如何使用 X-Forwarded-* headers 创建没有 context-path 的 URI？

Question

我正在尝试找到一个解决方案，它将使用 X-Forwarded-* headers.

构建一个新的 link

public class ApiUriBuilderTest {

    private MockHttpServletRequest request = new MockHttpServletRequest();
    private HttpRequest httpRequest = new ServletServerHttpRequest(request);


    @Before
    public void setUp() throws Exception {
        request.setScheme("http");
        request.setServerName("localhost");
        request.setServerPort(80);
        request.setRequestURI("/mvc-showcase");
        request.addHeader("X-Forwarded-Proto", "https");
        request.addHeader("X-Forwarded-Host", "84.198.58.199");
        request.addHeader("X-Forwarded-Port", "443");

        request.setContextPath("/mvc-showcase");
        request.setServletPath("/app");
        request.setRequestURI("/mvc-showcase/app/uri/of/request?hello=world&raw#my-frag");

        httpRequest = new ServletServerHttpRequest(request);

    }

    @Test
    public void test() {
        String uri = ForwardedContextPathServletUriComponentsBuilder.fromRequest(request).build().toUriString();
        assertThat(uri, is("https://84.198.58.199:443"));

    }

    @Test
    public void test_uri_components_builder() throws URISyntaxException {
        UriComponents result = UriComponentsBuilder.fromHttpRequest(httpRequest).build();
        assertEquals("https://84.198.58.199:443", result.toString());
    }

但返回值为“https://84.198.58.199/mvc-showcase/app/uri/of/request?hello=world&raw#my-frag”。我怎样才能摆脱 context-path、setvlet-path 和请求 uri？

Answer 1

@Test
public void test() {
    String uri = ServletUriComponentsBuilder.fromRequest(request).replacePath("relativePath").replaceQuery(null).build().toUriString();
    assertThat(uri, is("https://84.198.58.199:8080/relativePath"));

}

有帮助。

如何使用 X-Forwarded-* headers 创建没有 context-path 的 URI？

How to create URI without context-path using X-Forwarded-* headers?

java

spring

uri

builder

spring-mvc