In my flask application, I am using a function to upload file to Amazon s3, using Boto.
Its working fine most of the cases, but some times its uploading files as zero byte file with no extension.
Why its failing sometimes,
I am validating user image file in form.
FileField('Your photo',validators=[FileAllowed(['jpg', 'png'], 'Images only!')])
My image upload function.
def upload_image_to_s3(image_from_form):
#upload pic to amazon
source_file_name_photo = secure_filename(image_from_form.filename)
source_extension = os.path.splitext(source_file_name_photo)[1]
destination_file_name_photo = uuid4().hex + source_extension
s3_file_name = destination_file_name_photo
# Connect to S3 and upload file.
conn = boto.connect_s3('ASJHjgjkhSDJJHKJKLSDH','GKLJHASDJGFAKSJDGJHASDKJKJHbbvhjcKJHSD')
b = conn.get_bucket('mybucket')
# Connect to S3 and upload file.
sml = b.new_key("/".join(["myfolder",destination_file_name_photo]))
sml.set_contents_from_string(image_from_form.read())
acl='public-read'
sml.set_acl(acl)
return s3_file_name
How large are your assets? If there is too large of an upload, you may have to multipart/chunk it otherwise it will timeout.
bucketObject.initiate_multipart_upload('/local/object/as/file.ext')
it means you will not be using set_contents_from_string but rather store and upload. You may have to use something to chuck the file, like FileChuckIO.
An example is here if this applies to you : http://www.bogotobogo.com/DevOps/AWS/aws_S3_uploading_large_file.php
Also, you may want to edit your post above and alter your AWS keys.
Related
I was trying to open a file/image in python/django and upload it to s3 but I get different errors depending on what I try. I can get it to work when I send the image using the front end html form but not when opening the file on the back end. I get errors such as "'bytes' object has no attribute 'file'" Any ideas how to open an image and upload it to s3? I wasn't sure if I was using the correct upload function, but it worked when I received the file from an html form instead of opening it directly.
image = open(fileURL, encoding="utf-8")
S3_BUCKET = settings.AWS_BUCKET
session = boto3.Session(
aws_access_key_id=settings.AWS_ACCESS_KEY_ID,
aws_secret_access_key=settings.AWS_SECRET_ACCESS_KEY,
)
s3 = session.resource('s3')
s3.Bucket(S3_BUCKET).put_object(Key='folder/%s' % fileName, Body=image)
Thanks.
The open command return a file object. Therefore Body=image does not contain the actual contents of the object.
Since you want to upload an existing object, you could use:
Key = 'folder/' + fileName
s3.Object(S3_BUCKET, Key).upload_file(fileURL)
My code is to add watermark on video and upload that video directly to s3.
I am able to do add the watermark on video. Upload that video Directly to s3 need to be done.
import moviepy.editor as mp
import boto3
AWS_ACCESS_KEY_ID="aws_key_id"
AWS_SECRET_ACCESS_KEY="aws_secret_access_key"
s3_resource = boto3.resource("s3")
BUCKET = "bucket_name"
video = mp.VideoFileClip('path_of_ video_file_stored')
logo = (mp.ImageClip('logo_file_stored')
.set_duration(video.duration))
final = mp.CompositeVideoClip([video,logo])
final.subclip(0).write_videofile('localpath/filename.mp4')
Instead of writing to local from final.subclip(0).write_videofile(localpath/filename.mp4)
I need to write file directly on s3 instead of first writing on local and then uploading to s3. May I know how to write to s3 directly with code solution of above.
I have a project in hand to backup a website for some reasons. I use Requests in Python to crawl the contents and images (urls). The problem is, how can I save the image in the cloud, by using the url of that image, in Cloud services (Google Drive, Dropbox, etc.).
I know there is a way to first save the image locally and then upload the local image to the cloud. But I'm wondering if there are APIs that support uploading images by urls, not the local file.
