Question

我想要实现上传zip文件的api。

我的功能适用于文本文件但不适用于zip文件。 Zip文件已保存但无法打开。你知道这对此很好吗？

在客户端，我会在下一步行动中致电api：

 [HttpPost]
        public async Task<IActionResult> Upload(ICollection<IFormFile> files)
        {
            using (var client = new HttpClient())
            {
                foreach (var file in files)
                {
                    if (file.Length > 0)
                    {
                        var fileName = ContentDispositionHeaderValue.Parse(file.ContentDisposition).FileName.Trim('"');
                        var fileContent = new StreamContent(file.OpenReadStream());
                        fileContent.Headers.Add("X-FileName", fileName);
                        fileContent.Headers.Add("X-ContentType", file.ContentType);

                        var response = await client.PostAsync(url2, fileContent);
                    }
                }
            }
            return View(nameof(this.Index));

        }

这是我的api：

 [HttpPost]
    public async Task<IActionResult> Post()
    {

        var input = new StreamReader(Request.Body).ReadToEnd();
        var fileName = Request.Headers["X-FileName"];
        var fileType = Request.Headers["X-ContentType"];

        using (var sw = new StreamWriter(@"C:\" + fileName))
        {
            sw.Write(input);
        }

        await Task.FromResult(0);
        return new ObjectResult(true);
    }

Answer 1

这是我的解决方案：

在客户端API上

        [HttpPost("{lastModified}")]
        public async Task<string> Upload(long lastModified)
        {
            using (var client = new HttpClient())
            {

                foreach (var file in Request.Form.Files)
                {
                    if (file.Length > 0)
                    {
                        var fileName = ContentDispositionHeaderValue.Parse(file.ContentDisposition).FileName.Trim('"');
                        var fileContent = new StreamContent(file.OpenReadStream());
                        var archiveUrl = "path to api with 2 parameters {fileName}/{lastModified}";
                        var datasetResponse = await client.PostAsync(archiveUrl, fileContent);
                        var dataset = await datasetResponse.Content.ReadAsStringAsync();
                        return dataset;
                    }
                }

                throw new ApplicationException("Cannot updated dataset to archive");
            }
        }

在服务器API上

        [HttpPost("{fileName}/{lastModified}")]
        public async Task<IActionResult> Post(string fileName, long lastModified)
        {           
            var dataSet = getDataSet(); 

            return new ObjectResult(dataSet);
        }

如何上传文件mvc6？

1 个答案: