I have several very large zip files available to download on a website. I am using Flask microframework (based on Werkzeug) which uses Python.
Is there a way to show the contents of a zip file (i.e. file and folder names) - to someone on a webpage - without actually downloading it? As in doing the working out server side.
Assume that I do not know what are in the zip archives myself.
I apoligize that this post does not include code.
Thank you for helping.
Sure, have a look at zipfile.ZipFile.namelist(). Usage is pretty simple, as you'd expect: you just create a ZipFile object for the file you want, and then namelist() gives you a list of the paths of files stored in the archive.
with ZipFile('foo.zip', 'r') as f:
names = f.namelist()
print names
# ['file1', 'folder1/file2', ...]
http://docs.python.org/library/zipfile.html
Specifically, try using the ZipFile.namelist() method.
Related
I am trying to extract zip files using the zipfile module's extractall method.
My code snippet is
import zipfile
file_path = '/something/airway.zip'
dir_path = 'something/'
with zipfile.ZipFile(file_path, "r") as zip_ref:
zip_ref.extractall(dir_path)
I have two zip files named, test (1.1 mb) and airway (520 mb).
For test.zip the folder contains all the files but for airway.zip, it creates another folder inside my target folder named Airway, and then extracts all the files there. Even after renaming the airway.zip to any garbage name, the result was same.
Is there some workaround to get only the files extracted in my target folder? It is critical for me as I'm doing this extraction automated from django
Python version: 3.9.6;
Django version: 2.2
I ran your code and it seems to be only a problem of the zipfile itself. If you create a zipfile by selecting only the elements you get the result you got with test.zip. If you create it by selecting a folder holding the elements the folder will be there if you extract it again, no matter what you name your zip file.
I have two articles related to this:
https://www.kite.com/python/docs/zipfile.ZipFile.extractall
https://www.geeksforgeeks.org/working-zip-files-python/
Even if both of these articles do not solve your problem then I think that instead of zipping the files in the folder you just zipped the folder itself so try by zipping the files inside the folder.
I'm working with zipped files in python for the first time, and I'm stumped.
I read the documentation for zipfile, but I'm not sure what would be the best way to do what I'm trying to do. I have a zipped folder with CSV files inside, and I'd like to be able to open the zip file, and retrieve certain values from the csv files inside.
Do I use zipfile.extract(file name here) to bring it to the current working directory? And if I do that, do I just use the file name to work with the file, or does this index or list them differently?
Currently, I manually extract all files in the zipped folder to the current working directory for my project, and then use the csv module to read them. All I'm really trying to do is remove that step.
Any and all help would be greatly appreciated!
You are looking to avoid extracting to disk, in the zip docs for python there is ZipFile.open() which gives you a file-like object. That is an object that mostly behaves like a regular file on disk, but it is in memory. It gives a bytes array when read, at least in py3.
Something like this...
from zipfile import ZipFile
import csv
with ZipFile('abc.zip') as myzip:
print(myzip.filelist)
for mf in myzip.filelist:
with myzip.open(mf.filename) as myfile:
mc = myfile.read()
c = csv.StringIO(mc.decode())
for row in c:
print(row)
The documentation of Python is actually quite good once one has learned how to find things as well as some of the basic programming terms/descriptions used in the documentation.
For some reason csv.BytesIO is not implemented, hence the extra step via csv.StringIO.
I was wondering is there any way by which I can download only a part of a .rar or .zip file without downloading the whole file ? There is a zip file containing files A,B,C and D. I only need A. Can I somehow, use zipfile module so that i can only download 1 file ?
i am trying below code:
r = c.get(file)
z = ZipFile.ZipFile(BytesIO(r.content))
for file1 in z.namelist():
if 'time' not in file1:
print("hi")
z.extractall(file1,download_path + filename)
This code is downloading whole zip file and only extracting specific one. Can i somehow download only the file i Need.
There is similar question here but it shows only approch by command line in linux. That question dosent address how it can be done using python liabraries.
The question #Juggernaut mentioned in a comment is actually very helpful, as it points you in the direction of the solution.
You need to create a replacement for Bytes.IO that returns the necessary information to ZipFile. You will need to get the length of the file, and then get whatever sections ZipFile asks for.
How large are those file? Is it really worth the trouble?
Use remotezip: https://github.com/gtsystem/python-remotezip. You can install it using pip:
pip install remotezip
Usage example:
from remotezip import RemoteZip
with RemoteZip("https://path/to/zip/file.zip") as zip_file:
for file in zip_file.namelist():
if 'time' not in file:
print("hi")
zip_file.extract(file, path="/path/to/extract")
Note that to use this approach, the web server from which you receive the file needs to support the Range header.
I have a folder full of jar, html, css, exe type file. How can I check the file?
I already run "file" command on *NIX and using python-magic. but the result is all like this.
test : Zip archive data, at least v1.0 to extract
How can I get information specifically like test : jar only using using magic number.
How do I do like this?
While not required, most JAR files have a META-INF/MANIFEST.MF file contained within them. You could check for the existence of this file, after checking if it's a zip file:
import zipfile
def zipFileContains(zipFileName, pathName):
f = zipfile.ZipFile(zipFileName, "r")
result = any(x.startswith(pathName.rstrip("/")) for x in f.namelist())
f.close()
return result
print zipFileContains("test.jar", "META-INF/MANIFEST.MF")
However, it might be better to just check if it's a zip file that ends in .jar.
Magic alone won't do it for you, since a JAR is literally just a zip file. Read more about the format here.
I have a *.tar.gz compressed file that I would like to read in with Python 2.7. The file contains multiple h5 formatted files as well as a few text files. I'm a novice with Python. Here is the code I'm trying to adapt:
`subset_path='c:\data\grant\files'
f=gzip.open(filename,'subset_full.tar.gz')
subset_data_path=os.path.join(subset_path,'f')
The first statement identifies the path to the folder with the data. The second statement tells Python to open a specific compressed file and the third statement (hopefully) executes a join of the prior two statements.
Several lines below this code I get an error when Python tries to use the 'subset_data_path' assignment.
What's going on?
The gzip module will only open a single file that has been compressed, i.e. my_file.gz. You have a tar archive of multiple files that are also compressed. This needs to be both untarred and uncompressed.
Try using the tarfile module instead, see https://docs.python.org/2/library/tarfile.html#examples
edit: To add a bit more information on what has happened, you have successfully opened the zipped tarball into a gzip file object, which will work almost the same as a standard file object. For instance you could call f.readlines() as if f was a normal file object and it would return the uncompressed lines.
However, this did not actually unpack the archive into new files in the filesystem. You did not create a subdirectory 'c:\data\grant\files\f', and so when you try to use the path subset_data_path you are looking for a directory that does not exist.
The following ought to work:
import tarfile
subset_path='c:\data\grant\files'
tar = tarfile.open("subset_full.tar.gz")
tar.extractall(subset_path)
subset_data_path=os.path.join(subset_path,'subset_full')