I'm using Selenium Grid and I need to download files for the automated tests download. I'm running into an issue where the response is not the file contents itself. Instead, the contents are zipped first and then wrapped within JSON.
Here's the response I'm trying to unzip:
b'{\n "filename": "test.txt",\n "contents": "UEsDBBQACAgIAFV1TlYAAAAAAAAAAAAAAAAIAAAAdGVzdC50eHQDAFBLBwgAAAAAAgAAAAAAAABQSwECFAAUAAgICABVdU5WAAAAAAIAAAAAAAAACAAAAAAAAAAAAAAAAAAAAAAAdGVzdC50eHRQSwUGAAAAAAEAAQA2AAAAOAAAAAAA"\n}'
The data above should contain an empty file called test.txt.
According to the documentation, the contents are a zipped folder in Base64 encoding. I want to be able to unzip this string and read the contents of the test.txt file within, but I'm not sure how to unzip a byte string with an encoding of base 64.
I'm using Python so if anyone knows how to unzip the contents and read the test.txt file within that would be extremely helpful.
Here's the documentation on Selenium Grid for Downloading files:
https://www.selenium.dev/documentation/grid/configuration/cli_options/#important-information-when-dowloading-a-file
You will need to b64decode the contents after the json has been parsed, and then store the result in either a temporary file or a BytesIO object for it to be usable with ZipFile:
import json
from base64 import b64decode
from io import BytesIO
from pathlib import Path
from zipfile import ZipFile
data = b'{\n "filename": "test.txt",\n "contents": "UEsDBBQACAgIAFV1TlYAAAAAAAAAAAAAAAAIAAAAdGVzdC50eHQDAFBLBwgAAAAAAgAAAAAAAABQSwECFAAUAAgICABVdU5WAAAAAAIAAAAAAAAACAAAAAAAAAAAAAAAAAAAAAAAdGVzdC50eHRQSwUGAAAAAAEAAQA2AAAAOAAAAAAA"\n}'
contents = json.loads(data)['contents']
bio = BytesIO(b64decode(contents))
We can then see the metadata about the files stored within that zip file (this step is not required):
>>> with ZipFile(bio) as zip_file:
... zip_file.infolist()
...
[<ZipInfo filename='test.txt' compress_type=deflate file_size=0 compress_size=2>]
To extract all files, you can use the extractall method:
>>> path = Path('temp/extracted')
>>> path.mkdir()
>>> with ZipFile(bio) as zip_file:
... zip_file.extractall(path)
...
>>> for p in path.iterdir():
... print(p.as_posix())
...
temp/extracted/test.txt
It is also possible to extract individual files, if desired.
Related
I had uploaded zip file in my azure account as a blob in azure container.
Zip file contains .csv, .ascii files and many other formats.
I need to read specific file, lets say ascii file data containing in zip file. I am using python for this case.
How to read particular file data from this zip file without downloading it on local? I would like to handle this process in memory only.
I am also trying with jypyter notebook provided by azure for ML functionality
I am using ZipFile python package for this case.
Request you to assist in this matter to read the file
Please find following code snippet.
blob_service=BlockBlobService(account_name=ACCOUNT_NAME,account_key=ACCOUNT_KEY)
blob_list=blob_service.list_blobs(CONTAINER_NAME)
allBlobs = []
for blob in blob_list:
allBlobs.append(blob.name)
sampleZipFile = allBlobs[0]
print(sampleZipFile)
The below code should work. This example accesses an Azure Container using an Account URL and Key combination.
