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如何创建一个 Web API 控制器，该控制器生成并返回从内存中 JPEG 文件（MemoryStream 对象）集合流式传输的压缩 zip 文件。我正在尝试使用 DotNetZip 库。我找到了这个例子：http ://www.4guysfromrolla.com/articles/092910-1.aspx#postadlink 。但是 Response.OutputStream 在 Web API 中不可用，因此该技术不太适用。因此，我尝试将 zip 文件保存到新的 MemoryStream；但它扔了。最后，我尝试使用 PushStreamContent。这是我的代码：

    public HttpResponseMessage Get(string imageIDsList) {
        var imageIDs = imageIDsList.Split(',').Select(_ => int.Parse(_));
        var any = _dataContext.DeepZoomImages.Select(_ => _.ImageID).Where(_ => imageIDs.Contains(_)).Any();
        if (!any) {
            throw new HttpResponseException(new HttpResponseMessage(HttpStatusCode.NotFound));
        }
        var dzImages = _dataContext.DeepZoomImages.Where(_ => imageIDs.Contains(_.ImageID));
        using (var zipFile = new ZipFile()) {
            foreach (var dzImage in dzImages) {
                var bitmap = GetFullSizeBitmap(dzImage);
                var memoryStream = new MemoryStream();
                bitmap.Save(memoryStream, ImageFormat.Jpeg);
                var fileName = string.Format("{0}.jpg", dzImage.ImageName);
                zipFile.AddEntry(fileName, memoryStream);
            }
            var response = new HttpResponseMessage(HttpStatusCode.OK);
            var memStream = new MemoryStream();
            zipFile.Save(memStream); //Null Reference Exception
            response.Content = new ByteArrayContent(memStream.ToArray());
            response.Content.Headers.ContentType = new MediaTypeHeaderValue("application/zip");
            response.Content.Headers.ContentDisposition = new ContentDispositionHeaderValue("attachment") { FileName = string.Format("{0}_images.zip", dzImages.Count()) };
            return response;
        }
    }

zipFile.Save(memStream) 抛出空引用。但是 zipFile 和 memStream 都不是 null 并且没有内部异常。所以我不确定是什么导致了空引用。我对 Web API、内存流的经验很少，而且我以前从未使用过 DotNetZipLibrary。这是对这个问题的跟进：想要一个高效的 ASP.NET Web API 控制器，能够可靠地返回 30 到 50 ~3MB JPEG

有任何想法吗？谢谢！

4

4 回答 4

16

一个更通用的方法是这样工作的：

using Ionic.Zip; // from NUGET-Package "DotNetZip"

public HttpResponseMessage Zipped()
{
    using (var zipFile = new ZipFile())
    {
        // add all files you need from disk, database or memory
        // zipFile.AddEntry(...);

        return ZipContentResult(zipFile);
    }
}

protected HttpResponseMessage ZipContentResult(ZipFile zipFile)
{
    // inspired from http://stackoverflow.com/a/16171977/92756
    var pushStreamContent = new PushStreamContent((stream, content, context) =>
    {
        zipFile.Save(stream);
        stream.Close(); // After save we close the stream to signal that we are done writing.
    }, "application/zip");

    return new HttpResponseMessage(HttpStatusCode.OK) {Content = pushStreamContent};
}

该ZipContentResult方法也可以存在于基类中，并在任何 api 控制器中的任何其他操作中使用。

于 2014-04-11T15:59:17.743 回答
7

在这种情况下，可以使用PushStreamContent类来消除对 MemoryStream 的需求，至少对于整个 zip 文件是这样。它可以这样实现：

    public HttpResponseMessage Get(string imageIDsList)
    {
        var imageIDs = imageIDsList.Split(',').Select(_ => int.Parse(_));
        var any = _dataContext.DeepZoomImages.Select(_ => _.ImageID).Where(_ => imageIDs.Contains(_)).Any();
        if (!any)
        {
            throw new HttpResponseException(new HttpResponseMessage(HttpStatusCode.NotFound));
        }
        var dzImages = _dataContext.DeepZoomImages.Where(_ => imageIDs.Contains(_.ImageID));
        var streamContent = new PushStreamContent((outputStream, httpContext, transportContent) =>
            {
                try
                {
                    using (var zipFile = new ZipFile())
                    {
                        foreach (var dzImage in dzImages)
                        {
                            var bitmap = GetFullSizeBitmap(dzImage);
                            var memoryStream = new MemoryStream();
                            bitmap.Save(memoryStream, ImageFormat.Jpeg);
                            memoryStream.Position = 0;
                            var fileName = string.Format("{0}.jpg", dzImage.ImageName);
                            zipFile.AddEntry(fileName, memoryStream);
                        }
                        zipFile.Save(outputStream); //Null Reference Exception
                    }
                }
                finally
                {
                    outputStream.Close();
                }
            });
        streamContent.Headers.ContentType = new MediaTypeHeaderValue("application/zip");
        streamContent.Headers.ContentDisposition = new ContentDispositionHeaderValue("attachment")
        {
            FileName = string.Format("{0}_images.zip", dzImages.Count()),
        };

        var response = new HttpResponseMessage(HttpStatusCode.OK)
            {
                Content = streamContent
            };

        return response;
    }

理想情况下，使用ZipOutputStream类而不是使用 ZipFile 来动态创建 zip可以使其更加动态地创建。在这种情况下，将不需要每个位图的 MemoryStream。

于 2013-06-24T18:58:18.557 回答
1 
public HttpResponseMessage GetItemsInZip(int id)
    {           
            var itemsToWrite = // return array of objects based on id;

            // create zip file stream
            MemoryStream archiveStream = new MemoryStream();
            using (ZipArchive archiveFile = new ZipArchive(archiveStream, ZipArchiveMode.Create, true))
            {
                foreach (var item in itemsToWrite)
                {
                    // create file streams
                    // add the stream to zip file

                    var entry = archiveFile.CreateEntry(item.FileName);
                    using (StreamWriter sw = new StreamWriter(entry.Open()))
                    {
                        sw.Write(item.Content);
                    }
                }
            }

            // return the zip file stream to http response content                
            HttpResponseMessage responseMsg = new HttpResponseMessage(HttpStatusCode.OK);                
            responseMsg.Content = new ByteArrayContent(archiveStream.ToArray());
            archiveStream.Dispose();
            responseMsg.Content.Headers.ContentDisposition = new ContentDispositionHeaderValue("attachment") { FileName = "test.zip" };
            responseMsg.Content.Headers.ContentType = new MediaTypeHeaderValue("application/zip");

            return responseMsg;          
    }

使用带有 MVC 5 的 Framework .NET 4.6.1

于 2017-07-21T13:46:39.293 回答
0

我刚遇到和你一样的问题。

zipFile.Save(outputStream); //I got Null Reference Exception here.

问题是我正在从内存流中添加文件，如下所示：

zip.AddEntry(fileName, ms);

您所要做的就是将其更改为：

zip.AddEntry(fileName, ms.ToArray());

似乎当作者决定实际写入文件并尝试读取流时，流被垃圾收集或......

干杯!

于 2014-04-30T12:37:18.303 回答