A faster way to list files in a folder?

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pcapdown

• 2019年03月22日 17:57

I want to navigate a series of folders and pull file name & file size information, using the Java api.

I am using the recommended approach, which is basically to enumerate all the BoxItems found in each BoxFolder.

However it is extremely slow.

Is there not a single command that can be sent to return ALL of the BoxFolder's metadata contents in one go? i.e. the equivalent of the "ls" command?

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• 

  pcapdown

  • 2019年03月22日 18:00

  By the way, when I say slow, I mean it takes almost 1 second to get each ItemInfo:

  for (BoxItem.Info itemInfo : targetFolder) {
  if (itemInfo instanceof BoxFolder.Info) {
  ...
  etc.

   

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  billevans1963

  • 2019年04月05日 09:56

  Ok, I finally figured out why it is so slow:  By default, the method gets only one Box.FileInfo object at a time from the server.  To speed up you need to use pagination with max limit of 100.  Something like the following:

   

  
  int offset = 0; //Start
  int pageSize = 100;
  while (true) {
     Collection itemsInFolder = targetFolder.getChildrenRange(offset, pageSize, "name", "size");
     for (BoxItem.Info itemInfo : itemsInFolder) {
         if (itemInfo instanceof BoxFile.Info) {
               ....
          }
          else if (itemInfo instanceof BoxFolder.Info) {

              ...
          }
      }
      if (itemsInFolder.size() < pageSize) break;           //Less than max returned signals end
      offset += pageSize;
  }

   

   

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