It seems like Dropbox has a feature called /save_url that
lets app developers upload files to Dropbox by just providing a URL, without having to download the file first.
https://www.dropbox.com/developers-v1/core/docs#save-url
If you don't mind paying for the storage, you can save it to your own cloud storage. I occasionally have to do a similar action, and handle it as such:
def on_upload_image(self):
url = self.request.get('url')
result = urlfetch.fetch(url)
binary = result.content
blob_key = functions.get_blob_key_by_data(binary)
self.url = images.get_serving_url(blob_key, secure_url=True)
self.json()
from google.appengine.api import app_identity
def get_blob_key_by_data(data):
bucket = app_identity.get_default_gcs_bucket_name()
filename = hashlib.sha256(data).hexdigest()
mime_type = get_mime_type(data)
if not mime_type:
return None
gcs_filename = '/%s/image_%s' % (bucket, filename)
with gcs.open(gcs_filename, 'w', content_type=mime_type) as f:
f.write(data)
blob_key = blobstore.create_gs_key("/gs" + gcs_filename)
return blob_key
I'm attempting to save an image to S3 using boto. It does save a file, but it doesn't appear to save it correctly. If I try to open the file in S3, it just shows a broken image icon. Here's the code I'm using:
# Get and verify the file
file = request.FILES['file']
try:
img = Image.open(file)
except:
return api.error(400)
# Determine a filename
filename = file.name
# Upload to AWS and register
s3 = boto.connect_s3(aws_access_key_id=settings.AWS_KEY_ID,
aws_secret_access_key=settings.AWS_SECRET_ACCESS_KEY)
bucket = s3.get_bucket(settings.AWS_BUCKET)
f = bucket.new_key(filename)
f.set_contents_from_file(file)
I've also tried replacing the last line with:
f.set_contents_from_string(file.read())
But that didn't work either. Is there something obvious that I'm missing here? I'm aware django-storages has a boto backend, but because of complexity with this model, I do not want to use forms with django-storages.
Incase you don't want to go for django-storages and just want to upload few files to s3 rather then all the files then below is the code:
import boto3
file = request.FILES['upload']
s3 = boto3.resource('s3', aws_access_key_id=settings.AWS_ACCESS_KEY, aws_secret_access_key=settings.AWS_SECRET_ACCESS_KEY)
bucket = s3.Bucket('bucket-name')
bucket.put_object(Key=filename, Body=file)
You should use django-storages which uses boto internally.
You can either swap the default FileSystemStorage, or create a new storage instance and manually save files. Based on your code example I guess you really want to go with the first option.
Please consider using django's Form instead of directly accessing the request.
I want to store files and images that I get from an api in the blobstore (or rather so that they are accessible from the blobstore api). Since the file-api is deprecated, how do I do this?
One way is to store images in Cloud Storage (gcs) and access them via the blogstore api. Basically you call gcs.open() and write the file. Then when you need to use the blobstore api you call blobkey = blobstore.create_gs_key(). With that you can do things such as use the images api with calls like images.get_serving_url(blobkey, secure_url=False).
How you do that depends on what you're particular goals are. I am using to serve images in a gallery that I upload. To do that I have an file upload on an html form on the front end, which sends the file. On the backend I am doing this (these are just the broad strokes):
# inside the webapp2.RequestHandler get method:
import mimetypes
file_data = self.request.get("photoUpload", default_value = None)
filename = self.request.POST["photoUpload"].filename
folder = "someFolderName"
content_type = mimetypes.guess_type(self.filename)[0]
Then save the file data to GCS:
from google.appengine.api import app_identity
import cloudstorage as gcs
# gcs_filename must be unique so I'm using bucket/folder/file
# it would be smart to check uniqueness before proceeding
gcs_filename = '/%s%s/%s' % (bucket or app_identity.get_default_gcs_bucket_name(), folder, filename)
gcs.open(gcs_filename, 'w', content_type=content_type or b'binary/octet-stream', options={b'x-goog-acl': b'public-read'}) as f:
f.write(file_data)
Now I can access using the GCS api with calls like:
gcs.delete(gcs_filename)
Or use the Blobstore API by getting the previously mentioned blocky:
blobkey = blobstore.create_gs_key()