ZipArchive.Extract() Method

Extract all archive items as files into the current directory.

Namespace: DevExpress.Compression

Assembly: DevExpress.Docs.v23.2.dll

NuGet Package: DevExpress.Document.Processor

Declaration

C#

public void Extract()

VB.NET

Public Sub Extract

Remarks

If the archive contains directories, the directory structure will be recreated in the current directory, and the current directory is considered to be the root directory. The ZipItem.Name property value becomes the file name of the extracted file.

This example illustrates how to load the zip file and extract it to the specified directory. If the target directory is not specified, the current directory is the target.

View Example

ZipExamples.cs

using DevExpress.Compression;
        public void UnzipArchive() {
            string pathToZipArchive = "Documents\\Example.zip";
            string pathToExtract = "Documents\\!Extracted";
            using (ZipArchive archive = ZipArchive.Read(pathToZipArchive)) {
                foreach (ZipItem item in archive) {
                    item.Extract(pathToExtract);
                }
            }
        }

ZipExamples.vb

Imports DevExpress.Compression
        Public Sub UnzipArchive()
            Dim pathToZipArchive As String = "Documents\Example.zip"
            Dim pathToExtract As String = "Documents\!Extracted"
            Using archive As ZipArchive = ZipArchive.Read(pathToZipArchive)
                For Each item As ZipItem In archive
                    item.Extract(pathToExtract)
                Next item
            End Using
        End Sub