:
    request.getLocalName(); 
    // or this
    request.getLocalAddr();
}

From JSP:

Declare this on top of your document:

<c:set var="baseURL" value="${pageContext.request.localName}"/> //or ".localAddr"

Then, to use it, reference the variable:

<a href="http://${baseURL}">Go Home</a>

Jura

or, from the JSP you can set the value of the "baseURL" variable dirrectly with the "http://" prefix – Jura

he method getLocalAddr() is undefined for the type HttpServletRequest – Illogic

Tbh you shouldn't use this as it's quite an old answer and is not up to date anymore. Have a look at my answer #5013025 – Aldershot

Answer 2

I prefer to use

final String baseUrl = ServletUriComponentsBuilder.fromCurrentContextPath().build().toUriString();

It returns a completely built URL, scheme, server name and server port, rather than concatenating and replacing strings which is error prone.

Bushel

I was looking for this one because this can be used everywhere, even at app startup. – Cute

It's what I was looking for! – Obi

It does not work: java.lang.IllegalStateException: No current ServletRequestAttributes I'm using it in @EventListener(ApplicationReadyEvent.class) handler, trying to use app base url for something, but it does not work. SpringBoot 2.1.7 – Polite

Same here, I get that error anywhere but in the controller. I think it needs a request context to work. – Apostasy

This does not return the base URL. It instead returns the full request URL. – Westing

Answer 3

Explanation

I know this question is quite old but it's the only one I found about this topic, so I'd like to share my approach for future visitors.

If you want to get the base URL from a WebRequest you can do the following:

ServletUriComponentsBuilder.fromRequestUri(HttpServletRequest request);

This will give you the scheme ("http" or "https"), host ("example.com"), port ("8080") and the path ("/some/path"), while fromRequest(request) would give you the query parameters as well. But as we want to get the base URL only (scheme, host, port) we don't need the query params.

Now you can just delete the path with the following line:

ServletUriComponentsBuilder.fromRequestUri(HttpServletRequest request).replacePath(null);

---

TLDR

Finally our one-liner to get the base URL would look like this:

//request URL: "http://example.com:8080/some/path?someParam=42"

String baseUrl = ServletUriComponentsBuilder.fromRequestUri(HttpServletRequest request)
        .replacePath(null)
        .build()
        .toUriString();

//baseUrl: "http://example.com:8080"

---

Addition

If you want to use this outside a controller or somewhere, where you don't have the HttpServletRequest present, you can just replace

ServletUriComponentsBuilder.fromRequestUri(HttpServletRequest request).replacePath(null)

with

ServletUriComponentsBuilder.fromCurrentContextPath()

This will obtain the HttpServletRequest through spring's RequestContextHolder. You also won't need the replacePath(null) as it's already only the scheme, host and port.

Aldershot

Unfortunately, this works only in a bean or a controller. Searching for a solution to add the application server URL in generated PDFs for almost 2 days, no luck :( Getting java.lang.IllegalStateException: No current ServletRequestAttributes outside the controllers and beans. – Midweek

Be careful using this solution with the call replacePath(null) as it loses the application's root context (eg set via the server.servlet.context-path application property). The version that uses just ServletUriComponentsBuilder.fromCurrentContextPath() without the call to replacePath(null) automatically includes the app context root, so works as intended. – Blatant

Answer 4

You can also create your own method to get it:

public String getURLBase(HttpServletRequest request) throws MalformedURLException {

    URL requestURL = new URL(request.getRequestURL().toString());
    String port = requestURL.getPort() == -1 ? "" : ":" + requestURL.getPort();
    return requestURL.getProtocol() + "://" + requestURL.getHost() + port;

}

Galina

The problem with this is that the URL is taken from the request. Meaning, if a certificate is signed for *.example.com, which resolves to 192.168.42.1 and the request was made with the IP address instead of the name, it will cause issues. The best way is to configure (hard code) the name somewhere in the application's configuration. This way, when sending things like emails, your emails will be more legit and trusted. – Calkins

Answer 5

Simply :

/*
 * Returns the base URL from a request.
 *
 * @example: http://myhost:80/myapp
 * @example: https://mysecuredhost:443/
 */
String getBaseUrl(HttpServletRequest req) {
  return ""
    + req.getScheme() + "://"
    + req.getServerName()
    + ":" + req.getServerPort()
    + req.getContextPath();
}

Mabuse

Best answer in the world. – Cassimere

Really great answer – Advised

where to add this code – Assert

@vanisaladhagu you can add it in a utility class, just make the method static you could call it from anywhere like RequestUtil.getBaseUrl(httpRequest) – Mabuse

Answer 6

request.getRequestURL().toString().replace(request.getRequestURI(), request.getContextPath())

Sight

Whats the difference between getRequestURL and getRequestURI? – Transparency

take a look at this link coderanch.com/t/360343/java/request-getURL-request-getURI – Sight

Note: this doesn't work if you request URI is just "/" as it strips the slashes from the protocol. – Piecemeal

Answer 7

In controller, use HttpServletRequest.getContextPath().

