Develop Reference

Getting code from a .txt on a website and pasting it in a tempfile PYTHON

I was trying to make a script that gets a .txt from a websites, pastes the code into a python executable temp file but its not working. Here is the code:
from urllib.request import urlopen as urlopen
import os
import subprocess
import os
import tempfile
filename = urlopen("https://randomsiteeeee.000webhostapp.com/script.txt")
temp = open(filename)
temp.close()
# Clean up the temporary file yourself
os.remove(filename)
temp = tempfile.TemporaryFile()
temp.close()
If you know a fix to this please let me know. The error is :
File "test.py", line 9, in <module>
temp = open(filename)
TypeError: expected str, bytes or os.PathLike object, not HTTPResponse
I tried everything such as a request to the url and pasting it but didnt work as well. I tried the code that i pasted here and didnt work as well.
And as i said, i was expecting it getting the code from the .txt from the website, and making it a temp executable python script

you are missing a read:
from urllib.request import urlopen as urlopen
import os
import subprocess
import os
import tempfile
filename = urlopen("https://randomsiteeeee.000webhostapp.com/script.txt").read() # <-- here
temp = open(filename)
temp.close()
# Clean up the temporary file yourself
os.remove(filename)
temp = tempfile.TemporaryFile()
temp.close()
But if the script.txt contains the script and not the filename, you need to create a temporary file and write the content:
from urllib.request import urlopen as urlopen
import os
import subprocess
import os
import tempfile
content = urlopen("https://randomsiteeeee.000webhostapp.com/script.txt").read() #
with tempfile.TemporaryFile() as fp:
name = fp.name
fp.write(content)
If you want to execute the code you fetch from the url, you may also use exec or eval instead of writing a new script file.
eval and exec are EVIL, they should only be used if you 100% trust the input and there is no other way!
EDIT: How do i use exec?
Using exec, you could do something like this (also, I use requests instead of urllib here. If you prefer urllib, you can do this too):
import requests
exec(requests.get("https://randomsiteeeee.000webhostapp.com/script.txt").text)

Your trying to open a file that is named "the content of a website".
filename = "path/to/my/output/file.txt"
httpresponse = urlopen("https://randomsiteeeee.000webhostapp.com/script.txt").read()
temp = open(filename)
temp.write(httpresponse)
temp.close()
Is probably more like what you are intending

How to download a CSV file from the World Bank's dataset

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')

Working with a pdf from the web directly in Python?

