How copy file automatically bewteen 2 buckets with two different projects gcp? - google-cloud-platform

Actually i use that command , and it works well :
gsutil cp gs:/bucket1/file.xml gs://bucket2/destination_folder
(bucket1 is in project1 in GCP and bucket2 is in another project in GCP)
But i would like to do that command every day at 9am, how can i do that on my GCP project in a easy way ?
Edit : It will copy the file over and over each day from the source bucket to the destination bucket( the two buckets are in a different project each). (actually when the file arrive in the destination bucket, it is consume and ingest in bigquery automatically , i just want to trigg my command gsutil and stop to do it manually each morning )
(except the method with Data transfert because i have not the right of the source project so i cannot activate the service account for data transfert , i have only the rights on destination project.)
Bests regards,
Actually i can copy a file from a bucket into another bucket into a specfic folder (RQ : the 2 buckets are on the same gcp project)
I don't arrive to use the second method with a gs://
EDIT 2:
import base64
import sys
import urllib.parse
# Imports the Google Cloud client library , dont forget the requirement or else it's ko
from google.cloud import storage
def copy_blob(
bucket_name ="prod-data", blob_name="test.csv", destination_bucket_name = "prod-data-f", destination_blob_name ="channel_p"
):
"""Copies a blob from one bucket to another with a new name."""
bucket_name = "prod-data"
blob_name = "test.csv"
destination_bucket_name = "prod-data-f"
destination_blob_name = "channel_p/test.csv"
storage_client = storage.Client()
source_bucket = storage_client.bucket(bucket_name)
source_blob = source_bucket.blob("huhu/"+blob_name)
destination_bucket = storage_client.bucket(destination_bucket_name)
blob_copy = source_bucket.copy_blob(
source_blob, destination_bucket, destination_blob_name
)
# Second Method (KO)
#
# client = storage.Client()
# with open('gs://prod-data-f/channelp.xml','wb') as file_obj:
# client.download_blob_to_file(
# 'gs://pathsource/somefolder/channelp.xml', file_obj)
#
# End of second Method
print(
"Blob {} in bucket {} copied to blob {} in bucket {}.".format(
source_blob.name,
source_bucket.name,
blob_copy.name,
destination_bucket.name,
)
)

Data transfer is obviously the right tool for doing this, but since you cannot use it, there are alternative solutions.
One of them is to copy files using a Cloud Function (you can use this snippet), and trigger each day at 9am that Cloud Function using Cloud Scheduler. Cloud Function can also be triggered by a Pub/Sub message.

The solution that i was seeking (it works for me when i test):
Main.py
import base64
import os
import sys
import json
import uuid
import logging
from time import sleep
from flask import request
from random import uniform
from google.cloud import firestore
from google.cloud.exceptions import Forbidden, NotFound
from google.cloud import storage
# set retry deadline to 60s
DEFAULT_RETRY = storage.retry.DEFAULT_RETRY.with_deadline(60)
def Move2FinalBucket(data, context):
# if 'data' in event:
# name = base64.b64decode(event['data']).decode('utf-8')
# else:
# name = 'NO_DATA'
# print('Message {}!'.format(name))
# Get cache source bucket
cache_bucket = storage.Client().get_bucket('nameofmysourcebucket', timeout=540, retry=DEFAULT_RETRY)
# Get source file to copy
blob2transfer = cache_bucket.blob('uu/oo/pp/filename.csv')
# Get cache destination bucket
destination_bucket = storage.Client().get_bucket('nameofmydestinationbucket', timeout=540, retry=DEFAULT_RETRY)
# Get destination file
new_file = destination_bucket.blob('kk/filename.csv')
#rewrite into new_file
new_file.rewrite(blob2transfer, timeout=540, retry=DEFAULT_RETRY)
requirement.txt
# Function dependencies, for example:
# package>=version
#google-cloud-storage==1.22.0
google-cloud-storage
google-cloud-firestore
google-api-core
flask==1.1.4
Dont forget to add a service account with the right Storage admin on this CF and it will works.
Best regards,

