For a project in which we created an app that records certain scores throughout the day, we also created some graphs in R which we saved as jpegs on the Raspberry.
We want to upload the jpg to Firebase via Python (we uploaded a variable to Firebase and it worked)
We first tried this code:
from google.cloud import storage
client = storage.Client()
bucket = client.get_bucket('teddy-aztech-ehealth.appspot.com')
graphicBlob = bucket.get_blob('graph.jpeg')
graphBlob.upload_from_filename(filename='/home/pi/graph.jpeg')
But we get a long error from the client bucket part , telling us the bucket name must start and end with a number.
We also tried this code:
import sys
import requests
import firebase_admin
from firebase_admin import credentials
from firebase_admin import storage
sys.argv = "/home/pi/graph.jpeg"
image_url = sys.argv
cred = credentials.Certificate('teddy-aztech-ehealth-firebase-adminsdk-t0iz1-61f49237f4.json')
firebase_admin.initialize_app(cred, {
'storageBucket': 'https://teddy-aztech-ehealth.appspot.com'
})
bucket = storage.bucket()
image_data = requests.get(image_url).content
blob = bucket.blob('graph.jpg')
blob.upload_from_string(
image_data,
content_type='image/jpg'
)
print(blob.public_url)
But get an error at the part with initializeapp (again, because of the bucket...) Do we have to activate/give access from Firebase?
Your initial attempt is close to what you need.
import io
from google.cloud import storage
# Google Cloud Project ID. This can be found on the 'Overview' page at
# https://console.developers.google.com
PROJECT_ID = 'your-project-id'
CLOUD_STORAGE_BUCKET = 'your-bucket-name'
filename = "graph-filename.jpeg"
# Create unique filename to avoid name collisions in Google Cloud Storage
date = datetime.datetime.utcnow().strftime("%Y-%m-%d-%H%M%S")
basename, extension = filename.rsplit('.', 1)
unique_filename = "{0}-{1}.{2}".format(basename, date, extension)
# Instantiate a client on behalf of the project
client = storage.Client(project=PROJECT_ID)
# Instantiate a bucket
bucket = client.bucket(CLOUD_STORAGE_BUCKET)
# Instantiate a blob
blob = bucket.blob(unique_filename)
# Upload the file
with open(filename, "rb") as fp:
blob.upload_from_file(fp)
# The public URL for this blob
url = blob.public_url
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