from azure.storage.blob import BlobServiceClient
from io import BytesIO
from zipfile import ZipFile
key = r'my_key'
service = BlobServiceClient(account_url="my_account_url",
credential=key
)
container_client = service.get_container_client('container_name')
zipfilename = 'myzipfile.zip'
blob_data = container_client.download_blob(zipfilename)
blob_bytes = blob_data.content_as_bytes()
inmem = BytesIO(blob_bytes)
myzip = ZipFile(inmem)
otherfilename = 'mycontainedfile.csv'
filetoread = BytesIO(myzip.read(otherfilename))
Now all you have to do is pass filetoread into whatever method you would normally use to read a local file (eg. pandas.read_csv())
you could use below code for reading file inside .zip file without extracting in python
import zipfile
archive = zipfile.ZipFile('images.zip', 'r')
imgdata = archive.read('img_01.png')
For details , you can refer to ZipFile docs here
Alternatively, you can do something like this
-- coding: utf-8 --
"""
Created on Mon Apr 1 11:14:56 2019
#author: moverm
"""
import zipfile
zfile = zipfile.ZipFile('C:\\LAB\Pyt\sample.zip')
for finfo in zfile.infolist():
ifile = zfile.open(finfo)
line_list = ifile.readlines()
print(line_list)
Here is the output for the same
Hope it helps.
When I log the request in my frontend, I see characters like this:
PKS/L'�I�rrQUsers/my/local/dev_env/django_app/webapp/webapp/csvs/mailMagaOutput.csv会員id,e-mail,お名前(姓),お名前(名),購入回数
And the response logs the rest of the CSV files I am trying to log as the way they should look... Does this mean that I am not properly encrypting them?
The response I am sending (from Django):
zipped_file = zipfile.ZipFile("csvOutput.zip", 'w')
zipped_file.write(sms_file_path)
zipped_file.write(mail_maga_file_path)
zipped_file.close()
response_file = open('csvOutput.zip', 'rb')
response = HttpResponse(response_file, content_type="application/zip")
response['Content-Disposition'] = 'attachment; filename=csvOutput.zip"'
If I try to unzip the file that python generates on the server, it works just fine. When I try to unzip it locally, I am getting:
tar: Too-small extra data: Need at least 4 bytes, but only found 3 bytes
Note: the zip file contains two CSV files, those CSV files, are in two different Japanese encodings, shift_jis and shiftjisx0213
Instead of having the full path in the zipfile can you just zip the CSV files only?
EX:
import os
import zipfile
zipped_file = zipfile.ZipFile("csvOutput.zip", 'w')
os.chdir("/path/to/ CSV file path ")
zipped_file.write('mail_maga_file.csv')
zipped_file.write('sms_file.csv')
zipped_file.close()
I would like to automate the download of CSV files from the World Bank's dataset.
My problem is that the URL corresponding to a specific dataset does not lead directly to the desired CSV file but is instead a query to the World Bank's API. As an example, this is the URL to get the GDP per capita data: http://api.worldbank.org/v2/en/indicator/ny.gdp.pcap.cd?downloadformat=csv.
If you paste this URL in your browser, it will automatically start the download of the corresponding file. As a consequence, the code I usually use to collect and save CSV files in Python is not working in the present situation:
baseUrl = "http://api.worldbank.org/v2/en/indicator/ny.gdp.pcap.cd?downloadformat=csv"
remoteCSV = urllib2.urlopen("%s" %(baseUrl))
myData = csv.reader(remoteCSV)
How should I modify my code in order to download the file coming from the query to the API?
This will get the zip downloaded, open it and get you a csv object with whatever file you want.
import urllib2
import StringIO
from zipfile import ZipFile
import csv
baseUrl = "http://api.worldbank.org/v2/en/indicator/ny.gdp.pcap.cd?downloadformat=csv"
remoteCSV = urllib2.urlopen(baseUrl)
sio = StringIO.StringIO()
sio.write(remoteCSV.read())