Related

google cloud create Http function to list of objects from bucket in python

from google.cloud import storage
def list_blobs(bucket_name):
"""Lists all the blobs in the bucket."""
# bucket_name = "your-bucket-name"
storage_client = storage.Client()
When I am creating cloud function using http. Its not deploying.
# Note: Client.list_blobs requires at least package version 1.17.0.
blobs = storage_client.list_blobs(bucket_name)
for blob in blobs:
print(blob.name)
You need to put the google-cloud-storage in the requirements.txt and check if a function with the same name exists as the entry point. The main file need to have a function with the same name as the entry point

Download folder with Google Cloud Storage Api [duplicate]

At the following page
https://googlecloudplatform.github.io/google-cloud-python/latest/storage/blobs.html
there are all the API calls which can be used for Python & Google Cloud storage. Even in the "official" samples on github
https://github.com/GoogleCloudPlatform/python-docs-samples/blob/master/storage/cloud-client/snippets.py
don't have a related example.
Finally, downloading a directory with the same method used for download files gives the error
Error: [Errno 21] Is a directory:
You just have to first list all the files in a directory and then download them one by one:
bucket_name = 'your-bucket-name'
prefix = 'your-bucket-directory/'
dl_dir = 'your-local-directory/'
storage_client = storage.Client()
bucket = storage_client.get_bucket(bucket_name=bucket_name)
blobs = bucket.list_blobs(prefix=prefix) # Get list of files
for blob in blobs:
filename = blob.name.replace('/', '_')
blob.download_to_filename(dl_dir + filename) # Download
blob.name includes the entire directory structure + filename, so if you want the same file name as in the bucket, you might want to extract it first (instead of replacing / with _)
If you want to keep the same directory structure without renaming and also create nested folders. I have for python 3.5+ a solution based on #ksbg answer :
from pathlib import Path
bucket_name = 'your-bucket-name'
prefix = 'your-bucket-directory/'
dl_dir = 'your-local-directory/'
storage_client = storage.Client()
bucket = storage_client.get_bucket(bucket_name=bucket_name)
blobs = bucket.list_blobs(prefix=prefix) # Get list of files
for blob in blobs:
if blob.name.endswith("/"):
continue
file_split = blob.name.split("/")
directory = "/".join(file_split[0:-1])
Path(directory).mkdir(parents=True, exist_ok=True)
blob.download_to_filename(blob.name)
Lets say, we want to download FINALFOLDER from the storage path: gs://TEST_BUCKET_NAME/FOLDER1/FOLDER2/FINALFOLDER
After downloading, the final path will look like: D:\\my_blob_data\FINALFOLDER
from os import makedirs
from os.path import join, isdir, isfile, basename
from google.cloud import storage
# if your environment was authenticated, the default config will be picked up
storage_client = storage.Client() # comment this line if you want to use service account