I'm trying to use Python to read .pdf files from the web directly rather than save them all to my computer. All I need is the text from the .pdf and I'm going to be reading a lot (~60k) of them, so I'd prefer to not actually have to save them all.
I know how to save a .pdf from the internet using urllib and open it with PyPDF2. (example)
I want to skip the saving-to-file step.
import urllib, PyPDF2
urllib.urlopen('https://bitcoin.org/bitcoin.pdf')
wFile = urllib.urlopen('https://bitcoin.org/bitcoin.pdf')
lFile = PyPDF2.pdf.PdfFileReader(wFile.read())
I get an error that is fairly easy to understand:
Traceback (most recent call last):
File "<pyshell#6>", line 1, in <module>
fil = PyPDF2.pdf.PdfFileReader(wFile.read())
File "C:\Python27\lib\PyPDF2\pdf.py", line 797, in __init__
self.read(stream)
File "C:\Python27\lib\PyPDF2\pdf.py", line 1245, in read
stream.seek(-1, 2)
AttributeError: 'str' object has no attribute 'seek'
Obviously PyPDF2 doesn't like that I'm giving it the urllib.urlopen().read() (which appears to return a string). I know that this string is not the "text" of the .pdf but a string representation of the file. How can I resolve this?
EDIT: NorthCat's solution resolved my error, but when I try to actually extract the text, I get this:
>>> print lFile.getPage(0).extractText()
ˇˆ˘˘˙˘˘˝˘˛˘ˇ˘ˇ˚ˇˇˇ˘ˆ˘˘˘˚ˇˆ˘ˆ˘ˇ˜ˇ˝˚˘˛˘ˇ ˘˘˘ˇ˛˘˚˚ˆˇˇ!
˝˘˚ˇ˘˘˚"˘˘ˇ˘˚ˇ˘˘˚ˇ˘˘˘˙˘˘˘#˘˘˘ˆ˘˛˘˚˛˙ ˘˘˚˚˘˛˙#˘ˇ˘ˇˆ˘˘˛˛˘˘!˘˘˛˘˝˘˘˘˚ ˛˘˘ˇ˘ˇ˛$%&˘ˇ'ˆ˛
$%&˘ˇˇ˘˚ˆ˚˘˘˘˘ ˘ˆ(ˇˇ˘˘˘˘ˇ˘˚˘˘#˘˘˘ˇ˛!ˇ)˘˘˚˘˘˛ ˚˚˘ˇ˘˝˘˚'˘˘ˇˇ ˘˘ˇ˘˛˙˛˛˘˘˚ˇ˘˘ˆ˘˘ˆ˙
$˘˘˘*˘˘˘ˇˆ˘˘ˇˆ˛ˇ˘˝˚˚˘˘ˇ˘ˆ˘"˘ˆ˘ˇˇ˘˛ ˛˛˘˛˘˘˘˘˘˘˛˘˘˚˚˘$ˇ˘ˇˆ˙˘˝˘ˇ˘˘˘ˇˇˆˇ˘ ˘˛ˇ˝˘˚˚#˘˛˘˚˘˘
˘ˇ˘˚˛˛˘ˆ˛ˇˇˇ ˚˘˘˚˘˘ˇ˛˘˙˘˝˘ˇ˘ˆ˘˛˙˘˝˘ˇ˘˘˝˘"˘˛˘˝˘ˇ ˘˘˘˚˛˘˚)˘˘ˆ˛˘˘
˘˛˘˛˘ˆˇ˚˘˘˘˘˚˘˘˘˘˛˛˚˘˚˝˚ˇ˘#˘˘˚ˆ˘˘˘˝˘˚˘ˆˆˇ˘ˆ
˘˘˘ˆ˘˝˘˘˚"˘˘˚˘˚˘ˇ˘ˆ˘ˆ˘˚ˆ˛˚˛ˆ˚˘˘˘˘˘˘˚˛˚˚ˆ#˘ˇˇˆˇ˘˝˘˘ˇ˚˘ˇˇ˘˛˛˚ ˚˘˘˘ˇ˚˘˘ˇ˘˘˚ˆ˘*˘
˘˘ˇ˘˚ˇ˘˙˘˚ˇ˘˘˘˙˙˘˘˚˚˘˘˝˘˘˘˛˛˘ˇˇ˚˘˛#˘ˆ˘˘ˇ˘˚˘ˇˇ˘˘ˇˆˇ˘$%&˘ˆ˘˛˘˚˘,

Try this:
import urllib, PyPDF2
import cStringIO
wFile = urllib.urlopen('https://bitcoin.org/bitcoin.pdf')
lFile = PyPDF2.pdf.PdfFileReader( cStringIO.StringIO(wFile.read()) )
Because PyPDF2 does not work, there are a couple of solutions, however, require saving the file to disk.
Solution 1
You can use ps2ascii (if you are using linux or mac ) or xpdf (Windows). Example of using xpdf:
import os
os.system('C:\\xpdfbin-win-3.03\\bin32\\pdftotext.exe C:\\xpdfbin-win-3.03\\bin32\\bitcoin.pdf bitcoin1.txt')
or
import subprocess
subprocess.call(['C:\\xpdfbin-win-3.03\\bin32\\pdftotext.exe', 'C:\\xpdfbin-win-3.03\\bin32\\bitcoin.pdf', 'bitcoin2.txt'])
Solution 2
You can use one of online pdf to txt converter. Example of using pdf.my-addr.com
import MultipartPostHandler
import urllib2
def pdf2text( absolute_path ):
url = 'http://pdf.my-addr.com/pdf-to-text-converter-tool.php'
params = { 'file' : open( absolute_path, 'rb' ),
'encoding': 'UTF-8',
}
opener = urllib2.build_opener( MultipartPostHandler.MultipartPostHandler )
return opener.open( url, params ).read()
print pdf2text('bitcoin.pdf')
Code of MultipartPostHandler you can find here. I tried to use the cStringIO instead open(), but it did not work.
Maybe it will be helpful for you.

