如何下载zip文件
如何下载zip文件
提问于 2015-05-11 08:36:09
我正在尝试从我的web api控制器下载zip文件。它正在返回文件，但当我尝试打开时，我收到一条消息，该zipfile无效。我看过其他关于这方面的帖子，回复是添加了responseType：'arraybuffer‘。对我还是不起作用。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
回答 3
Stack Overflow用户
回答已采纳
发布于 2015-05-14 13:45:00
我认为您在错误的位置设置了responseType，而不是这样：
$http.post('/api/apiZipPipeLine/', model)试试这个：
$http.post('/api/apiZipPipeLine/', model, {responseType:'arraybuffer'})有关更多详细信息，请查看this answer。
Stack Overflow用户
发布于 2016-03-12 14:18:08
事实上，添加responseType:'arraybuffer'是正确的。当接收到来自ajax的响应时，添加到以下代码中将提示下载一个文件：
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();Stack Overflow用户
发布于 2020-03-16 15:03:53
下面的代码对我来说下载zip文件很好，
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;
}页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/30158115
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