# uncomment the line below if you have a service account json
# storage_client = storage.Client.from_service_account_json('creds/sa.json')
bucket_name = 'TEST_BUCKET_NAME'
prefix = 'FOLDER2'
dst_path = 'D:\\my_blob_data'
if isdir(dstPath) == False:
makedirs(dstPath)
bucket = storage_client.bucket(bucket_name=bucket_name)
blobs = bucket.list_blobs(prefix=prefix) # Get list of files
for blob in blobs:
blob_name = blob.name
dst_file_name = blob_name.replace('FOLDER1/FOLDER2', dst_path) #.replace('FOLDER1/FOLDER2', 'D:\\my_blob_data')
# extract the final directory and create it in the destination path if it does not exist
dst_dir = dst_file_name.replace('/' + basename(dst_file_name), '')
if isdir(dst_dir) == False:
makedirs(dst_dir)
# download the blob object
blob.download_to_filename(dst_file_name)
Using tensoflow gfile package, here is a recursive function.
root_dir is the GCS parent folder.
local_base_dir is the parent folder created at local
def copy_recursively(root_dir, local_base_dir):
if tf.io.gfile.exists(local_base_dir):
tf.io.gfile.rmtree(local_base_dir)
tf.io.gfile.mkdir(local_base_dir)
file_list = tf.io.gfile.glob(root_dir+'/**')
for item in file_list:
if not tf.io.gfile.isdir(item):
fname = item.rsplit('/',1)[-1]
if not fname.startswith('.'):
tf.io.gfile.copy(item,
os.path.join(local_base_dir,fname),
overwrite=False)
else:
child_dir= item.rsplit('/',1)[-1]
full_dir_path = os.path.join(local_base_dir,child_dir)
print(f"Setting up child directory: {full_dir_path}")
copy_recursively(item,full_dir_path)
root_dir = 'gs://.../.../..'
local_base_dir = root_dir.rsplit('/',1)[-1]
copy_recursively(root_dir, local_base_dir)
Refer This Link- https://medium.com/#sandeepsinh/multiple-file-download-form-google-cloud-storage-using-python-and-gcs-api-1dbcab23c44
1 - Add Your Credential Json
2 - List Bucket Items
3 - Download
import logging
import os
from google.cloud import storage
global table_id
global bucket_name
logging.basicConfig(format=’%(levelname)s:%(message)s’, level=logging.DEBUG)
bucket_name = ‘mybucket’
table_id = ‘shakespeare’
storage_client = storage.Client.from_service_account_json(‘/google-cloud/keyfile/service_account.json’)
# The “folder” where the files you want to download are
folder=’/google-cloud/download/{}’.format(table_id)
delimiter=’/’
bucket=storage_client.get_bucket(bucket_name)
blobs=bucket.list_blobs(prefix=table_id, delimiter=delimiter) #List all objects that satisfy the filter.
# Download the file to a destination
def download_to_local():
logging.info(‘File download Started…. Wait for the job to complete.’)
# Create this folder locally if not exists
if not os.path.exists(folder):
os.makedirs(folder)
# Iterating through for loop one by one using API call
for blob in blobs:
logging.info(‘Blobs: {}’.format(blob.name))
destination_uri = ‘{}/{}’.format(folder, blob.name)
blob.download_to_filename(destination_uri)
logging.info(‘Exported {} to {}’.format(
blob.name, destination_uri))
if __name__ == ‘__main__’:
download_to_local()

