How to get URL given path in Spring boot Controller?

Question

In thymeleaf we have:

<a th:href="@{/somepath}">Link</a>

It becomes:

<a href="http://hostname:port/somepath">Link</a>

I want to get the full URL given path like that in controller, like:

@GetMapping(path="/")
public String index(SomeInjectedClass cls) {
    String link = cls.someMethod('/somepath');
    // expected, link = http://hostname:port/somepath
    return "index";
}

@GetMapping(path="/home")
public String home(SomeInjectedClass cls) {
    String link = cls.someMethod('/somepath');
    // expected, link = http://hostname:port/somepath
    return "home";
}

EDITED This question can be interpreted as this:

public static String APPLICATION_BASE_URL = "http://hostname:port";
function someMethod(String method){
    return APPLICATION_BASE_URL + method;
}

I think that APPLICATION_BASE_URL is ugly since i can deploy everywhere. I wonder there is a pretty function in spring boot or even java to get the base URL of my application.

This [post](https://stackoverflow.com/questions/1629102/root-url-of-the-servlet/37908450#37908450) answers my question, although looks ugly. — smftr, Mar 13 '18 at 17:31

score 4 · Answer 1 · answered Mar 13 '18 at 13:55

4

The way to do it is using HttpServletRequest

@GetMapping(path="/")
public String index(HttpServletRequest httpServletRequest) {
    String link = httpServletRequest.getRequestURL();
    return "index";
}

answered Mar 13 '18 at 13:55

pvpkiran

25,582
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this will return my current url in these mapping. i want to get another path instead – smftr Mar 13 '18 at 14:17
what do u meant by another path? – pvpkiran Mar 13 '18 at 14:17
edited my question, those links in both mapping will have same value – smftr Mar 13 '18 at 14:39
I upvoted this answer, because it answers the question as it is worded today (although the poster wanted in fact something else). *PS In fact it answers the posters question!* It helped me solve my issue. Thanks @pvpkiran – t0r0X May 02 '19 at 15:59

Sastrija · Answer 2 · 2021-09-03T17:05:05.520

0

Below one line code should work.

ServletUriComponentsBuilder.fromCurrentContextPath().path("/newPath").toUriString();

or

ServletUriComponentsBuilder.fromCurrentContextPath().replacePath("/newPath").toUriString();

edited Sep 03 '21 at 17:05

answered Dec 24 '20 at 19:38

Sastrija

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This will only append a path. I think the TO want a replacing procedure. – Snickbrack Sep 02 '21 at 10:07
@Snickbrack, Please check the Edited section of the question and sample there. This solution gives same result which OP is asking – Sastrija Sep 02 '21 at 15:24
No, I have tested it and the `.path`-Method appends the `/newPath` instead of replacing it. – Snickbrack Sep 03 '21 at 14:14
@Snickbrack, I'm getting the expected results when I tried from my side. You can try `.replacePath()` instead of `.path()` – Sastrija Sep 03 '21 at 16:38

How to get URL given path in Spring boot Controller?

2 Answers2