# We create a StringIO object so that we can work on the results of the request (a string) as though it is a file.
z = ZipFile(sio, 'r')
# We now create a ZipFile object pointed to by 'z' and we can do a few things here:
print z.namelist()
# A list with the names of all the files in the zip you just downloaded
# We can use z.namelist()[1] to refer to 'ny.gdp.pcap.cd_Indicator_en_csv_v2.csv'
with z.open(z.namelist()[1]) as f:
# Opens the 2nd file in the zip
csvr = csv.reader(f)
for row in csvr:
print row
For more information see ZipFile Docs and StringIO Docs
import os
import urllib
import zipfile
from StringIO import StringIO
package = StringIO(urllib.urlopen("http://api.worldbank.org/v2/en/indicator/ny.gdp.pcap.cd?downloadformat=csv").read())
zip = zipfile.ZipFile(package, 'r')
pwd = os.path.abspath(os.curdir)
for filename in zip.namelist():
csv = os.path.join(pwd, filename)
with open(csv, 'w') as fp:
fp.write(zip.read(filename))
print filename, 'downloaded successfully'
From here you can use your approach to handle CSV files.
We have a script to automate access and data extraction for World Bank World Development Indicators like: https://data.worldbank.org/indicator/GC.DOD.TOTL.GD.ZS
The script does the following:
Downloading the metadata data
Extracting metadata and data
Converting to a Data Package
The script is python based and uses python 3.0. It has no dependencies outside of the standard library. Try it:
python scripts/get.py
python scripts/get.py https://data.worldbank.org/indicator/GC.DOD.TOTL.GD.ZS
You also can read our analysis about data from World Bank:
https://datahub.io/awesome/world-bank
Just a suggestion than a solution. You can use pd.read_csv to read any csv file directly from a URL.
import pandas as pd
data = pd.read_csv('http://url_to_the_csv_file')
I am trying to download and open a zipped file and seem to be having trouble using a file type handle with zipfile. I'm getting the error "AttributeError: addinfourl instance has no attribute 'seek'" when running this:
import zipfile
import urllib2
def download(url,directory,name):
webfile = urllib2.urlopen('http://www.sec.gov'+url)
webfile2 = zipfile.ZipFile(webfile)
content = zipfile.ZipFile.open(webfile2).read()
localfile = open(directory+name, 'w')
localfile.write(content)
localfile.close()
return()
download(link.get("href"),'./fails_data', link.text)
Putting things together, the following retrieves the content of the first file within a zipped file from a website:
import urllib.request
import zipfile
url = 'http://www.gutenberg.lib.md.us/4/8/8/2/48824/48824-8.zip'
filehandle, _ = urllib.request.urlretrieve(url)
zip_file_object = zipfile.ZipFile(filehandle, 'r')
first_file = zip_file_object.namelist()[0]
file = zip_file_object.open(first_file)
content = file.read()
As of 2020, you can use dload to download and unzip a file, i.e.:
import dload
dload.save_unzip("https://file-examples.com/wp-content/uploads/2017/02/zip_2MB.zip")
By default it extracts to a dir on the script path with the zip file name, but you can specify the extract location:
dload.save_unzip("https://file-examples.com/wp-content/uploads/2017/02/zip_2MB.zip", "/extract/here")
install using pip install dload
You can't seek on a urllib2.urlopened file. The methods it supports are listed here: http://docs.python.org/library/urllib.html#urllib.urlopen.
You'll have to retrieve the file (possibly with urllib.urlretrieve, http://docs.python.org/library/urllib.html#urllib.urlretrieve), then use zipfile on it.
Alternatively, you could read() the urlopened file, then put it into a StringIO, then use zipfile on that, if you wanted the zipped data in memory. Also check out the extract and extract_all methods of zipfile if you just want to extract the file, instead of using read.
I do not have enough rep to comment but regarding Marius's answer above please note that for Python3 there is a slight modification needed regarding import and urlretrieve call, since urllib has been split into several modules.
import urllib
Becomes:
import urllib.request
And
filehandle, _ = urllib.urlretrieve(url)
Becomes
filehandle, _ = urllib.request.urlretrieve(url)
Iterating on #Marius answer (which reads a single file directly from the zip), if you want to extract all files to a directory, do this:
import urllib
import zipfile
url = "http://www.gutenberg.lib.md.us/4/8/8/2/48824/48824-8.zip"
extract_dir = "example"
zip_path, _ = urllib.request.urlretrieve(url)
with zipfile.ZipFile(zip_path, "r") as f:
f.extractall(extract_dir)
This stores the zip file in a temporary dir. If you want to keep it around, you can pass a filename to urlretrieve, e.g. urllib.request.urlretrieve(url, "my_zip_file.zip").