Creating a dmarc parser using parsedmarc in python3 for use in AWS s3

I am very new to programming. I am working on a pipeline to analyze DMARC report files that are sent to my email account, that I am manually placing in an s3 bucket. The goal of this task is to download, extract, and analyze files using parsedmarc: https://github.com/domainaware/parsedmarc The part I'm having difficulty with is setting a conditional statement to extract .gz files if the target file is not a .zip file. I'm assuming the gzip library will be sufficient for this purpose. Here is the code I have so far. I'm using python3 and the boto3 library for AWS. Any help is appreciated!
import parsedmarc
import pprint
import json
import boto3
import zipfile
import gzip
pp = pprint.PrettyPrinter(indent=2)
def main():
#Set default session profile and region for sandbox account. Access keys are pulled from /.aws/config and /.aws/credentials.
#The 'profile_name' value comes from the header for the account in question in /.aws/config and /.aws/credentials
boto3.setup_default_session(region_name="aws-region-goes-here")
boto3.setup_default_session(profile_name="aws-account-profile-name-goes-here")
#Define the s3 resource, the bucket name, and the file to download. It's hardcoded for now...
s3_resource = boto3.resource(s3)
s3_resource.Bucket('dmarc-parsing').download_file('source-dmarc-report-filename.zip' '/home/user/dmarc/parseme.zip')
#Use the zipfile python library to extract the file into its raw state.
with zipfile.ZipFile('/home/user/dmarc/parseme.zip', 'r') as zip_ref:
zip_ref.extractall('/home/user/dmarc')
#Ingest all locations for xml file source
dmarc_report_directory = '/home/user/dmarc/'
dmarc_report_file = 'parseme.xml'
"""I need an if statement here for extracting .gz files if the file type is not .zip. The contents of every archive are .xml files"""
#Set report output variables using functions in parsedmarc. Variable set to equal the output
pd_report_output=parsedmarc.parse_aggregate_report_file(_input=f"{dmarc_report_directory}{dmarc_report_file}")
#use jsonify to make the output in json format
pd_report_jsonified = json.loads(json.dumps(pd_report_output))
dkim_status = pd_report_jsonified['records'][0]['policy_evaluated']['dkim']
spf_status = pd_report_jsonified['records'][0]['policy_evaluated']['spf']
if dkim_status == 'fail' or spf_status == 'fail':
print(f"{dmarc_report_file} reports failure. oh crap. report:")
else:
print(f"{dmarc_report_file} passes. great. report:")
pp.pprint(pd_report_jsonified['records'][0]['auth_results'])
if __name__ == "__main__":
main()
Here is the code using the parsedmarc.parse_aggregate_report_xml method I found. Hope this helps others in parsing these reports:
import parsedmarc
import pprint
import json
import boto3
import zipfile
import gzip
pp = pprint.PrettyPrinter(indent=2)
def main():
#Set default session profile and region for account. Access keys are pulled from ~/.aws/config and ~/.aws/credentials.
#The 'profile_name' value comes from the header for the account in question in ~/.aws/config and ~/.aws/credentials
boto3.setup_default_session(profile_name="aws_profile_name_goes_here", region_name="region_goes_here")
source_file = 'filename_in_s3_bucket.zip'
destination_directory = '/tmp/'
destination_file = 'compressed_report_file'
#Define the s3 resource, the bucket name, and the file to download. It's hardcoded for now...
s3_resource = boto3.resource('s3')
s3_resource.Bucket('bucket-name-for-dmarc-report-files').download_file(source_file, f"{destination_directory}{destination_file}")
#Extract xml
outputxml = parsedmarc.extract_xml(f"{destination_directory}{destination_file}")
#run parse dmarc analysis & convert output to json
pd_report_output = parsedmarc.parse_aggregate_report_xml(outputxml)
pd_report_jsonified = json.loads(json.dumps(pd_report_output))
#loop through results and find relevant status info and pass fail status
dmarc_report_status = ''
for record in pd_report_jsonified['records']:
if False in record['alignment'].values():
dmarc_report_status = 'Failed'
#************ add logic for interpreting results
#if fail, publish to sns
if dmarc_report_status == 'Failed':
message = "Your dmarc report failed a least one check. Review the log for details"
sns_resource = boto3.resource('sns')
sns_topic = sns_resource.Topic('arn:aws:sns:us-west-2:112896196555:TestDMARC')
sns_publish_response = sns_topic.publish(Message=message)
if __name__ == "__main__":
main()

How to extract files in S3 on the fly with boto3?

