angularjs - 如何下载一个zip文件

Question

我正在尝试从Web API Controller 下载一个zip文件。它正在返回文件，但是当我尝试打开时，我收到一条消息，指出zipfile无效。我还看到了其他有关此的帖子，并且响应中添加了responseType:'arraybuffer'。仍然不能为我工作。我在控制台中也没有收到任何错误。

  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');
            notificationFactory.success();
    });

api Controller

 [HttpPost]
    [ActionName("ZipFileAction")]
    public HttpResponseMessage ZipFiles([FromBody]int[] id)
    {
        if (id == null)
        {//Required IDs were not provided
            throw new HttpResponseException(new HttpResponseMessage(HttpStatusCode.BadRequest));
        }

        List<Document> documents = new List<Document>();
        using (var context = new ApplicationDbContext())
        {
            foreach (int NextDocument in id)
            {
                Document document = context.Documents.Find(NextDocument);

                if (document == null)
                {
                    throw new HttpResponseException(new HttpResponseMessage(HttpStatusCode.NotFound));
                }

                documents.Add(document);
            }
            var streamContent = new PushStreamContent((outputStream, httpContext, transportContent) =>
            {
                try
                {
                    using (var zipFile = new ZipFile())
                    {
                        foreach (var d in documents)
                        {
                            var dt = d.DocumentDate.ToString("y").Replace('/', '-').Replace(':', '-');
                            string fileName = String.Format("{0}-{1}-{2}.pdf", dt, d.PipeName, d.LocationAb);
                            zipFile.AddEntry(fileName, d.DocumentUrl);
                        }
                        zipFile.Save(outputStream); //Null Reference Exception
                    }
                }

                finally
                {
                    outputStream.Close();
                }
            });
            streamContent.Headers.ContentType = new MediaTypeHeaderValue("application/octet-stream");
            streamContent.Headers.ContentDisposition = new ContentDispositionHeaderValue("attachment");
            streamContent.Headers.ContentDisposition.FileName = "reports.zip";

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

更新资料

Answer 1

我认为您在错误的地方设置了responseType，而不是这样:

$http.post('/api/apiZipPipeLine/', model)

尝试这个:

$http.post('/api/apiZipPipeLine/', model, {responseType:'arraybuffer'})

查看 this answer了解更多详细信息。