如何下载zip文件

Question

我正在尝试从我的Web API控制器下载一个zip文件。它返回了文件，但是当我尝试打开时，收到了无效的zip文件消息。我已经看到其他有关此问题的帖子，响应是添加responseType：'arraybuffer'。对我来说仍然不起作用。控制台中也没有任何错误。

  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控制器

 [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

- texas697

---

saveAs 函数中包含了什么？ - miensol

这是一个文件保存器（filesaver.js）的方法。 - texas697

Answer 1

我认为你设置了错误的responseType位置，应该修改为：

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

试一下这个：

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

查看此答案以获取更多细节。

- Buddy

---

那个确实起作用了！但是PDF文件损坏了？我需要在我的API控制器中更改一些东西吗？ - texas697

我在我的API中发布了一个异常的截图。我知道这可能是一个单独的问题，但如果您能看一下，我会很感激。 - texas697

似乎不支持get_Length()，也许您需要先下载整个流？ - Buddy

Answer 2

事实上，您是正确的，添加responseType:'arraybuffer'。将其添加到以下代码中，在从Ajax接收响应时，将提示下载文件：

var a = document.createElement('a');
var blob = new Blob([responseData], {'type':"application/octet-stream"});
a.href = URL.createObjectURL(blob);
a.download = "filename.zip";
a.click();

- David Villacis C.

Answer 3

以下代码对于我来说可以很好地下载zip文件，

控制器

$scope.downloadExport = function (id,filename,frmdata) {
         frmdata = {};
        var data = tdioServices.downloadexport(id,filename,frmdata);
         data.success(function(success) {
                var blob = new Blob([success], { type:"arraybuffer" });           
                var downloadLink = angular.element('<a></a>');
                downloadLink.attr('href',window.URL.createObjectURL(blob));
                downloadLink.attr('download', filename);
                downloadLink[0].click();
                $COMMON_ACCEPT = "Download Successfully";
                $COMMON_ACCEPT_TEXT = "Success";
                toastr.success($COMMON_ACCEPT, $COMMON_ACCEPT_TEXT);

        });
    };

服务

 this.downloadexport = function(id,filename,frmdata){
       var request = $http({method:'get', url:APP_URL+'/exports/'+id+'/download/'+filename, data:frmdata,responseType:'arraybuffer'});
       return request;        
    }

- HEMAL