小编Uma*_*mar的帖子

在Java 8中,有没有办法根据条件在流上应用过滤器,

例

我有这个流

if (isAccessDisplayEnabled) {
     src = (List < Source > ) sourceMeta.getAllSources.parallelStream()
         .filter(k - > isAccessDisplayEnabled((Source) k))
         .filter(k - > containsAll((Source) k, substrings, searchString))
         .collect(Collectors.toList());
 } else {
     src = (List < Source > ) sourceMeta.getAllSources.parallelStream()
         .filter(k - > containsAll((Source) k, substrings, searchString))
         .collect(Collectors.toList());
 }

我正在添加过滤器

.filter(k - > isAccessDisplayEnabled((Source) k)))

在基于if-else条件的流上.有没有办法避免if-else,因为如果有更多的过滤器出现,那么它将很难维护.

请告诉我

这是我的Rest模板配置,

    @Bean
    @Qualifier("myRestService")
    public RestTemplate createRestTemplate(@Value("${connection.timeout}") String maxConn) {
        PoolingHttpClientConnectionManager connectionManager = new PoolingHttpClientConnectionManager();
         connectionManager.setMaxTotal(maxTotalConn);
         connectionManager.setDefaultMaxPerRoute(maxPerChannel);

        RequestConfig config = RequestConfig.custom().setConnectTimeout(100000).build();
        CloseableHttpClient httpClient = HttpClientBuilder.create().setConnectionManager(connectionManager)
                .setDefaultRequestConfig(config).build();
        ClientHttpRequestFactory factory = new HttpComponentsClientHttpRequestFactory(httpClient);

        RestTemplate restTemplate = new RestTemplate(factory);

        restTemplate.setErrorHandler(new RestResponseErrorHandler());
         restTemplate.setMessageConverters(createMessageConverters());

        return restTemplate;
    }

我使用PoolingHttpClientConnectionManager来管理连接.

它可以通过以下代码访问,

ResponseEntity<String> response = restClient.exchange( url, HttpMethod.GET, entity , String.class );

我是否需要在上述调用之后释放连接,或者是否由RestTemplate处理.如果我们需要照顾释放连接.

请有人解释/说明如何释放连接.

我正在尝试一个简单的弹簧启动应用程序,它总是自动关闭

 :: Spring Boot ::        (v1.4.1.RELEASE)
2016-10-23 13:05:21.681  INFO 16532 --- [           main] com.example.RestBootApplication          : No active profile set, falling back to default profiles: default
2016-10-23 13:05:21.766  INFO 16532 --- [           main] s.c.a.AnnotationConfigApplicationContext : Refreshing org.springframework.context.annotation.AnnotationConfigApplicationContext@6e20b53a: startup date [Sun Oct 23 13:05:21 EDT 2016]; root of context hierarchy
2016-10-23 13:05:23.682  INFO 16532 --- [           main] o.s.j.e.a.AnnotationMBeanExporter        : Registering beans for JMX exposure on startup
2016-10-23 13:05:23.704  INFO 16532 --- [           main] com.example.RestBootApplication          : Started RestBootApplication in 2.632 seconds (JVM running …

有两个地图,我试图将它们合并到一个地图(finalResp).

Map<String, String[]> map1 = new HashMap<>();
Map<String, String> map2 = new HashMap<>();

HashMap<String, String> finalResp = new HashMap<String, String>();

解决方案 - 在Java 8之前 - 实现如下:

for (Map.Entry<String, String[]> entry : map1.entrySet()) {
    if (map2.containsKey(entry.getKey())) {
        String newValue  = changetoAnother(map1.get(entry.getKey()), map2.get(entry.getKey()));
        finalResp.put(entry.getKey(), newValue);
    }
}

使用Java 8,我坚持这样:

HashMap<String, String> map3 = new HashMap<>(map2);
map1.forEach((k, v) -> map3.merge(k, v, (i, j) -> mergeValue(i, j) ));

如何检查地图1中是否没有地图2键并修改值？