1

我正在尝试制作一个包含 pdf 的 zip 文件。当我提取 zip 文件时，pdf 文件已损坏。我在“outputStream”上放了一只手表。这是第一个例外

“outputStream.Length”引发了“System.NotSupportedException”类型的异常 long

这是整个手表 手表

代码

[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;
        }
4

2 回答 2

1

PushStreamContent用于数据流。我认为您应该创建一个临时 zip 文件，并将其传递给Response.

例子

public FileResult ZipFiles(...)
{
   // create a temporary file

    return File("path/to/file.zip", System.Net.Mime.MediaTypeNames.Application.Octet);
}
于 2015-05-14T11:14:28.080 回答
1

问题原来是我没有在pdf中保存任何内容。我只是保存documentUrl。所以通过使用 FileInfo 来改变事情。工作代码

[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);
                            FileInfo fi = new FileInfo(d.DocumentUrl);
                            var fileReadStream = fi.OpenRead();
                            var fileSize = (int)fi.Length;
                            var fileContent = new byte[fileSize];
                            fileReadStream.Read(fileContent, 0, fileSize);
                            zipFile.AddEntry(fileName, fileContent);
                        }
                        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;
        }
    }
}
于 2015-05-16T03:58:29.327 回答