Javascript 尝试使用AngularJs和c#webapi从服务器下载zip文件

我知道有类似标题的帖子存在，但这对我不起作用，我是如何做到这一点的：

WebApi

public async Task<HttpResponseMessage> ExportAnalyticsData([FromODataUri] int siteId, [FromODataUri] string start, [FromODataUri] string end) {
    DateTime startDate = Date.Parse(start);
    DateTime endDate = Date.Parse(end);

    using (ZipFile zip = new ZipFile()) {
        using (var DailyLogLanguagesCsv = new CsvWriter(new StreamWriter("src"))) {
            var dailyLogLanguages = await _dbContext.AggregateDailyLogSiteObjectsByDates(siteId, startDate, endDate).ToListAsync();
            DailyLogLanguagesCsv.WriteRecords(dailyLogLanguages);
            zip.AddFile("src");
        }


        using (var DailyLogSiteObjectsCsv = new CsvWriter(new StreamWriter("src"))) {
            var dailyLogSiteObjects = await _dbContext.AggregateDailyLogSiteObjectsByDates(siteId, startDate, endDate).ToListAsync();
            DailyLogSiteObjectsCsv.WriteRecords(dailyLogSiteObjects);
            zip.AddFile("src");
        }

        zip.Save("src");
        HttpResponseMessage result = null;
        var localFilePath = HttpContext.Current.Server.MapPath("src");

        if (!File.Exists(localFilePath)) {
            result = Request.CreateResponse(HttpStatusCode.Gone);
        } else {
            // Serve the file to the client
            result = Request.CreateResponse(HttpStatusCode.OK);
            result.Content = new StreamContent(new FileStream(localFilePath, FileMode.Open, FileAccess.Read));
            result.Content.Headers.ContentDisposition = new System.Net.Http.Headers.ContentDispositionHeaderValue("attachment");
            result.Content.Headers.ContentDisposition.FileName = "Analytics";
        }
        return result;
    }
}

当我下载一个文件时，我会得到该文件已损坏的信息。它只在我返回

zip

文件时发生。它适用于

csv

在@wannadream建议和编辑我的代码之后

                else
            {
                // Serve the file to the client
                result = Request.CreateResponse(HttpStatusCode.OK);
                result.Content = new StreamContent(new FileStream(localFilePath, FileMode.Open, FileAccess.Read));
                result.Content.Headers.ContentDisposition = new System.Net.Http.Headers.ContentDispositionHeaderValue("attachment");
                result.Content.Headers.ContentDisposition.FileName = "Analytics";
                result.Content.Headers.ContentType = new MediaTypeHeaderValue("application/octet-stream");
            }

当我试图打开下载的

zip

时，我遇到了这样的问题。 zip.AddFile（“src”）；然后zip.Save（“src”）？这没有道理

您正在压缩目标名称为“src”的“src”。请为zip文件尝试其他名称

zip.Save("target")

var localFilePath = HttpContext.Current.Server.MapPath("target");

尝试设置此选项：

result.Content.Headers.ContentType = new MediaTypeHeaderValue("application/octet-stream");

zip.AddFile（“src”）；然后zip.Save（“src”）？这没有道理

您正在压缩目标名称为“src”的“src”。请为zip文件尝试其他名称

zip.Save("target")

var localFilePath = HttpContext.Current.Server.MapPath("target");

尝试设置此选项：

result.Content.Headers.ContentType = new MediaTypeHeaderValue("application/octet-stream");

---

尝试通过普通浏览器访问WebAPI控制器操作，看看它下载的ZIP文件是否可以打开。如果不能，那么问题出在WebAPI上。

尝试通过普通浏览器访问WebAPI控制器操作，看看它下载的ZIP文件是否可以打开。如果不能，那么问题就出在WebAPI中。

我通过设置类型responseType来解决它

{ type: "application/octet-stream", responseType: 'arraybuffer' }

在我的服务中也是一样

$http.get(serviceBase + path), {responseType:'arraybuffer'});

---

我通过设置类型responseType来解析它

{ type: "application/octet-stream", responseType: 'arraybuffer' }

在我的服务中也是一样

$http.get(serviceBase + path), {responseType:'arraybuffer'});

