Unzip a file to s3 - amazon-s3

Question

I am looking at a simple way to extract a zip/gzip present in s3 bucket to the same bucket location and delete the parent zip/gzip file post extraction.
I am unable to achieve this with any of the API's currently.
Have tried native boto, pyfilesystem(fs), s3fs.
The source and destination links seem to be an issue for these functions.
(Using with Python 2.x/3.x & Boto 2.x )
I see there is an API for node.js(unzip-to-s3) to do this job , but none for python.
Couple of implementations i can think of:
A simple API to extract the zip file within the same bucket.
Use s3 as a filesystem and manipulate data
Use a data pipeline to achieve this
Transfer the zip to ec2 , extract and copy back to s3.
The option 4 would be the least preferred option, to minimise the architecture overhead with ec2 addon.
Need support in getting this feature implementation , with integration to lambda at a later stage. Any pointers to these implementations are greatly appreciated.
Thanks in Advance,
Sundar.

Answer 1

You could try https://www.cloudzipinc.com/ that unzips/expands several different formats of archives from S3 into a destination in your bucket. I used it to unzip components of a digital catalog into S3.

Answer 2

Have solved by using ec2 instance.
Copy the s3 files to local dir in ec2
and copy that directory back to S3 bucket.

Answer 3

Sample to unzip to local directory in ec2 instance
def s3Unzip(srcBucket,dst_dir):
'''
function to decompress the s3 bucket contents to local machine
Args:
srcBucket (string): source bucket name
dst_dir (string): destination location in the local/ec2 local file system
Returns:
None
'''
#bucket = s3.lookup(bucket)
s3=s3Conn
path=''
bucket = s3.lookup(bucket_name)
for key in bucket:
path = os.path.join(dst_dir, key.name)
key.get_contents_to_filename(path)
if path.endswith('.zip'):
opener, mode = zipfile.ZipFile, 'r'
elif path.endswith('.tar.gz') or path.endswith('.tgz'):
opener, mode = tarfile.open, 'r:gz'
elif path.endswith('.tar.bz2') or path.endswith('.tbz'):
opener, mode = tarfile.open, 'r:bz2'
else:
raise ValueError ('unsuppported format')
try:
os.mkdir(dst_dir)
print ("local directories created")
except Exception:
logger_s3.warning ("Exception in creating local directories to extract zip file/ folder already existing")
cwd = os.getcwd()
os.chdir(dst_dir)
try:
file = opener(path, mode)
try: file.extractall()
finally: file.close()
logger_s3.info('(%s) extracted successfully to %s'%(key ,dst_dir))
except Exception as e:
logger_s3.error('failed to extract (%s) to %s'%(key ,dst_dir))
os.chdir(cwd)
s3.close
sample code to upload to mysql instance
Use the "LOAD DATA LOCAL INFILE" query to upload to mysql directly
def upload(file_path,timeformat):
'''
function to upload a csv file data to mysql rds
Args:
file_path (string): local file path
timeformat (string): destination bucket to copy data
Returns:
None
'''
for file in file_path:
try:
con = connect()
cursor = con.cursor()
qry="""LOAD DATA LOCAL INFILE '%s' INTO TABLE xxxx FIELDS TERMINATED BY ',' LINES TERMINATED BY '\n' (col1 , col2 ,col3, #datetime , col4 ) set datetime = str_to_date(#datetime,'%s');""" %(file,timeformat)
cursor.execute(qry)
con.commit()
logger_rds.info ("Loading file:"+file)
except Exception:
logger_rds.error ("Exception in uploading "+file)
##Rollback in case there is any error
con.rollback()
cursor.close()
# disconnect from server
con.close()

Answer 4

Lambda function:
You can use a Lambda function where you read zipped files into the buffer, gzip the individual files, and reupload them to S3. Then you can either archive the original files or delete them using boto.
You can also set an event based trigger that runs the lambda automatically everytime there is a new zipped file in S3. Here's a full tutorial for the exact thing here: https://betterprogramming.pub/unzip-and-gzip-incoming-s3-files-with-aws-lambda-f7bccf0099c9