zipfile

I am downloading a zip file using c# program and I get the error at System.IO.Compression.ZipArchive.ReadEndOfCentralDirectory() at System.IO.Compression.ZipArchive.Init(Stream stream, ZipArchiveMode mode, Boolean leaveOpen) at System.IO.Compression.ZipArchive..ctor(Stream stream, ZipArchiveMode mode, Boolean leaveOpen, Encoding entryNameEncoding) at System.IO.Compression.ZipFile.Open(String archiveFileName, ZipArchiveMode mode, Encoding entryNameEncoding) at System.IO.Compression.ZipFile.ExtractToDirectory(String sourceArchiveFileN ame, String destinationDirectoryName, Encoding

I try to extract all files from .zip containing subfolders in one folder. I want all the files from subfolders extract in only one folder without keeping the original structure. At the moment, I extract all, move the files to a folder, then remove previous subfolders. The files with same names are overwrited. Is it possible to do it before writing files? Here is a structure for example: my_zip/file1.txt my_zip/dir1/file2.txt my_zip/dir1/dir2/file3.txt my_zip/dir3/file4.txt At the end I whish this: my_dir/file1.txt my_dir/file2.txt my_dir/file3.txt my_dir/file4.txt What can I add to this code ?

How do I extract a zip to memory? My attempt (returning None on .getvalue() ): from zipfile import ZipFile from StringIO import StringIO def extract_zip(input_zip): return StringIO(ZipFile(input_zip).extractall()) mata extractall extracts to the file system, so you won't get what you want. To extract a file in memory, use the ZipFile.read() method. If you really need the full content in memory, you could do something like: def extract_zip(input_zip): input_zip=ZipFile(input_zip) return {name: input_zip.read(name) for name in input_zip.namelist()} Deviacium Frequently working with in-memory

I have a flask server that grabs binary data for several different files from a database and puts them into a python 'zipfile' object. I want to send the generated zip file with my code using flask's "send_file" method. I was originally able to send non-zip files successfully by using the BytesIO(bin) as the first argument to send_file, but for some reason I can't do the same thing with my generated zip file. It gives the error: 'ZipFile' does not have the buffer interface. How do I send this zip file object to the user with Flask? This is my code: @app.route("/getcaps",methods=['GET','POST'])

I have two files in two different directories, one is '/home/test/first/first.pdf' , the other is '/home/text/second/second.pdf' . I use following code to compress them: import zipfile, StringIO buffer = StringIO.StringIO() first_path = '/home/test/first/first.pdf' second_path = '/home/text/second/second.pdf' zip = zipfile.ZipFile(buffer, 'w') zip.write(first_path) zip.write(second_path) zip.close() After I open the zip file that I created, I have a home folder in it, then there are two sub-folders in it, first and second , then the pdf files. I don't know how to include only two pdf files