从服务器下载zip文件并解析它

Question

I am trying to download a zipped file from the server and trying to show the content of each files in zipped folder to the view.

I wrote a separate code where the file is on my laptop and I ran across each file and dislpayed the content such as

static void Main(string[] args)
{
     string filePath = "C:\\ACL Data\\New folder\\files.zip";
     var zip= new ZipInputStream(File.OpenRead(filePath));
     var filestream=new FileStream(filePath,FileMode.Open,FileAccess.Read);
     ZipFile zipfile = new ZipFile(filestream);
     ZipEntry item;
     while ((item = zip.GetNextEntry()) != null)
     {
         Console.WriteLine(item.Name);
         using (StreamReader s = new StreamReader(zipfile.GetInputStream(item)))
         {
             Console.WriteLine(s.ReadToEnd());
         }
     }
     Console.Read();
 }

I am using sharplibzip library to implement this This is the case when the zip file is located locally in the system. My next task scenario is what if the zipped file is located on the server. I am figuring out the way to implement it, below is the code what I assume should work

static void Main(string[] args)
{
    string url = "https://test/code/304fd9c6-7e53-42a2-845a-624608bfd2ce.zip";
    WebRequest webRequest = WebRequest.Create(url);
    webRequest.Method = "GET";
    WebResponse webResponse = webRequest.GetResponse();
    var zip = new ZipInputStream(webResponse.GetResponseStream());
    ZipEntry item1;

    //var zip= new ZipInputStream(File.OpenRead(filePath));
    var filestream = new FileStream(filepath, FileMode.Open, FileAccess.Read);
    ZipFile zipfile = new ZipFile(filestream);
    ZipEntry item;
    while ((item = zip.GetNextEntry()) != null)
    {
        Console.WriteLine(item.Name);
        using (StreamReader s = new StreamReader(zipfile.GetInputStream(item)))
        {
            Console.WriteLine(s.ReadToEnd());
        }
    }
    Console.Read();
} 

I am stuck at this part: var filestream = new FileStream(filepath, FileMode.Open, FileAccess.Read);

This expect the first parameter to be path of the zip file. Since in the new scenario zip file is located remotely on the server. What should be the parameter in this case?

Answer 1

Your original code opens the stream twice on the following rows, which I think is causing some confusion:

var zip= new ZipInputStream(File.OpenRead(filePath));
var filestream=new FileStream(filePath,FileMode.Open,FileAccess.Read);

There is an overload to the ZipFile constructor that takes "any" Stream rather than specifically a FileStream, which you - unsurprisingly - can only create for files.

However, you cannot use the stream returned by GetResponseStream directly, because it's CanSeek property is false . This is because it's a NetworkStream, which can only be read once from beginning to end. SharpZipLib needs random access to read the file contents.

Depending on the size of the ZIP file, loading it in memory may be an option. If you expect large files, writing it to a temporary file may be better.

This should do the trick, without using both ZipInputStream and ZipFile , by enumerating through ZipFile instead:

string url = "https://test/code/304fd9c6-7e53-42a2-845a-624608bfd2ce.zip";
WebRequest webRequest = WebRequest.Create(url);
webRequest.Method = "GET";
WebResponse webResponse = webRequest.GetResponse();

using (var responseStream = webResponse.GetResponseStream())
using (var ms = new MemoryStream())
{
    // Copy entire file into memory. Use a file if you expect a lot of data
    responseStream.CopyTo(ms);

    var zipFile = new ZipFile(ms);

    foreach (ZipEntry item in zipFile)
    {
        Console.WriteLine(item.Name);
        using (var s = new StreamReader(zipFile.GetInputStream(item)))
        {
            Console.WriteLine(s.ReadToEnd());
        }
    }
}
Console.Read();

PS: starting .NET 4.5, there is support for ZIP files built in. See the ZipArchive class.