I know this question is old, but I had the same issue and here is how I solved it.
In the newer docs of Py2PDF there is a section about streaming data
The example there looks like this:
from io import BytesIO
# Prepare example
with open("example.pdf", "rb") as fh:
bytes_stream = BytesIO(fh.read())
# Read from bytes_stream
reader = PdfReader(bytes_stream)
Therefore, what I did instead was this:
import urllib
from io import BytesIO
from PyPDF2 import PdfReader
NEW_PATH = 'https://example.com/path/to/pdf/online?id=123456789&date=2022060'
wFile = urllib.request.urlopen(NEW_PATH)
bytes_stream = BytesIO(wFile.read())
reader = PdfReader(bytes_stream)

Python - Download File Using Requests, Directly to Memory

The goal is to download a file from the internet, and create from it a file object, or a file like object without ever having it touch the hard drive. This is just for my knowledge, wanting to know if its possible or practical, particularly because I would like to see if I can circumvent having to code a file deletion line.
This is how I would normally download something from the web, and map it to memory:
import requests
import mmap
u = requests.get("http://www.pythonchallenge.com/pc/def/channel.zip")
with open("channel.zip", "wb") as f: # I want to eliminate this, as this writes to disk
f.write(u.content)
with open("channel.zip", "r+b") as f: # and his as well, because it reads from disk
mm = mmap.mmap(f.fileno(), 0)
mm.seek(0)
print mm.readline()
mm.close() # question: if I do not include this, does this become a memory leak?

r.raw (HTTPResponse) is already a file-like object (just pass stream=True):
#!/usr/bin/env python
import sys
import requests # $ pip install requests
from PIL import Image # $ pip install pillow
url = sys.argv[1]
r = requests.get(url, stream=True)
r.raw.decode_content = True # Content-Encoding
im = Image.open(r.raw) #NOTE: it requires pillow 2.8+
print(im.format, im.mode, im.size)
In general if you have a bytestring; you could wrap it as f = io.BytesIO(r.content), to get a file-like object without touching the disk:
#!/usr/bin/env python
import io
import zipfile
from contextlib import closing
import requests # $ pip install requests
r = requests.get("http://www.pythonchallenge.com/pc/def/channel.zip")
with closing(r), zipfile.ZipFile(io.BytesIO(r.content)) as archive:
print({member.filename: archive.read(member) for member in archive.infolist()})
You can't pass r.raw to ZipFile() directly because the former is a non-seekable file.
I would like to see if I can circumvent having to code a file deletion line
tempfile can delete files automatically f = tempfile.SpooledTemporaryFile(); f.write(u.content). Until .fileno() method is called (if some api requires a real file) or maxsize is reached; the data is kept in memory. Even if the data is written on disk; the file is deleted as soon as it closed.

Your answer is u.content. The content is in the memory. Unless you write it to a file, it won’t be stored on disk.

This is what I ended up doing.
import zipfile
import requests
import StringIO
u = requests.get("http://www.pythonchallenge.com/pc/def/channel.zip")
f = StringIO.StringIO()
f.write(u.content)
def extract_zip(input_zip):
input_zip = zipfile.ZipFile(input_zip)
return {i: input_zip.read(i) for i in input_zip.namelist()}
extracted = extract_zip(f)

Downloading and unzipping a .zip file without writing to disk

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.