We currently use blobstore.create_upload_url to create upload urls to be used on the frontend see Uploading a blob.
However, with the push toward Google Cloud Storage (GCS) by Google, I'd like to use GCS instead of the blobstore. We use currently blobstore.create_upload_url but I can't find anything equivalent in the GCS documentation. Am I missing something? Is there a better way to upload files to GCS from the frontend?
Thanks
Rob
If you will provide gs_bucket_name to blobstore.create_upload_url file will be stored in GCS instead of blobstore, this is described in official documentation: Using the Blobstore API with Google Cloud Storage
blobstore.create_upload_url(
success_path=webapp2.uri_for('upload'),
gs_bucket_name="mybucket/dest/location")
You can take a look at simple upload handler implementation made in webapp2
from google.appengine.ext import blobstore
from google.appengine.ext.webapp import blobstore_handlers
import webapp2
import cloudstorage as gcs
class Upload(blobstore_handlers.BlobstoreUploadHandler):
"""Upload handler
To upload new file you need to follow those steps:
1. send GET request to /upload to retrieve upload session URL
2. send POST request to URL retrieved in step 1
"""
def post(self):
"""Copy uploaded files to provided bucket destination"""
fileinfo = self.get_file_infos()[0]
uploadpath = fileinfo.gs_object_name[3:]
stat = gcs.stat(uploadpath)
# remove auto generated filename from upload path
destpath = "/".join(stat.filename.split("/")[:-1])
# copy file to desired location with proper filename
gcs.copy2(uploadpath, destpath)
# remove file from uploadpath
gcs.delete(uploadpath)
def get(self):
"""Returns URL to open upload session"""
self.response.write(blobstore.create_upload_url(
success_path=uri_for('upload'),
gs_bucket_name="mybucket/subdir/subdir2/filename.ext"))
Related
I'm building an API that uploads images to Firebase storage, everything works as expected in that regard, the problem is that the syntax makes me specify the file name in each upload, and in production mode the API will receive upload requests from multiple devices, so I need to make to code so it checks for an available id, set it for the "blob()" object, and then do a normal upload, but I have no idea how to do that. or a random name I don't care as long as it doesn't overwrite another picture
Here is my current code:
from flask_pymongo import PyMongo
import firebase_admin
from firebase_admin import credentials, auth, storage, firestore
import os
import io
cred = credentials.Certificate('service_account_key.json')
firebase_admin.initialize_app(cred, {'storageBucket': 'MY-DATABASE-NAME.appspot.com'})
bucket = storage.bucket()
blob = bucket.blob("images/newimage.png") #here is where im guessing i #should put the next available name
# "apple.png" is a sample image #for testing in my directory
with open("apple.png", "rb") as f:
blob.upload_from_file(f)
As "Klaus D."'s comment said the solution was to implement the "uuid" module
import uuid
.....
.....
blob = bucket.blob("images/" + str(uuid.uuid4()))
On a regular (non-flexible) instance of Google App Engine, you can use the Blobstore API and create a URL to allow a user to upload a file directly into your Blobstore. When it is uploaded, your app engine application is notified of the location of the file and can process it. An example of the python code is:
from google.appengine.ext import blobstore
upload_url = blobstore.create_upload_url('/upload_photo')
See the Blobstore docs.
Switching to Google App Engine Flexible Environment, usage of the Blobstore has been largely replaced by Cloud Storage. In such a case, is there an equivalent of create_upload_url?
My current implementation takes a standard file upload to a python Flask application. Then proceeds with something like:
from flask import request
from google.cloud import storage
uploaded_file = request.files.get('file')
gcs = storage.Client()
bucket = gcs.get_bucket(bucket_name)
blob = bucket.blob(blob_name)
blob.upload_from_string(
uploaded_file.read(),
content_type=uploaded_file.content_type
)
This seems like it is doubling the network load compared with create_upload_url because the file is coming into my app engine instance and then immediately being copied out. So the uploader will be made to wait extra time whilst this is happening. Presumably I will also incur extra App Engine charges for this. Is there a better way?
I have workers that later process the uploaded file, but I tend to download the file from Cloud Storage again in their code because I don't think you can assume that the worker will still have access to a file stored in the instance file system. Therefore I don't get any benefit of having the file uploaded to my instance rather than direct to it's storage location.
I have started using create_resumable_upload_session to create a signed URL that our client side application can upload a file to. Something like:
gcs = storage.Client()
bucket = gcs.get_bucket(BUCKET)
blob = bucket.blob(blob_name)
signed_url = blob.create_resumable_upload_session(content_type=content_type)
Then when the client has successfully uploaded a file to our storage, I subscribe to a Pub/Sub notification of the creation using this Cloud Pub/Sub Notifications for Cloud Storage.
