extract.zip: Unzip archives in a specified directory

Description Usage Arguments Value Author(s)

Description

extract.zip extracts the files from a .zip archive in a specific directory.

Usage

1
extract.zip(file, extractpath = dirname(file)[1])

Arguments

file

A file name.

extractpath

A path to define where the files are to be extracted.

Value

Success is indicated by returning the directory in which the files have been extracted. If it fails, it returns an empty character string.

Author(s)

Audrey Kauffmann

Maintainer: <[email protected]>


ArrayExpress documentation built on May 6, 2019, 3:40 a.m.