unzip
Unpack inner zips in zip with Maven
You can unzip any files using ant task runner plugin: <plugin> <groupId>org.apache.maven.plugins</groupId> <artifactId>maven-antrun-plugin</artifactId> <version>1.6</version> <executions> <execution> <id>prepare</id> <phase>validate</phase> <configuration> <tasks> <echo message=”prepare phase” /> <unzip src=”https://stackoverflow.com/questions/3264064/zips/archive.zip” dest=”output/” /> <unzip src=”output/inner.zip” dest=”output/” /> <unzip dest=”output”> <fileset dir=”archives”> <include name=”prefix*.zip” /> </fileset> </unzip> </tasks> </configuration> <goals> <goal>run</goal> </goals> </execution> </executions> </plugin>
Unzip all zipped files in a folder to that same folder using Python 2.7.5
Below is the code that worked for me: import os, zipfile dir_name=”C:\\SomeDirectory” extension = “.zip” os.chdir(dir_name) # change directory from working dir to dir with files for item in os.listdir(dir_name): # loop through items in dir if item.endswith(extension): # check for “.zip” extension file_name = os.path.abspath(item) # get full path of files zip_ref = zipfile.ZipFile(file_name) … Read more
How to zip and unzip the files?
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
How can unrar a file with python
Late, but I wasn’t satisfied with any of the answers. pip install patool import patoolib patoolib.extract_archive(“foo_bar.rar”, outdir=”path here”) Works on Windows and linux without any other libraries needed.
How to read from a zip file within zip file in Python? [duplicate]
When you use the .open() call on a ZipFile instance you indeed get an open file handle. However, to read a zip file, the ZipFile class needs a little more. It needs to be able to seek on that file, and the object returned by .open() is not seekable in your case. Only Python 3 … Read more
How to unzip a piped zip file (from “wget -qO-“)?
The ZIP file format includes a directory (index) at the end of the archive. This directory says where, within the archive each file is located and thus allows for quick, random access, without reading the entire archive. This would appear to pose a problem when attempting to read a ZIP archive through a pipe, in … Read more
easy way to unzip file
Slight rework of the OP’s solution to create the containing directory dest if it doesn’t exist, and to wrap the file extraction/writing in a closure to eliminate stacking of defer .Close() calls per @Nick Craig-Wood’s comment: func Unzip(src, dest string) error { r, err := zip.OpenReader(src) if err != nil { return err } defer … Read more
gradle – download and unzip file from url
Let’s say you want to download this zip file as a dependency: https://github.com/jmeter-gradle-plugin/jmeter-gradle-plugin/archive/1.0.3.zip You define your ivy repo as: repositories { ivy { url ‘https://github.com/’ patternLayout { artifact ‘/[organisation]/[module]/archive/[revision].[ext]’ } // This is required in Gradle 6.0+ as metadata file (ivy.xml) // is mandatory. Docs linked below this code section metadataSources { artifact() } } … Read more
Simple way to unzip a .zip file using zlib [duplicate]
zlib handles the deflate compression/decompression algorithm, but there is more than that in a ZIP file. You can try libzip. It is free, portable and easy to use. UPDATE: Here I attach quick’n’dirty example of libzip, with all the error controls ommited: #include <zip.h> int main() { //Open the ZIP archive int err = 0; … Read more