Each blob created with the new Google Cloud Storage Client has a public_url property:
from flask import request
from google.cloud import storage
uploaded_file = request.files.get('file')
gcs = storage.Client()
bucket = gcs.get_bucket(bucket_name)
blob = bucket.blob('blob_name')
blob.upload_from_string(
uploaded_file.read(),
content_type=uploaded_file.content_type
)
url = blob.public_url
--
With the Blobstore, a GAE system handler in your instance takes care of the uploaded file you pass to the upload url created. I'm not sure if it's an issue handling it yourself in your code. If your current approach is problematic, you might want to consider doing the upload client side and not pass the file through App Engine at all. GCS has a REST API and the cloud storage client uses it underneath, so you can read and upload the file directly to GCS on the client side if it's more convenient. There's firebase.google.com/docs/storage/web/upload-files to ease you through the process
I have a server application written in python/django (REST api) for accepting a file upload from the client application. I want this uploaded file to be stored in AWS S3. I also want the file to be uploaded from client as multipart form / data . How can i achieve this. Any sample code application will help me to understand the way it should be done. Please assist.
class FileUploadView(APIView):
parser_classes = (FileUploadParser,)
def put(self, request, filename, format=None):
file_obj = request.data['file']
self.handle_uploaded_file(file_obj)
return self.get_response("", True, "", {})
def handle_uploaded_file(self, f):
destination = open('<path>', 'wb+')
for chunk in f.chunks():
destination.write(chunk)
destination.close()
Thanks in advance
If you want to your uploads to go directly to AWS S3, you can use django-storages and set your Django file storage backend to use AWS S3.
django-storages
django-storages documentation
This will allow your Django project to handle storage transparently to S3 without your having to manually re-upload your uploaded files to S3.
Storage Settings
You will need to add at least these configurations to your Django settings:
# default remote file storage
DEFAULT_FILE_STORAGE = 'storages.backends.s3boto.S3BotoStorage'
# aws access keys
AWS_ACCESS_KEY_ID = 'YOUR-ACCESS-KEY'
AWS_SECRET_ACCESS_KEY = 'YOUR-SECRET-ACCESS-KEY'
AWS_BUCKET_NAME = 'your-bucket-name'
AWS_STORAGE_BUCKET_NAME = AWS_BUCKET_NAME
Example Code to Store Upload to Remote Storage
This is a modified version of your view with a the handle_uploaded_file method using Django's storage backend to save the uploade file to the remote destination (using django-storages).
Note: Be sure to define the DEFAULT_FILE_STORAGE and AWS keys in your settings so django-storage can access your bucket.
from django.core.files.storage import default_storage
from django.core.files import File
# set file i/o chunk size to maximize throughput
FILE_IO_CHUNK_SIZE = 128 * 2**10
class FileUploadView(APIView):
parser_classes = (FileUploadParser,)
def put(self, request, filename, format=None):
file_obj = request.data['file']
self.handle_uploaded_file(file_obj)
return self.get_response("", True, "", {})
def handle_uploaded_file(self, f):
"""
Write uploaded file to destination using default storage.
"""
# set storage object to use Django's default storage
storage = default_storage
# set the relative path inside your bucket where you want the upload
# to end up
fkey = 'sub-path-in-your-bucket-to-store-the-file'
# determine mime type -- you may want to parse the upload header
# to find out the exact MIME type of the upload file.
content_type = 'image/jpeg'
# write file to remote server
# * "file" is a File storage object that will use your
# storage backend (in this case, remote storage to AWS S3)
# * "media" is a File object created with your upload file
file = storage.open(fkey, 'w')
storage.headers.update({"Content-Type": content_type})
f = open(path, 'rb')
media = File(f)
for chunk in media.chunks(chunk_size=FILE_IO_CHUNK_SIZE):
file.write(chunk)
file.close()
media.close()
f.close()
See more explanation and examples on how to access the remote storage here:
django-storages: Amazon S3
Take a look at boto package which provides AWS APIs:
from boto.s3.connection import S3Connection
s3 = S3Connection(access_key, secret_key)
b = s3.get_bucket('<bucket>')
mp = b.initiate_multipart_upload('<object>')
for i in range(1, <parts>+1):
io = <receive-image-part> # E.g. StringIO
mp.upload_part_from_file(io, part_num=i)
mp.complete_upload()
All the examples I can find on google endpoint api (e.g., tic-tac-toe sample) show strings, integers, enums, etc fields. None of the examples say anything about how to specify document (e.g., image or zip files) uploads or downloads using the API. Is this not possible?
If this is possible, can anyone share a code snippet on how to define google endpoint api on the server to allow downloads and uploads of files? For example, is there a way to set HTTPResponse headers to specify that an endpoint response will serve a zip file? How do we include the zip file in the response?