I'm trying to find a way to extract .gz files in S3 on the fly, that is no need to download it to locally, extract and then push it back to S3.
With boto3 + lambda, how can i achieve my goal?
I didn't see any extract part in boto3 document.
You can use BytesIO to stream the file from S3, run it through gzip, then pipe it back up to S3 using upload_fileobj to write the BytesIO.
# python imports
import boto3
from io import BytesIO
import gzip
# setup constants
bucket = '<bucket_name>'
gzipped_key = '<key_name.gz>'
uncompressed_key = '<key_name>'
# initialize s3 client, this is dependent upon your aws config being done
s3 = boto3.client('s3', use_ssl=False) # optional
s3.upload_fileobj( # upload a new obj to s3
Fileobj=gzip.GzipFile( # read in the output of gzip -d
None, # just return output as BytesIO
'rb', # read binary
fileobj=BytesIO(s3.get_object(Bucket=bucket, Key=gzipped_key)['Body'].read())),
Bucket=bucket, # target bucket, writing to
Key=uncompressed_key) # target key, writing to
Ensure that your key is reading in correctly:
# read the body of the s3 key object into a string to ensure download
s = s3.get_object(Bucket=bucket, Key=gzip_key)['Body'].read()
print(len(s)) # check to ensure some data was returned
The above answers are for gzip files, for zip files, you may try
import boto3
import zipfile
from io import BytesIO
bucket = 'bucket1'
s3 = boto3.client('s3', use_ssl=False)
Key_unzip = 'result_files/'
prefix = "folder_name/"
zipped_keys = s3.list_objects_v2(Bucket=bucket, Prefix=prefix, Delimiter = "/")
file_list = []
for key in zipped_keys['Contents']:
file_list.append(key['Key'])
#This will give you list of files in the folder you mentioned as prefix
s3_resource = boto3.resource('s3')
#Now create zip object one by one, this below is for 1st file in file_list
zip_obj = s3_resource.Object(bucket_name=bucket, key=file_list[0])
print (zip_obj)
buffer = BytesIO(zip_obj.get()["Body"].read())
z = zipfile.ZipFile(buffer)
for filename in z.namelist():
file_info = z.getinfo(filename)
s3_resource.meta.client.upload_fileobj(
z.open(filename),
Bucket=bucket,
Key='result_files/' + f'{filename}')
This will work for your zip file and your result unzipped data will be in result_files folder. Make sure to increase memory and time on AWS Lambda to maximum since some files are pretty large and needs time to write.
Amazon S3 is a storage service. There is no in-built capability to manipulate the content of files.
However, you could use an AWS Lambda function to retrieve an object from S3, decompress it, then upload content back up again. However, please note that there is default limit of 500MB in temporary disk space for Lambda, so avoid decompressing too much data at the same time.
You could configure the S3 bucket to trigger the Lambda function when a new file is created in the bucket. The Lambda function would then:
Use boto3 to download the new file
Use the gzip Python library to extract files
Use boto3 to upload the resulting file(s)
Sample code:
import gzip
import io
import boto3
bucket = '<bucket_name>'
key = '<key_name>'
s3 = boto3.client('s3', use_ssl=False)
compressed_file = io.BytesIO(
s3.get_object(Bucket=bucket, Key=key)['Body'].read())
uncompressed_file = gzip.GzipFile(None, 'rb', fileobj=compressed_file)
s3.upload_fileobj(Fileobj=uncompressed_file, Bucket=bucket, Key=key[:-3])

django boto3: NoCredentialsError -- Unable to locate credentials

I am trying to use boto3 in my django project to upload files to Amazon S3. Credentials are defined in settings.py:
AWS_ACCESS_KEY = xxxxxxxx
AWS_SECRET_KEY = xxxxxxxx
S3_BUCKET = xxxxxxx
In views.py:
import boto3
s3 = boto3.client('s3')
path = os.path.dirname(os.path.realpath(__file__))
s3.upload_file(path+'/myphoto.png', S3_BUCKET, 'myphoto.png')
The system complains about Unable to locate credentials. I have two questions:
(a) It seems that I am supposed to create a credential file ~/.aws/credentials. But in a django project, where do I have to put it?
(b) The s3 method upload_file takes a file path/name as its first argument. Is it possible that I provide a file stream obtained by a form input element <input type="file" name="fileToUpload">?
This is what I use for a direct upload, i hope it provides some assistance.
import boto
from boto.exception import S3CreateError
from boto.s3.connection import S3Connection
conn = S3Connection(settings.AWS_ACCESS_KEY,
settings.AWS_SECRET_KEY,
is_secure=True)
try:
bucket = conn.create_bucket(settings.S3_BUCKET)
except S3CreateError as e:
bucket = conn.get_bucket(settings.S3_BUCKET)
k = boto.s3.key.Key(bucket)
k.key = filename
k.set_contents_from_filename(filepath)
Not sure about (a) but django is very flexible with file management.
Regarding (b) you can also sign the upload and do it directly from the client to reduce bandwidth usage, its quite sneaky and secure too. You need to use some JavaScript to manage the upload. If you want details I can include them here.