本文介绍了如何下载一个zip文件的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我正在尝试从我的 web api 控制器下载一个 zip 文件.它正在返回文件,但当我尝试打开时收到一条消息,zipfile 无效.我看过其他关于这个的帖子,回复是添加了 responseType: 'arraybuffer'.仍然不适合我.我也没有在控制台中收到任何错误.
I am trying to download a zip file from my web api controller. It is returning the file but I am getting a message the zipfile is invalid when i try to open. I have seen other posts about this and the response was adding the responseType: 'arraybuffer'. Still isn't working for me. I'm not getting any errors in the console either.
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
推荐答案
我认为你在错误的地方设置了 responseType,而不是这个:
I think you're setting the responseType in the wrong place, instead of this:
$http.post('/api/apiZipPipeLine/', model)
试试这个:
$http.post('/api/apiZipPipeLine/', model, {responseType:'arraybuffer'})
查看此答案了解更多详情.
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