I have managed to get my first python script to work which downloads a list of .ZIP files from a URL and then proceeds to extract the ZIP files and writes them to disk.
I am now at a loss to achieve the next step.
My primary goal is to download and extract the zip file and pass the contents (CSV data) via a TCP stream. I would prefer not to actually write any of the zip or extracted files to disk if I could get away with it.
Here is my current script which works but unfortunately has to write the files to disk.
import urllib, urllister
import zipfile
import urllib2
import os
import time
import pickle
# check for extraction directories existence
if not os.path.isdir('downloaded'):
os.makedirs('downloaded')
if not os.path.isdir('extracted'):
os.makedirs('extracted')
# open logfile for downloaded data and save to local variable
if os.path.isfile('downloaded.pickle'):
downloadedLog = pickle.load(open('downloaded.pickle'))
else:
downloadedLog = {'key':'value'}
# remove entries older than 5 days (to maintain speed)
# path of zip files
zipFileURL = "http://www.thewebserver.com/that/contains/a/directory/of/zip/files"
# retrieve list of URLs from the webservers
usock = urllib.urlopen(zipFileURL)
parser = urllister.URLLister()
parser.feed(usock.read())
usock.close()
parser.close()
# only parse urls
for url in parser.urls:
if "PUBLIC_P5MIN" in url:
# download the file
downloadURL = zipFileURL + url
outputFilename = "downloaded/" + url
# check if file already exists on disk
if url in downloadedLog or os.path.isfile(outputFilename):
print "Skipping " + downloadURL
continue
print "Downloading ",downloadURL
response = urllib2.urlopen(downloadURL)
zippedData = response.read()
# save data to disk
print "Saving to ",outputFilename
output = open(outputFilename,'wb')
output.write(zippedData)
output.close()
# extract the data
zfobj = zipfile.ZipFile(outputFilename)
for name in zfobj.namelist():
uncompressed = zfobj.read(name)
# save uncompressed data to disk
outputFilename = "extracted/" + name
print "Saving extracted file to ",outputFilename
output = open(outputFilename,'wb')
output.write(uncompressed)
output.close()
# send data via tcp stream
# file successfully downloaded and extracted store into local log and filesystem log
downloadedLog[url] = time.time();
pickle.dump(downloadedLog, open('downloaded.pickle', "wb" ))
Below is a code snippet I used to fetch zipped csv file, please have a look:
Python 2:
from StringIO import StringIO
from zipfile import ZipFile
from urllib import urlopen
resp = urlopen("http://www.test.com/file.zip")
myzip = ZipFile(StringIO(resp.read()))
for line in myzip.open(file).readlines():
print line
Python 3:
from io import BytesIO
from zipfile import ZipFile
from urllib.request import urlopen
# or: requests.get(url).content
resp = urlopen("http://www.test.com/file.zip")
myzip = ZipFile(BytesIO(resp.read()))
for line in myzip.open(file).readlines():
print(line.decode('utf-8'))
Here file is a string. To get the actual string that you want to pass, you can use zipfile.namelist(). For instance,
resp = urlopen('http://mlg.ucd.ie/files/datasets/bbc.zip')
myzip = ZipFile(BytesIO(resp.read()))
myzip.namelist()
# ['bbc.classes', 'bbc.docs', 'bbc.mtx', 'bbc.terms']
My suggestion would be to use a StringIO object. They emulate files, but reside in memory. So you could do something like this:
# get_zip_data() gets a zip archive containing 'foo.txt', reading 'hey, foo'
import zipfile
from StringIO import StringIO
zipdata = StringIO()
zipdata.write(get_zip_data())
myzipfile = zipfile.ZipFile(zipdata)
foofile = myzipfile.open('foo.txt')
print foofile.read()
# output: "hey, foo"
Or more simply (apologies to Vishal):
myzipfile = zipfile.ZipFile(StringIO(get_zip_data()))
for name in myzipfile.namelist():
[ ... ]
In Python 3 use BytesIO instead of StringIO:
import zipfile
from io import BytesIO
filebytes = BytesIO(get_zip_data())
myzipfile = zipfile.ZipFile(filebytes)
for name in myzipfile.namelist():
[ ... ]
I'd like to offer an updated Python 3 version of Vishal's excellent answer, which was using Python 2, along with some explanation of the adaptations / changes, which may have been already mentioned.