In JSP use Spring's tag library: or jstl

Sedan

Answer 8

Either inject a UriCompoenentsBuilder:

@RequestMapping(yaddie yadda)
public void doit(UriComponentBuilder b) {
  //b is pre-populated with context URI here
}

. Or make it yourself (similar to Salims answer):

// Get full URL (http://user:[email protected]/root/some?k=v#hey)
URI requestUri = new URI(req.getRequestURL().toString());
// and strip last parts (http://user:[email protected]/root)
URI contextUri = new URI(requestUri.getScheme(), 
                         requestUri.getAuthority(), 
                         req.getContextPath(), 
                         null, 
                         null);

You can then use UriComponentsBuilder from that URI:

// http://user:[email protected]/root/some/other/14
URI complete = UriComponentsBuilder.fromUri(contextUri)
                                   .path("/some/other/{id}")
                                   .buildAndExpand(14)
                                   .toUri();

Smaragd

Thanks for the intro to UriComponentBuilder - I've never used it before. Very useful. – Franctireur

No problem, glad I could help. – Smaragd

Answer 9

In JSP

<c:set var="scheme" value="${pageContext.request.scheme}"/>
<c:set var="serverPort" value="${pageContext.request.serverPort}"/>
<c:set var="port" value=":${serverPort}"/>

<a href="${scheme}://${pageContext.request.serverName}${port}">base url</a>

reference https://github.com/spring-projects/greenhouse/blob/master/src/main/webapp/WEB-INF/tags/urls/absoluteUrl.tag

Cottonmouth

Answer 10

     @RequestMapping(value="/myMapping",method = RequestMethod.POST)
      public ModelandView myAction(HttpServletRequest request){

       //then follow this answer to get your Root url
     }

Root URl of the servlet

If you need it in jsp then get in in controller and add it as object in ModelAndView.

Alternatively, if you need it in client side use javascript to retrieve it: http://www.gotknowhow.com/articles/how-to-get-the-base-url-with-javascript

Butlery

Personally, I would use this answer to get it in a JSP: #4278945 – Figurative

Answer 11

I think the answer to this question: Finding your application's URL with only a ServletContext shows why you should use relative url's instead, unless you have a very specific reason for wanting the root url.

Tocci

Answer 12

If you just interested in the host part of the url in the browser then directly from request.getHeader("host")) -

import javax.servlet.http.HttpServletRequest;

@GetMapping("/host")
public String getHostName(HttpServletRequest request) {

     request.getLocalName() ; // it will return the hostname of the machine where server is running.

     request.getLocalName() ; // it will return the ip address of the machine where server is running.


    return request.getHeader("host"));

}

If the request url is https://localhost:8082/host

localhost:8082

Sunbeam

Answer 13

Here:

In your .jsp file inside the [body tag]

<input type="hidden" id="baseurl" name="baseurl" value=" " />

In your .js file

var baseUrl = windowurl.split('://')[1].split('/')[0]; //as to split function 
var xhr = new XMLHttpRequest();
var url='http://'+baseUrl+'/your url in your controller';
xhr.open("POST", url); //using "POST" request coz that's what i was tryna do
xhr.send(); //object use to send```

Voorhis

Welcome to SO! Please read the tour, and How do I write a good answer? – Techy

Answer 14

I had the exact requirement and reached to below solution:

String baseUrl = ServletUriComponentsBuilder.fromCurrentContextPath()
                    .replacePath(null).replaceQuery(null).fragment(null).build().toUriString();

For this code to work, it should run inside a thread bound to a Servlet request.

Westing

Answer 15

The following worked for me:

In the controller method, add a parameter of type HttpServletRequest. You can have this parameter and still have an @RequestBody parameter, which is what all the previous answers fail to mention.

@PostMapping ("/your_endpoint")
public ResponseEntity<Object> register(
   HttpServletRequest servletRequest,
   @RequestBody RegisterRequest request
) {

    String url = servletRequest.getRequestURL().toString();
    String contextPath = servletRequest.getRequestURI();
    String baseURL = url.replace(contextPath,"");

    /// .... Other code 

}

I tested this on Spring Boot 3.0.6.

Kindred