use*_*306 10 java long-polling instagram instagram-api
我正在尝试实时下载贴有特定标签的照片.我发现实时api很无用,所以我使用长轮询策略.下面是伪代码,其中包含sublte错误的注释
newMediaCount = getMediaCount();
delta = newMediaCount - mediaCount;
if (delta > 0) {
// if mediaCount changed by now, realDelta > delta, so realDelta - delta photos won't be grabbed and on next poll if mediaCount didn't change again realDelta - delta would be duplicated else ...
// if photo posted from private account last photo will be duplicated as counter changes but nothing is added to recent
recentMedia = getRecentMedia(delta);
// persist recentMedia
mediaCount = newMediaCount;
}
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第二个问题可以通过Set of some sort来解决.但首先真的困扰我.我已尽可能接近两次打电话给instagram api,但这还够吗?
编辑
正如Amir建议我使用min/max_tag_id
s 重写代码.但它仍然会跳过照片.我找不到更好的方法来测试这个比在磁盘上保存图像一段时间并将结果与之比较instagram.com/explore/tags/
.
public class LousyInstagramApiTest {
@Test
public void testFeedContinuity() throws Exception {
Instagram instagram = new Instagram(Settings.getClientId());
final String TAG_NAME = "portrait";
String id = instagram.getRecentMediaTags(TAG_NAME).getPagination().getMinTagId();
HashtagEndpoint endpoint = new HashtagEndpoint(instagram, TAG_NAME, id);
for (int i = 0; i < 10; i++) {
Thread.sleep(3000);
endpoint.recentFeed().forEach(d -> {
try {
URL url = new URL(d.getImages().getLowResolution().getImageUrl());
BufferedImage img = ImageIO.read(url);
ImageIO.write(img, "png", new File("D:\\tmp\\" + d.getId() + ".png"));
} catch (Exception e) {
e.printStackTrace();
}
});
}
}
}
class HashtagEndpoint {
private final Instagram instagram;
private final String hashtag;
private String minTagId;
public HashtagEndpoint(Instagram instagram, String hashtag, String minTagId) {
this.instagram = instagram;
this.hashtag = hashtag;
this.minTagId = minTagId;
}
public List<MediaFeedData> recentFeed() throws InstagramException {
TagMediaFeed feed = instagram.getRecentMediaTags(hashtag, minTagId, null);
List<MediaFeedData> dataList = feed.getData();
if (dataList.size() == 0) return Collections.emptyList();
String maxTagId = feed.getPagination().getNextMaxTagId();
if (maxTagId != null && maxTagId.compareTo(minTagId) > 0) dataList.addAll(paginateFeed(maxTagId));
Collections.reverse(dataList);
// dataList.removeIf(d -> d.getId().compareTo(minTagId) < 0);
minTagId = feed.getPagination().getMinTagId();
return dataList;
}
private Collection<? extends MediaFeedData> paginateFeed(String maxTagId) throws InstagramException {
System.out.println("pagination required");
List<MediaFeedData> dataList = new ArrayList<>();
do {
TagMediaFeed feed = instagram.getRecentMediaTags(hashtag, null, maxTagId);
maxTagId = feed.getPagination().getNextMaxTagId();
dataList.addAll(feed.getData());
} while (maxTagId.compareTo(minTagId) > 0);
return dataList;
}
}
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使用标签端点获取带有所需标签的最新媒体,它会返回min_tag_id
分页信息,该信息与您调用时最近标记的媒体相关联。由于 API 还接受min_tag_id
参数,因此您可以传递上次查询中的该数字,以便仅接收上次查询后标记的媒体。
因此,根据您拥有的任何轮询机制,您只需调用 API 来获取新的最近媒体(如果有的话)基于上次接收的min_tag_id
。
当标记速度比轮询速度快时,您还需要传递一个大count
参数并遵循响应的分页来接收所有数据而不会丢失任何数据。
更新:
根据您更新的代码:
public List<MediaFeedData> recentFeed() throws InstagramException {
TagMediaFeed feed = instagram.getRecentMediaTags(hashtag, minTagId, null, 100000);
List<MediaFeedData> dataList = feed.getData();
if (dataList.size() == 0) return Collections.emptyList();
// follow the pagination
MediaFeed recentMediaNextPage = instagram.getRecentMediaNextPage(feed.getPagination());
while (recentMediaNextPage.getPagination() != null) {
dataList.addAll(recentMediaNextPage.getData());
recentMediaNextPage = instagram.getRecentMediaNextPage(recentMediaNextPage.getPagination());
}
Collections.reverse(dataList);
minTagId = feed.getPagination().getMinTagId();
return dataList;
}
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