zipfile

I have folder named "data". This "data" folder contains a file "filecontent.txt" and another folder named "Files". The "Files" folder contains a "info.txt" file. So it is a folder inside folder structure. I have to zip this folder "data"(using php) along with the file and folder inside it, and download the zipped file. I have tried the examples available at http://www.php.net/manual/en/zip.examples.php These examples did not work. My PHP version is 5.2.10 Please help. I have written this code. <?php $zip = new ZipArchive; if ($zip->open('check/test2.zip',ZIPARCHIVE::CREATE) === TRUE) { if($zip

Could anyone post a simple snippet that does this? Files are text files, so compression would be nice rather than just archive the files. I have the filenames stored in an iterable. There's not currently any way to do this kind of thing from the standard Scala library, but it's pretty easy to use java.util.zip : def zip(out: String, files: Iterable[String]) = { import java.io.{ BufferedInputStream, FileInputStream, FileOutputStream } import java.util.zip.{ ZipEntry, ZipOutputStream } val zip = new ZipOutputStream(new FileOutputStream(out)) files.foreach { name => zip.putNextEntry(new ZipEntry

If I make the url for a zip file the href of a link and click the link, my zip file gets downloaded and opening it gets the contents as I expect. Here's that HTML: <a href="http://mysite.com/uploads/my-archive.zip">download zip</a> The problem is I'd like the link to point to my application such that I could determine whether the user is authorized to access this zip file. so I'd like my HTML to be this: <a href="/canDownload">download zip</a> and my PHP for the /canDownload page: //business logic to determine if user can download if($yesCanDownload){ $archive='https://mysite.com/uploads/my

I am trying to convert an array of bytes into a ZIP file. I got bytes using the following code: byte[] originalContentBytes= new Verification().readBytesFromAFile(new File("E://file.zip")); private byte[] readBytesFromAFile(File file) { int start = 0; int length = 1024; int offset = -1; byte[] buffer = new byte[length]; try { //convert the file content into a byte array FileInputStream fileInuptStream = new FileInputStream(file); BufferedInputStream bufferedInputStream = new BufferedInputStream( fileInuptStream); ByteArrayOutputStream byteArrayOutputStream = new ByteArrayOutputStream(); while

I am trying to download a zip file from my web api controller. It is returning the file but I am getting a message the zipfile is invalid when i try to open. I have seen other posts about this and the response was adding the responseType: 'arraybuffer'. Still isn't working for me. I'm not getting any errors in the console either. var model = $scope.selection; var res = $http.post('/api/apiZipPipeLine/', model) res.success(function (response, status, headers, config) { saveAs(new Blob([response], { type: "application/octet-stream", responseType: 'arraybuffer' }), 'reports.zip');

I have a folder containing .ZIP files . Now, I want to Extract the ZIP Files to specific folders using C#, but without using any external assembly or the .Net Framework 4.5. I have searched, but not found any solution for unpacking *.zip files using Framework 4.0 or below. I tried GZipStream, but it only supports .gz and not .zip files. Denys Denysenko Here is example from msdn . System.IO.Compression.ZipFile is made just for that: using System; using System.IO; using System.IO.Compression; namespace ConsoleApplication { class Program { static void Main(string[] args) { string startPath = @"c: