Flask 将图像上传到 S3 而不将其保存到本地文件系统

Question

我需要将用户提交的照片上传到 s3 存储桶。但是我不断收到以下错误：

TypeError: expected str, bytes or os.PathLike object, not FileStorage

我如何能够将文件存储为字符串/字节而不是 FileStorage？相关代码如下：

@user_api.route('upload-profile-photo', methods=['PUT'])
@Auth.auth_required
def upload_profile_photo():
  """
  Upload User Profile Photo
  """
  key = Auth.auth_user()
  bucket = 'profile-photos'
  content_type = request.mimetype
  image_file = request.files['file']
  client = boto3.client('s3',
                        region_name='sfo2',
                        endpoint_url='https://example.xxx.amazonaws.com',
                        aws_access_key_id=os.environ['ACCESS_KEY'],
                        aws_secret_access_key=os.environ['SECRET_KEY'])
  with open(image_file, "rb") as f:
    client.upload_fileobj(
        bucket,
        f,
        key,
        ExtraArgs={'ACL': 'public-read', 'ContentType': content_type}
    )

  return custom_response({'message': 'image uploaded'}, 200)

Answer 1

iTo 使用 a 实现这一点FileStorage，我使用以下方法put_object()：

from werkzeug import secure_filename

@user_api.route('upload-profile-photo', methods=['PUT'])
@Auth.auth_required
def upload_profile_photo():
    """
    Upload User Profile Photo
    """
    key = Auth.auth_user()
    bucket = 'profile-photos'
    content_type = request.mimetype
    image_file = request.files['file']

    client = boto3.client('s3',
                          region_name='sfo2',
                          endpoint_url='https://example.xxx.amazonaws.com',
                          aws_access_key_id=os.environ['ACCESS_KEY'],
                          aws_secret_access_key=os.environ['SECRET_KEY'])

    filename = secure_filename(image_file.filename)  # This is convenient to validate your filename, otherwise just use file.filename

    client.put_object(Body=image_file,
                      Bucket=bucket,
                      Key=filename,
                      ContentType=content_type)

    return custom_response({'message': 'image uploaded'}, 200)

请注意，对 的调用secure_filename()是可选的（您可以简单地传递image_file.filename），但可以非常方便地验证文件名。否则，添加一些异常处理会很好，但粗略的想法在这里：不需要open()文件（需要存储在本地）。

我鼓励查看这里的文档，以了解与upload_fileobj()