java - Twitter4j， "Too many terms specified in query"

Question

我使用以下方法获取所有关注已验证用户的用户的屏幕名称。

private void getFollowing() {   
     Twitter t = new TwitterFactory().getInstance();
     t.setOAuthConsumer(OAUTH.CONSUMER_KEY, OAUTH.CONSUMER_SECRET);
     aToken = getToken();
     t.setOAuthAccessToken(aToken);
     try {
        long[] friendsID = t.getFriendsIDs(userID, -1).getIDs();
        ResponseList<User> userName = t.lookupUsers(friendsID);
        int count = 0;
        for (User u : userName) {
            count++;
            Log.d("USERNAME : "+ Integer.toString(count), u.getScreenName());
        }
     } catch (TwitterException e) {
        e.printStackTrace();
    }
}

t.lookupUsers(friendsID) 导致以下错误。

W/System.err(16076): {"errors":[{"code":18,"message":"Too many terms specified in query"}]}

据我所知，lookupUsers() 方法将一次返回最多 100 个用户的信息。我提供的不止于此。这可能是为什么？如果是这样，我如何限制原始请求并遍历其余用户以获取他们的所有屏幕名称？

如果我对错误原因的理解是错误的，那么我还做错了什么？

回答

    private void getFollowing() {   
         Twitter t = new TwitterFactory().getInstance();
         t.setOAuthConsumer(OAUTH.CONSUMER_KEY, OAUTH.CONSUMER_SECRET);
         aToken = getToken();
         t.setOAuthAccessToken(aToken);
         ArrayList<String> names = new ArrayList<String>();
         try {
            int start = 0;
            int finish = 100;
            ArrayList<Long> IDS = new ArrayList<Long>();
            long[] friendsID =  t.getFriendsIDs(userID, -1).getIDs();
            boolean check = true;
            while (check) {
                for (int i=start;i<finish;i++) {
//get first 100     
                    IDS.add(friendsID[i]);
//if at the end, stop
                    if (friendsID.length-1 == i) {
                        check = false;
                        break;                      
                    }
                }
//set values for next 100
                start = start+100;
                finish = finish+100;
                long[] ids = Longs.toArray(IDS);
                ResponseList<User> userName = t.lookupUsers(ids);
//clear so long[] holds max 100 at any given time
                IDS.clear();
                for (User u : userName) {
                    names.add(u.getScreenName());
                }
            }
            String[] screenNames = (String[]) names.toArray(new String[names.size()]);

            ArrayAdapter<String> adapter = new ArrayAdapter<String>(this,android.R.layout.simple_dropdown_item_1line, screenNames);
            mPreview.setAdapter(adapter);
         } catch (TwitterException e) {
            e.printStackTrace();
        }
    }

Answer 1

private void getFollowing() {   
     Twitter t = new TwitterFactory().getInstance();
     t.setOAuthConsumer(OAUTH.CONSUMER_KEY, OAUTH.CONSUMER_SECRET);
     aToken = getToken();
     t.setOAuthAccessToken(aToken);
     ArrayList<String> names = new ArrayList<String>();
     try {
        int start = 0;
        int finish = 100;
        ArrayList<Long> IDS = new ArrayList<Long>();
        long[] friendsID =  t.getFriendsIDs(userID, -1).getIDs();
        boolean check = true;
        while (check) {
            for (int i=start;i<finish;i++) {
                //get first 100     
                IDS.add(friendsID[i]);
                //if at the end, stop
                if (friendsID.length-1 == i) {
                    check = false;
                    break;                      
                }
            }
            //set values for next 100
            start = start+100;
            finish = finish+100;
            long[] ids = Longs.toArray(IDS);
            ResponseList<User> userName = t.lookupUsers(ids);
            //clear so long[] holds max 100 at any given time
            IDS.clear();
            for (User u : userName) {
                names.add(u.getScreenName());
            }
        }
        String[] screenNames = (String[]) names.toArray(new String[names.size()]);

        ArrayAdapter<String> adapter = new ArrayAdapter<String>(this,android.R.layout.simple_dropdown_item_1line, screenNames);
        mPreview.setAdapter(adapter);
     } catch (TwitterException e) {
        e.printStackTrace();
    }
}