使用 Delphi 将文件上传到 Amazon S3

Question

using code I found in another topic. I'm always getting a signature mismatch when authenticating, although accessKeyID and secretAccessKey are verified correct. Also the storage endpoint is correct for the bucket. Using Delphi 10 Seattle. Possible that the CloudComponents have a problem with the region or (local) time ?

Function Amazon_Upload (fileName, bucket, accessKeyID, secretAccessKey : String) : TCallResult;
var
  Service: TAmazonStorageService;
  ConAmazon: TAmazonConnectionInfo;
  info : TCloudResponseInfo;
  upload_stream : TFileStream;
  bytes : TBytes;
begin
  // create file
  upload_stream := TFileStream.Create(fileName,fmOpenRead);
  try
    // filestream to tBytes
    upload_stream.Position := 0;
    SetLength(bytes, upload_stream.Size);
    upload_stream.Write(bytes[0], upload_stream.Size);
    ConAmazon := TAmazonConnectionInfo.Create(nil);
    try
      // amazon connection parameters
      ConAmazon.AccountKey := secretAccessKey;
      ConAmazon.AccountName := accessKeyID;
      ConAmazon.QueueEndpoint := 'queue.amazonaws.com';
      ConAmazon.StorageEndpoint := 's3-eu-central-1.amazonaws.com';
      ConAmazon.TableEndpoint := 'sdb.amazonaws.com';
      ConAmazon.UseDefaultEndpoints := False;
      // storage objects
      info := TCloudResponseInfo.Create;
      Service := TAmazonStorageService.Create(ConAmazon);
      try
        // upload document to storage
        Service.UploadObject(Bucket, fileName, bytes, TRUE, nil, nil, amzbaPrivate, info);
        // get results
        Result.Callstatus := info.StatusCode;
        Result.Success := info.StatusCode in [Ord(rrOK),ord(rrCreated),ord(rrNoContent)];
        Result.Response := TJSONObject.ParseJSONValue(info.StatusMessage);
        If Assigned(OnLog) Then
          FOnlog(info.StatusCode, ConAmazon.StorageEndpoint + #13#10 + bucket + #13#10 + accessKeyID + #13#10 + secretAccessKey, info.StatusMessage, '');
      finally
        info.Free;
        Service.Free;
      end;
    finally
      ConAmazon.Free;
    end;
  finally
    upload_stream.Free;
  end;
end;

Answer 1

I have modified slightly your code to use a TBytesStream instead of a TFileStream to set the file content on a TBytes buffer. But mostly you have to read the stream, not write it.

Now it doesn't corrupt its content.

Function Amazon_Upload (fileName, bucket, accessKeyID, secretAccessKey: String) : TCallResult;
var
  Service: TAmazonStorageService;
  ConAmazon: TAmazonConnectionInfo;
  info : TCloudResponseInfo;
  upload_stream : TBytesStream;
  bytes : TBytes;
begin
  // create file
  upload_stream := TBytesStream.Create;
  upload_stream.LoadFromFile(filename);
  try
    // filestream to tBytes
    upload_stream.Position := 0;
    SetLength(bytes, upload_stream.Size);
    upload_stream.ReadBuffer(bytes, upload_stream.Size);
    ConAmazon := TAmazonConnectionInfo.Create(nil);
    try
      // amazon connection parameters
      ConAmazon.AccountKey := secretAccessKey;
      ConAmazon.AccountName := accessKeyID;
      ConAmazon.QueueEndpoint := 'queue.amazonaws.com';
      ConAmazon.StorageEndpoint := 's3-eu-central-1.amazonaws.com';
      ConAmazon.TableEndpoint := 'sdb.amazonaws.com';
      ConAmazon.UseDefaultEndpoints := False;
      // storage objects
      info := TCloudResponseInfo.Create;
      Service := TAmazonStorageService.Create(ConAmazon);
      try
        // upload document to storage
        Service.UploadObject(Bucket, ExtractFileName(fileName), bytes, TRUE, nil, nil, amzbaPrivate, info);
        // get results
        Result.Callstatus := info.StatusCode;
        Result.Success := info.StatusCode in [Ord(rrOK),ord(rrCreated),ord(rrNoContent)];
        Result.Response := TJSONObject.ParseJSONValue(info.StatusMessage);
        If Assigned(OnLog) Then
          FOnlog(info.StatusCode, ConAmazon.StorageEndpoint + #13#10 + bucket + #13#10 + accessKeyID + #13#10 + secretAccessKey, info.StatusMessage, '');
      finally
        info.Free;
        Service.Free;
      end;
    finally
      ConAmazon.Free;
    end;
  finally
    upload_stream.Free;
  end;
end;

Answer 2

Running your code I have found the problem. Just change this line :

Service.UploadObject(Bucket, fileName, bytes, TRUE, nil, nil, amzbaPrivate, info);

By this line :

Service.UploadObject(Bucket, ExtractFileName(fileName), bytes, TRUE, nil, nil, amzbaPrivate, info);

The local path of your file must not be present on the remote call (I guess it confuses the server, thinking that you are trying to reach a resource you have not permissions to).