在 Java 中将列表作为参数传递

Question

我有一个清单。

[Attachments(filename=a.json,  id=ATT-mXRJB-BmVzs, contentType=application/json),
 Attachments(filename=b.pdf,  id=ATT-y7Qr2-8RqkW, contentType=application/pdf ),
 Attachments(filename=c.docx,  id=ATT-mYh3z-3YJ37, contentType=application/vnd.openxmlformats-officedocument.wordprocessingml.document)]

我需要迭代并从此列表中获取 id 并将每个 id 作为参数传递给 API 调用 - attachments/{{attachmentid}}/retrieve 。

我能够检索 ids 并将它们存储在 Map 中。 （不确定在这里使用 Map 是否正确？）这是我编写的代码片段，但之后我无法继续。

  Map<String, String> attachmentMap = new HashMap<String, String>();
    Optional.ofNullable(attList)
                .orElse(Collections.emptyList())
                .forEach(aList -> {
                     attachmentMap.put("id", aList.getId());
                    LOGGER.debug("Attachment ids : {}",attachmentMap);
                });
    

        

2020-08-27 22:21:49.967 调试 18644 --- [nio-8081-exec-2] cfphsiPlServiceImpl：附件 ID：{id=ATT-mXRJB-BmVzs} 2020-08-27：294G9166 --- [nio-8081-exec-2] cfphsiPlServiceImpl : 附件 ID : {id=ATT-y7Qr2-8RqkW} 2020-08-27 22:21:49.967 DEBUG 18644 --- [nio-8081-exec-2] cfphsiPlServiceImpl：附件 ID：{id=ATT-mYh3z-3YJ37}

  URI uri = UriComponentsBuilder.
                    .pathSegment(ATTACHMENTS)
                    .pathSegment(?)   //what to give here? 
                    .pathSegment(RETRIEVE)
                    .build().toUri();
        LOGGER.debug("URL to Retrieve  : {}", uri.toString());

我能够打印附件 id - 但是我如何将它们作为参数传递给这样的 api 调用 -attachments/{{attachmentid}}/retrieve

Answer 1

请使用foreach遍历map并调用http调用方法

Map<String, String> attachmentMap = new HashMap<String, String>();
    Optional.ofNullable(attList)
                .orElse(Collections.emptyList())
                .forEach(aList -> {
                     attachmentMap.put("id", aList.getId());
                     httpCall(aList.getId());
                    LOGGER.debug("Attachment ids : {}",attachmentMap);
                });

使用参数 id（以及您需要的其他参数）创建 Http 调用方法并使用方法进行调用。

public static void httpCall(id){
  URI uri = UriComponentsBuilder.
                    .pathSegment(ATTACHMENTS)
                    .pathSegment(id)   //what to give here? 
                    .pathSegment(RETRIEVE)
                    .build().toUri();
        LOGGER.debug("URL to Retrieve  : {}", uri.toString());
}

Answer 2

为什么不直接遍历Attachments列表并从每个Attachment检索 id 并将其传递给 UriComponentsBuilder 以构建 URI。 您不需要额外的数据结构来保存 Id。

List<URI> uriList = new ArrayList<>();

for(Attachment attachment: Attachments){
   String id = attachment.getId();
   if (id != null) {
       URI uri = UriComponentsBuilder.
                .pathSegment(ATTACHMENTS)
                .pathSegment(id)
                .pathSegment(RETRIEVE)
                .build().toUri();
       uriList.add(uri); // or you can call the uri directly from here using your http client.
   }
}