An example with python or php would be appreciated. If anyone from the endpoints-proto-datastore team is watching this discussion, please say whether or not file downloads are supported in endpoints at the moment. We hate to waste our time trying to figure this out if it is simply impossible. Thanks.
We are seeking a complete example for upload and download. We need to store the key for the uploaded file in our database during upload and retrieve it for download. The client app sends a token that the API needs to use to figure out what file to download. Hence, we would need to store the blob key generated during the upload process in our database. Our database would have the mapping between the token and the blob file's key.
class BlobDataFile(models.Model):
data_code = models.CharField(max_length=10) # Key used by client app to request file
blob_key = models.CharField()
By the way, our app is written in Django 1.7 with a mysql (modeled with models.Model) database. It is infuriating that all the examples for Google App Engine upload I can find is written for a standalone webapp Handlers (no urls.py/views.py solutions could be found anywhere). Hence, building a standalone uploader is as much of a challenge as writing the API code. If your solution has full urls.py/views.py example for uploading files and saving the blob_key in our BlobDataFile, it would be good enough for us.
f you use the blobstore use the get_serving_url function to read the images from url in the client, or use the messages.ByteField in the ResourceContainer and serialize the image with base64.b64decode
#the returned class
class Img(messages.Message):
message = messages.BytesField (1)
#The api class
#endpoints.api(name='helloImg', version='v1')
class HelloImgApi(remote.Service):
ID_RESOURCE = endpoints.ResourceContainer(
message_types.VoidMessage,
id=messages.StringField(1, variant=messages.Variant.STRING))
#endpoints.method(ID_RESOURCE, Img,
path='serveimage/{id}', http_method='GET', #ID is the blobstore key
name='greetings.getImage')
def image_get(self, request):
try:
blob_reader = blobstore.BlobReader(blob_key)
value = blob_reader.read()
return Img(message=value)
except:
raise endpoints.NotFoundException('image %s not found.' %
(request.id,))
APPLICATION = endpoints.api_server([HelloImgApi])
And this is the response (save it in the client with the proper format)
{
"message": 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}
in the client you can do this (in python for continuity)
import base64
myFile = open("mock.jpg", "wb")
img = base64.b64decode(value) #value is the returned string
myFile.write(img)
myFile.close()
Did you try converting the image to base64 string and send it as an argument of your request on the client side?
So you will be able to do that on the server side :
#strArg is the Base64 string sent from the client
img = base64.b64decode(strArg)
filename = 'someFileName.jpg'
with open(filename, 'wb') as f:
f.write(img)
#then you can save the file to your BlobStore
I am trying to get a simple image upload app working on Heroku using Flask. I'm following the tutorial here: http://flask.pocoo.org/docs/patterns/fileuploads/
However, I want to use S3 to store the file instead of a temporary directory, since Heroku does not let you write to disk. I cannot find any examples of how to do this specifically for Heroku and Flask.
It seems to me that in the example code that stores the uploaded file to a temporary file, you would just replace file.save(os.path.join(app.config['UPLOAD_FOLDER'], filename)) with code that uploads the file to S3 instead.
For example, from the linked page:
def upload_file():
if request.method == 'POST':
file = request.files['file']
if file and allowed_file(file.filename):
filename = secure_filename(file.filename)
s3 = boto.connect_s3()
bucket = s3.create_bucket('my_bucket')
key = bucket.new_key(filename)
key.set_contents_from_file(file, headers=None, replace=True, cb=None, num_cb=10, policy=None, md5=None)
return 'successful upload'
return ..
Or if you want to upload to S3 asynchrnously, you could use whatever queuing mechanism is provided by Heroku.
A bit of an old question, but I think ever since Amazon introduced CORS support to S3, the best approach is to upload directly to S3 from the user's browser - without the bits ever touching your server.
This is a very simple flask project that shows exactly how to do that.
Using boto library it will look something like this:
import boto
from boto.s3.connection import S3Connection
from boto.s3.key import Key
def upload_file():
if request.method == 'POST':
file = request.files['file']
if file and allowed_file(file.filename):
filename = secure_filename(file.filename)
conn = S3Connection('credentials', '')
bucket = conn.create_bucket('bucketname')
k = Key(bucket)
k.key = 'foobar'
k.set_contents_from_string(file.readlines())
return "Success!"
Instead of storing the file on the disk directly, you could also store its data in the database (base64 encoded for example).
Anyway, to interact with Amazon S3 using Python, you should consider using the boto library (the same is true for any other Amazon service).
To know how to use it, you could have a lookat the related documentation.
I'm working on something similar for a website I'm developing now. Users will be uploading very large files. I'm looking at using Plupload to upload directly to S3 following the advice here.
An alternative is to use the direct-to-S3 uploader in Boto.