这可以通过使用并将响应类型设置为arraybuffer来完成，请检查下面的代码以获得完整的理解

1.WebApi控制器

        [HttpPost]
        [Route("GetContactFileLink")]
        public HttpResponseMessage GetContactFileLink([FromBody]JObject obj)
        {
                string exportURL = "d:\\xxxx.text";//replace with your filepath

                var fileName =  obj["filename"].ToObject<string>();

                exportURL = exportURL+fileName;

                var resullt = CreateZipFile(exportURL);


                return resullt;
        }

private HttpResponseMessage CreateZipFile(string directoryPath)
        {
            try
            {
                HttpResponseMessage response = Request.CreateResponse(HttpStatusCode.OK);

                using (ZipFile zip = new ZipFile())
                {
                    zip.AlternateEncodingUsage = ZipOption.AsNecessary;
                    zip.AddFile(directoryPath, "");
                    //Set the Name of Zip File.
                    string zipName = String.Format("Zip_{0}.zip", DateTime.Now.ToString("yyyy-MMM-dd-HHmmss"));
                    using (MemoryStream memoryStream = new MemoryStream())
                    {
                        //Save the Zip File to MemoryStream.
                        zip.Save(memoryStream);

                        //Set the Response Content.
                        response.Content = new ByteArrayContent(memoryStream.ToArray());

                        //Set the Response Content Length.
                        response.Content.Headers.ContentLength = memoryStream.ToArray().LongLength;

                        //Set the Content Disposition Header Value and FileName.
                        response.Content.Headers.ContentDisposition = new ContentDispositionHeaderValue("attachment");
                        response.Content.Headers.ContentDisposition.FileName = zipName;

                        //Set the File Content Type.
                        response.Content.Headers.ContentType = new MediaTypeHeaderValue("application/zip");
                        return response;
                    }
                }
            }
            catch(Exception ex)
            {
                throw new ApplicationException("Invald file path or file not exsist");
            }
        }

这可以通过使用并将响应类型设置为arraybuffer来完成，请检查下面的代码以获得完整的理解

1.WebApi控制器

        [HttpPost]
        [Route("GetContactFileLink")]
        public HttpResponseMessage GetContactFileLink([FromBody]JObject obj)
        {
                string exportURL = "d:\\xxxx.text";//replace with your filepath

                var fileName =  obj["filename"].ToObject<string>();

                exportURL = exportURL+fileName;

                var resullt = CreateZipFile(exportURL);


                return resullt;
        }

private HttpResponseMessage CreateZipFile(string directoryPath)
        {
            try
            {
                HttpResponseMessage response = Request.CreateResponse(HttpStatusCode.OK);

                using (ZipFile zip = new ZipFile())
                {
                    zip.AlternateEncodingUsage = ZipOption.AsNecessary;
                    zip.AddFile(directoryPath, "");
                    //Set the Name of Zip File.
                    string zipName = String.Format("Zip_{0}.zip", DateTime.Now.ToString("yyyy-MMM-dd-HHmmss"));
                    using (MemoryStream memoryStream = new MemoryStream())
                    {
                        //Save the Zip File to MemoryStream.
                        zip.Save(memoryStream);

                        //Set the Response Content.
                        response.Content = new ByteArrayContent(memoryStream.ToArray());

                        //Set the Response Content Length.
                        response.Content.Headers.ContentLength = memoryStream.ToArray().LongLength;

                        //Set the Content Disposition Header Value and FileName.
                        response.Content.Headers.ContentDisposition = new ContentDispositionHeaderValue("attachment");
                        response.Content.Headers.ContentDisposition.FileName = zipName;

                        //Set the File Content Type.
                        response.Content.Headers.ContentType = new MediaTypeHeaderValue("application/zip");
                        return response;
                    }
                }
            }
            catch(Exception ex)
            {
                throw new ApplicationException("Invald file path or file not exsist");
            }
        }

---

我有src到我自己的文件，我只是不想显示它。我在返回单个csv文件时工作。但当我将所有这些文件都包含到zip中，然后将其返回时，我会得到文件已损坏的信息。请从服务器端打开zip文件。检查是否可以从服务器上打开。如果我从服务器端将此文件保存在光盘上，我可以打开此文件确定我稍后再试，因为我现在无法访问代码，并将提供我在那里的信息src到我自己的文件，我只是不想显示它。我在返回单个csv文件时工作。但当我将所有这些文件都包含到zip中，然后将其返回时，我会得到文件已损坏的信息。请从服务器端打开zip文件。检查是否可以从服务器打开。如果我从服务器端将此文件保存在光盘上，我可以打开此确定我稍后再试，因为我现在无法访问代码，并将提供信息如果我从服务器端将此文件保存在光盘上，我可以打开此尝试设置“{type:“application/zip”}”到“{type:“application/octet stream”}”如果我从服务器端将此文件保存在光盘上，我可以打开此尝试设置“{type:“application/zip”}”到“{type:“application/octet stream”}”