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 answers my question, although looks ugly.

smftr
– smftr

2018-03-13 17:31:05 +00:00
Commented Mar 13, 2018 at 17:31 — smftr
– smftr, Commented Mar 13, 2018 at 17:31

pvpkiran · Accepted Answer · 2018-03-13 13:55:55Z

5

The way to do it is using HttpServletRequest

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

answered Mar 13, 2018 at 13:55

pvpkiran

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4 Comments

smftr Over a year ago

this will return my current url in these mapping. i want to get another path instead

pvpkiran Over a year ago

what do u meant by another path?

smftr Over a year ago

edited my question, those links in both mapping will have same value

t0r0X Over a year ago

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

Sastrija · Accepted Answer · 2021-09-03 17:05:05Z

0

Below one line code should work.

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

or

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

edited Sep 3, 2021 at 17:05

answered Dec 24, 2020 at 19:38

Sastrija

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4 Comments

Gadawadara Over a year ago

This will only append a path. I think the TO want a replacing procedure.

Sastrija Over a year ago

@Snickbrack, Please check the Edited section of the question and sample there. This solution gives same result which OP is asking

Gadawadara Over a year ago

No, I have tested it and the .path-Method appends the /newPath instead of replacing it.

Sastrija Over a year ago

@Snickbrack, I'm getting the expected results when I tried from my side. You can try .replacePath() instead of .path()

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