使用 MemoryStream 和 ZipArchive return 将文件压缩到 asp.net web api 中的客户端

Question

我正在尝试 return 使用以下代码将文件从 asp.net 网络 api 压缩到客户端：

private byte[] CreateZip(string data)
{
    using (var ms = new MemoryStream())
    {
        using (var ar = new ZipArchive(ms, ZipArchiveMode.Create, true))
        {
            var file = archive.CreateEntry("file.html");

            using (var entryStream = file.Open())
            using (var sw = new StreamWriter(entryStream))
            {
                sw .Write(value);
            }
        }
        return memoryStream.ToArray();
    }
}

public HttpResponseMessage Post([FromBody] string data)
{
    HttpResponseMessage result = new HttpResponseMessage(HttpStatusCode.OK);
    result.Content = new ByteArrayContent(CreateZip(data));
    result.Content.Headers.ContentType = new MediaTypeHeaderValue("application/zip, application/octet-stream");
    return result;
}

当我 运行 此代码时，我收到以下错误：

ExceptionMessage":"The format of value 'application/zip, application/octet-stream' is invalid."

这是JS代码：

$.ajax({
  type: "POST",
  url: url,
  data: data,
  dataType: application/x-www-form-urlencoded
});

任何解释为什么会这样？我真的很感谢你们的帮助

Answer 1

您传递给 MediaTypeHeaderValue 的构造函数的值的格式无效。您还试图将多种内容类型添加到 header 值。

内容类型header采用单个type/subtype，后跟用semi-colons;

分隔的可选参数

例如：

Content-Type: text/html; charset=ISO-8859-4

对于您的结果，您需要决定要使用哪一个。 application/zip 或 application/octet-stream

result.Content.Headers.ContentType = new MediaTypeHeaderValue("application/zip");

此外，为避免异常，您可以使用 MediaTypeHeaderValue.TryParse 方法

var contentTypeString = "application/zip";
MediaTypeHeaderValue contentType = null;
if(MediaTypeHeaderValue.TryParse(contentTypeString, out contentType)) {
    result.Content.Headers.ContentType = contentType;
}

Answer 2

$.ajax 处理文本响应并将尝试 (utf-8) 解码内容：您的 zip 文件不是文本，您将得到损坏的内容。 jQuery 不支持二进制内容，因此您需要使用 this link 并在 jQuery 上添加 ajax 传输或直接使用 XmlHttpRequest。使用 xhr，您需要设置 xhr.responseType = "blob" 并从 xhr.response 读取 blob。

// with xhr.responseType = "arraybuffer"
var arraybuffer = xhr.response;
var blob = new Blob([arraybuffer], {type:"application/zip"});
saveAs(blob, "example.zip");

// with xhr.responseType = "blob"
var blob = xhr.response;
saveAs(blob, "example.zip");
Edit: examples:

with jquery.binarytransport.js（任何允许您下载 Blob 或 ArrayBuffer 的库都可以）

$.ajax({
  url: url,
  type: "POST",
  contentType: "application/json",
  dataType: "binary", // to use the binary transport
  // responseType:'blob', this is the default
  data: data,
  processData: false,
  success: function (blob) {
    // the result is a blob, we can trigger the download directly
    saveAs(blob, "example.zip");
  }
  // [...]
});

使用原始 XMLHttpRequest，您可以看到 this 问题，您只需添加一个 xhr.responseType = "blob" 即可获得一个 blob。

我个人建议你在jQuery上使用ajax传输，这很简单，你必须下载一个库，将其包含在项目中并写：dataType: "binary".

这是 API 代码，使用 DotNetZip (Ionic.Zip):

   [HttpPost]
    public HttpResponseMessage ZipDocs([FromBody] string[] docs)
    {
        using (ZipFile zip = new ZipFile())
        {
            //this code takes an array of documents' paths and Zip them
            zip.AddFiles(docs, false, "");
            return ZipContentResult(zip);
        }
    }

    protected HttpResponseMessage ZipContentResult(ZipFile zipFile)
    {
        var pushStreamContent = new PushStreamContent((stream, content, context) =>
        {
          zipFile.Save(stream);
            stream.Close(); 
        }, "application/zip");

        return new HttpResponseMessage(HttpStatusCode.OK) { Content = pushStreamContent };
    }

Answer 3

这是对我有用的解决方案

C#端

public IActionResult GetZip([FromBody] List<DocumentAndSourceDto> documents)
{
    List<Document> listOfDocuments = new List<Document>();

    foreach (DocumentAndSourceDto doc in documents)
        listOfDocuments.Add(_documentService.GetDocumentWithServerPath(doc.Id));

    using (var ms = new MemoryStream())
    {
        using (var zipArchive = new ZipArchive(ms, ZipArchiveMode.Create, true))
        {
            foreach (var attachment in listOfDocuments)
            {
                var entry = zipArchive.CreateEntry(attachment.FileName);

                using (var fileStream = new FileStream(attachment.FilePath, FileMode.Open))
                using (var entryStream = entry.Open())
                {
                    fileStream.CopyTo(entryStream);
                }
            }

        }
        ms.Position = 0;
        return File(ms.ToArray(), "application/zip");
    }

    throw new ErrorException("Can't zip files");
}

不要错过这里的ms.Position = 0;

正面 (Angular 4) :

downloadZip(datas: any) {
    const headers = new Headers({
        'Content-Type': 'application/json',
        'Accept': 'application/zip'
    });

    const options = new RequestOptions({ headers: headers, withCredentials: true, responseType: ResponseContentType.ArrayBuffer });
    const body = JSON.stringify(datas);
    return this.authHttp.post(`${environment.apiBaseUrl}api/documents/zip`, body, options)
        .map((response: Response) => {
            const blob = new Blob([response.blob()], { type: 'application/zip' });
            FileSaver.saveAs(blob, 'logs.zip');
        })
        .catch(this.handleError);
}

现在我可以将多个文件下载到 zip。

Answer 4

这个适合asp.net核心版本

    [HttpGet("api/DownloadZip")]
    public async Task<IActionResult> Download()
    {
        var path = "C:\test.zip";
        var memory = new MemoryStream();
        using (var stream = new FileStream(path, FileMode.Open))
        {
            await stream.CopyToAsync(memory);
        }

        memory.Position = 0;
        return File(memory, GetContentType(path), Path.GetFileName(path));
    }

然后使用Web客户端调用

      class Program
    {

        static string url = "http://localhost:5000/api/DownloadZip";

        static async Task Main(string[] args)
        {
            var p = @"c:\temp1\test.zip";

            WebClient webClient = new WebClient();

            webClient.DownloadFile(new Uri(url), p);                       

            Console.WriteLine("ENTER to exit...");
            Console.ReadLine();
        }
    }