Question

I’m making a zip file and i want to return a downloadable file but I cant figure out the return file part even tho i tried some examples i found online.

this is the code i use for this now, it does work without the $.post() but i need it to work with jquery. all ideas are welcome

$('.zipFiles').live('click', function () {
            $.post('/Home/ZipFiles');
});


//return a file
public FileResult ZipFiles()
    {
        var filesToZip = Session["DownloadQue"] as List<string>;
        var savedZipFile = Server.MapPath("~/App_Data/") + DateTime.Now.Minute + ".zip";

        if (filesToZip != null && filesToZip.Count > 0)
            using (var zip = new ZipFile(savedZipFile))
            {                    
                foreach (string item in filesToZip)
                {
                    var path = Server.MapPath(Path.Combine("~/Pics/", item));
                    zip.AddFile(path, @"\cf");
                }

                zip.Comment = "this was made online";                    
                zip.Save();
            }

        return File(savedZipFile, System.Net.Mime.MediaTypeNames.Application.Zip);
    }

Answer 1

You may return the content of file or result from an action via jQuery. Here in your code /home/zipfiles return .zip content and I think that you don’t want to show/print zip content.

$('.zipFiles').live('click', function () {
            $.post('/Home/ZipFiles',"",function(data){
               alert(data);
            });
});

To download a file try this:

$('.zipFiles').live('click', function () {
   window.location.href="/home/zipfiles";              
 });