from io import BytesIO
from zipfile import ZipFile
import urllib.request
url = urllib.request.urlopen("http://www.unece.org/fileadmin/DAM/cefact/locode/loc162txt.zip")
with ZipFile(BytesIO(url.read())) as my_zip_file:
for contained_file in my_zip_file.namelist():
# with open(("unzipped_and_read_" + contained_file + ".file"), "wb") as output:
for line in my_zip_file.open(contained_file).readlines():
print(line)
# output.write(line)
Necessary changes:
There's no StringIO module in Python 3 (it's been moved to io.StringIO). Instead, I use io.BytesIO]2, because we will be handling a bytestream -- Docs, also this thread.
urlopen:
"The legacy urllib.urlopen function from Python 2.6 and earlier has been discontinued; urllib.request.urlopen() corresponds to the old urllib2.urlopen.", Docs and this thread.
Note:
In Python 3, the printed output lines will look like so: b'some text'. This is expected, as they aren't strings - remember, we're reading a bytestream. Have a look at Dan04's excellent answer.
A few minor changes I made:
I use with ... as instead of zipfile = ... according to the Docs.
The script now uses .namelist() to cycle through all the files in the zip and print their contents.
I moved the creation of the ZipFile object into the with statement, although I'm not sure if that's better.
I added (and commented out) an option to write the bytestream to file (per file in the zip), in response to NumenorForLife's comment; it adds "unzipped_and_read_" to the beginning of the filename and a ".file" extension (I prefer not to use ".txt" for files with bytestrings). The indenting of the code will, of course, need to be adjusted if you want to use it.
Need to be careful here -- because we have a byte string, we use binary mode, so "wb"; I have a feeling that writing binary opens a can of worms anyway...
I am using an example file, the UN/LOCODE text archive:
What I didn't do:
NumenorForLife asked about saving the zip to disk. I'm not sure what he meant by it -- downloading the zip file? That's a different task; see Oleh Prypin's excellent answer.
Here's a way:
import urllib.request
import shutil
with urllib.request.urlopen("http://www.unece.org/fileadmin/DAM/cefact/locode/2015-2_UNLOCODE_SecretariatNotes.pdf") as response, open("downloaded_file.pdf", 'w') as out_file:
shutil.copyfileobj(response, out_file)
I'd like to add my Python3 answer for completeness:
from io import BytesIO
from zipfile import ZipFile
import requests
def get_zip(file_url):
url = requests.get(file_url)
zipfile = ZipFile(BytesIO(url.content))
files = [zipfile.open(file_name) for file_name in zipfile.namelist()]
return files.pop() if len(files) == 1 else files
write to a temporary file which resides in RAM
it turns out the tempfile module ( http://docs.python.org/library/tempfile.html ) has just the thing:
tempfile.SpooledTemporaryFile([max_size=0[,
mode='w+b'[, bufsize=-1[, suffix=''[,
prefix='tmp'[, dir=None]]]]]])
This
function operates exactly as
TemporaryFile() does, except that data
is spooled in memory until the file
size exceeds max_size, or until the
file’s fileno() method is called, at
which point the contents are written
to disk and operation proceeds as with
TemporaryFile().
The resulting file has one additional
method, rollover(), which causes the
file to roll over to an on-disk file
regardless of its size.
The returned object is a file-like
object whose _file attribute is either
a StringIO object or a true file
object, depending on whether
rollover() has been called. This
file-like object can be used in a with
statement, just like a normal file.
New in version 2.6.
or if you're lazy and you have a tmpfs-mounted /tmp on Linux, you can just make a file there, but you have to delete it yourself and deal with naming
Adding on to the other answers using requests:
# download from web
import requests
url = 'http://mlg.ucd.ie/files/datasets/bbc.zip'
content = requests.get(url)
# unzip the content
from io import BytesIO
from zipfile import ZipFile
f = ZipFile(BytesIO(content.content))
print(f.namelist())
# outputs ['bbc.classes', 'bbc.docs', 'bbc.mtx', 'bbc.terms']
Use help(f) to get more functions details for e.g. extractall() which extracts the contents in zip file which later can be used with with open.
All of these answers appear too bulky and long. Use requests to shorten the code, e.g.:
import requests, zipfile, io
r = requests.get(zip_file_url)
z = zipfile.ZipFile(io.BytesIO(r.content))
z.extractall("/path/to/directory")
Vishal's example, however great, confuses when it comes to the file name, and I do not see the merit of redefing 'zipfile'.
Here is my example that downloads a zip that contains some files, one of which is a csv file that I subsequently read into a pandas DataFrame:
from StringIO import StringIO
from zipfile import ZipFile
from urllib import urlopen
import pandas
url = urlopen("https://www.federalreserve.gov/apps/mdrm/pdf/MDRM.zip")
zf = ZipFile(StringIO(url.read()))
for item in zf.namelist():
print("File in zip: "+ item)
# find the first matching csv file in the zip:
match = [s for s in zf.namelist() if ".csv" in s][0]
# the first line of the file contains a string - that line shall de ignored, hence skiprows
df = pandas.read_csv(zf.open(match), low_memory=False, skiprows=[0])
(Note, I use Python 2.7.13)
This is the exact solution that worked for me. I just tweaked it a little bit for Python 3 version by removing StringIO and adding IO library
Python 3 Version
from io import BytesIO
from zipfile import ZipFile
import pandas
import requests
url = "https://www.nseindia.com/content/indices/mcwb_jun19.zip"
content = requests.get(url)
zf = ZipFile(BytesIO(content.content))
for item in zf.namelist():
print("File in zip: "+ item)
# find the first matching csv file in the zip:
match = [s for s in zf.namelist() if ".csv" in s][0]
# the first line of the file contains a string - that line shall de ignored, hence skiprows
df = pandas.read_csv(zf.open(match), low_memory=False, skiprows=[0])
It wasn't obvious in Vishal's answer what the file name was supposed to be in cases where there is no file on disk. I've modified his answer to work without modification for most needs.
from StringIO import StringIO
from zipfile import ZipFile
from urllib import urlopen
def unzip_string(zipped_string):
unzipped_string = ''
zipfile = ZipFile(StringIO(zipped_string))
for name in zipfile.namelist():
unzipped_string += zipfile.open(name).read()
return unzipped_string
Use the zipfile module. To extract a file from a URL, you'll need to wrap the result of a urlopen call in a BytesIO object. This is because the result of a web request returned by urlopen doesn't support seeking:
from urllib.request import urlopen
from io import BytesIO
from zipfile import ZipFile
zip_url = 'http://example.com/my_file.zip'
with urlopen(zip_url) as f:
with BytesIO(f.read()) as b, ZipFile(b) as myzipfile:
foofile = myzipfile.open('foo.txt')
print(foofile.read())
If you already have the file downloaded locally, you don't need BytesIO, just open it in binary mode and pass to ZipFile directly:
from zipfile import ZipFile
zip_filename = 'my_file.zip'
with open(zip_filename, 'rb') as f:
with ZipFile(f) as myzipfile:
foofile = myzipfile.open('foo.txt')
print(foofile.read().decode('utf-8'))
Again, note that you have to open the file in binary ('rb') mode, not as text or you'll get a zipfile.BadZipFile: File is not a zip file error.
It's good practice to use all these things as context managers with the with statement, so that they'll be closed properly.