Updating a file in a ZIP is not supported. You need to rebuild a new archive without the file, then add the updated version. import os import zipfile import tempfile def updateZip(zipname, filename, data): # generate a temp file tmpfd, tmpname = tempfile.mkstemp(dir=os.path.dirname(zipname)) os.close(tmpfd) # create a temp copy of the archive without filename with … Read more
This is a base class for a component that needs to handle data that is either uncompressed, PKZIP compressed (sharpziplib) or GZip compressed (built in .net). Perhaps a bit more than you need but should get you going. This is an example of using @PhonicUK’s suggestion to parse the header of the data stream. The … Read more
>>> map(list.__add__, L1, L2) [[1, 2, ‘a’, ‘b’], [3, 4, ‘c’, ‘d’], [5, 6, ‘e’, ‘f’]]
Java 7 introduced a much simpler way for doing zip archive manipulations – FileSystems API, which allows to access contents of a file as a file system. Besides much more straightforward API, it is doing the modification in-place and doesn’t require to rewrite other (irrelevant) files in a zip archive (as done in the accepted … Read more
You can pass ZipFile.open() to pandas.read_csv() to construct a pandas.DataFrame from a csv-file packed into a multi-file zip. Code: pd.read_csv(zip_file.open(‘file3.txt’)) Example to read all .csv into a dict: from zipfile import ZipFile zip_file = ZipFile(‘textfile.zip’) dfs = {text_file.filename: pd.read_csv(zip_file.open(text_file.filename)) for text_file in zip_file.infolist() if text_file.filename.endswith(‘.csv’)}
Take a look at java.util.zip.* classes for zip functionality. I’ve done some basic zip/unzip code, which I’ve pasted below. Hope it helps. public static void zip(String[] files, String zipFile) throws IOException { BufferedInputStream origin = null; ZipOutputStream out = new ZipOutputStream(new BufferedOutputStream(new FileOutputStream(zipFile))); try { byte data[] = new byte[BUFFER_SIZE]; for (int i = 0; … Read more
ZipOutputStream can handle empty directories by adding a forward-slash / after the folder name. Try (from) public class Test { public static void main(String[] args) { try { FileOutputStream f = new FileOutputStream(“test.zip”); ZipOutputStream zip = new ZipOutputStream(new BufferedOutputStream(f)); zip.putNextEntry(new ZipEntry(“xml/”)); zip.putNextEntry(new ZipEntry(“xml/xml”)); zip.close(); } catch(Exception e) { System.out.println(e.getMessage()); } } }
Unfortunately w/ PHP 5.3.4-dev and Zend Engine v2.3.0 on CentOS 5.x I couldn’t get the code above to work. An “Invalid or unitialized Zip object” error message was all I could get. So, in order to make it work, I had to use following snippet (taken from the example by Jonathan Baltazar on PHP.net manual, … Read more
2 things. First, if you keep the code design you have, you need to perform a Seek() on the MemoryStream before writing it into the entry. dt.TableName = “Declaration”; MemoryStream stream = new MemoryStream(); dt.WriteXml(stream); stream.Seek(0,SeekOrigin.Begin); // <– must do this after writing the stream! using (ZipFile zipFile = new ZipFile()) { zipFile.AddEntry(“Report.xml”, “”, stream); … Read more
A related post here provides the information needed to grab the .git directory and simplify the answer that umläute provided: Grab the .git directory by cloning a bare repository $ mkdir repo $ git clone –bare http://github/user/repo repo Make the .git directory and move the cloned files $ mkdir repo/.git $ mv repo/* repo/.git Unzip … Read more