Error Handling Python (ie. 409 error)

回答済み

フォローする

新規投稿

Tunnelvisie

• 2020年02月20日 00:12

Hi,

 

I've been working on a simple script to help upload files to box. 

 

I've noticed that when I try to upload a file that already exists I get a response back saying 

Message: Item with the same name already exists
Status: 409
Code: item_name_in_use

etc.

 

How do I properly handle these exceptions in Python?

 

For instance if I use boto3 to connect to s3 I can do:

 

try:
   bucket.download_file(file_loc, local_storage_file_name)
except botocore.exceptions.ClientError as e:
   if e.response['Error']['Code'] == "404":
      print(f"{s3_file_name} does not exist on s3.")
else:
      raise

How do I do something similar for Box API? 

0

• 

  jcleblanc

  • 2020年02月20日 12:29

  Hi ,

   

  Here's what you'll want to do:

  1. Listen for the 409 error with the item name in use description coming back whenever you're trying to upload a file (try / catch block) 
  2. When that error occurs, you can do a few things. You can instead call the update file endpoint to push a new file version up, or you can change the file name (e.g. filename(2).txt) and try the upload again.  

  I don't have the code in Python, but I do in Node and Java, if that helps provide some guidance:

   

  Node

  client.files.uploadFile('0', 'tempdoc.txt', stream).then((err, metadata) => {
    console.log("file uploaded");
  }).catch(function (err) {
    if (err.response && err.response.body && err.response.body.code === 'item_name_in_use') {
      console.log('duplicate file detected');
    }
  });  

  Java

  String filePath = "/localdir/temp.txt";
  String fileName = "current.txt";
  String folderId = "0";
  
  // Select Box folder
  BoxFolder folder = new BoxFolder(client, folderId);
  		
  try {
      // Upload file
      FileInputStream stream = new FileInputStream(filePath);
      folder.uploadFile(stream, fileName);
      stream.close();
  } catch (Exception e) {
      if (e.getMessage().contains("item_name_in_use")) {
          System.out.println("Duplicate file name detected");
      }
  }

  0

  コメントアクション パーマリンク

サインインしてコメントを残してください。