Question

Spring Cloud Gateway确实是新来的-但它“似乎”足够容易。我真的很挣扎的一个问题。我的要求是在路径前添加前缀，检查标头变量，基于该变量查找URI，然后进行调试。

问题是，即使我在idResolvingGatewayFilter中进行了更改，uri也始终是下面的DEFAULT_IMPLEMENTION。我该怎么做？可以随时添加新ID-这就是“动态”部分。因此，在网关过滤器中，我正在读取标头并查找uri（我正在查看的数据源可以随时更新）。但是下面的代码似乎覆盖了我在过滤器中分配的所有内容-如果不提供URI，则无法执行此操作。例如：

header-id = 123
uri = http://www.somedestination.com/something/services/v1.0

header-id = 999
uri = http://www.anotherdestination.com/something/services/v1.0

  @Bean
  public RouteLocator rosterRouteLocator( RouteLocatorBuilder builder )
  {
    log.info( "Establishing Gateway Routes" );
    return builder.routes()
        .route( r -> r.path( "/**" ).filters( f -> f.prefixPath( "/something/services/v1.0" ).filter( idResolvingGatewayFilter() ) )
            .uri( resolver.buildDestinationEndpoint( IdUrlResolver.DEFAULT_IMPLEMENTATION ) ) )
        .build();
  }

然后在idResolvingGatewatFilter中进行更改（并且log语句看起来是正确的……它只是不行了！

public Mono<Void> filter( ServerWebExchange exchange, GatewayFilterChain chain )
  {
    try
    {
      URI newUri = buildURI( exchange );
      ServerHttpRequest request = exchange.getRequest().mutate().uri( newUri ).build();
      exchange = exchange.mutate().request( request ).build();
      log.debug( "Modified URI: " + exchange.getRequest().getURI() );

Answer 1

@SpringBootApplication
public class SpringCloudGatewayApplication {

    @Autowired
    private CustomerFilter filter;

    public static void main(String[] args) {
        SpringApplication.run(SpringCloudGatewayApplication.class, args);
    }

    @Bean
    public RouteLocator myRoutes(RouteLocatorBuilder builder) {

        return builder.routes().route(p -> p.path("/**").filters(f -> f.filter(filter)).uri("no://op")).build();
    }
}

class CustomerFilter implements GatewayFilter, Ordered {

    @Override
    public int getOrder() {
        return RouteToRequestUrlFilter.ROUTE_TO_URL_FILTER_ORDER + 1;
    }

    @Override
    public Mono<Void> filter(ServerWebExchange exchange, GatewayFilterChain chain) {
        String newUrl = null;

        if (exchange.getRequest().getHeaders().getHost().toString().equals("localhost:8080")) {
            newUrl = "http://ip1/path1";
        } else {
            newUrl = "http://ip2/path2";
        }

        exchange.getAttributes().put(ServerWebExchangeUtils.GATEWAY_REQUEST_URL_ATTR, new URI(newUrl));
        return chain.filter(exchange);
    }
}

Spring Cloud Gateway动态路由

1 个答案: