我有后端端点Task<ActionResult> Post(IFormFile csvFile)
,需要从HttpClient调用此端点。目前,我正在获取Unsupported media type error
。
这是我的代码:
var filePath = Path.Combine("IntegrationTests", "file.csv");
var gg = File.ReadAllBytes(filePath);
var byteArrayContent = new ByteArrayContent(gg);
var postResponse = await _client.PostAsync("offers", new MultipartFormDataContent
{
{byteArrayContent }
});
答案 0 :(得分:3)
您需要在MultipartFormDataContent
集合中指定匹配操作参数名称(csvFile
)的参数名称和随机文件名
var multipartContent = new MultipartFormDataContent();
multipartContent.Add(byteArrayContent, "csvFile", "filename");
var postResponse = await _client.PostAsync("offers", multipartContent);
或同等
var postResponse = await _client.PostAsync("offers", new MultipartFormDataContent {
{ byteArrayContent, "csvFile", "filename" }
});
答案 1 :(得分:2)
使用以下代码段:
const string url = "https://localhost:5001/api/Upload";
const string filePath = @"C:\Path\To\File.png";
using (var httpClient = new HttpClient())
{
using (var form = new MultipartFormDataContent())
{
using (var fs = File.OpenRead(filePath))
{
using (var streamContent = new StreamContent(fs))
{
using (var fileContent = new ByteArrayContent(await streamContent.ReadAsByteArrayAsync()))
{
fileContent.Headers.ContentType = MediaTypeHeaderValue.Parse("multipart/form-data");
// "file" parameter name should be the same as the server side input parameter name
form.Add(fileContent, "file", Path.GetFileName(filePath));
HttpResponseMessage response = await httpClient.PostAsync(url, form);
}
}
}
}
}
答案 2 :(得分:1)
这对我来说是通用的
public static Task<HttpResponseMessage> PostFormDataAsync<T>(this HttpClient httpClient, string url, string token, T data)
{
var content = new MultipartFormDataContent();
foreach (var prop in data.GetType().GetProperties())
{
var value = prop.GetValue(data);
if (value is FormFile)
{
var file = value as FormFile;
content.Add(new StreamContent(file.OpenReadStream()), prop.Name, file.FileName);
content.Headers.ContentDisposition = new ContentDispositionHeaderValue("form-data") { Name = prop.Name, FileName = file.FileName };
}
else
{
content.Add(new StringContent(JsonConvert.SerializeObject(value)), prop.Name);
}
}
if (!string.IsNullOrWhiteSpace(token))
httpClient.DefaultRequestHeaders.Authorization = new AuthenticationHeaderValue("Bearer", token);
return httpClient.PostAsync(url, content);
}
答案 3 :(得分:0)
通过使用以下代码解决:
const string fileName = "csvFile.csv";
var filePath = Path.Combine("IntegrationTests", fileName);
var bytes = File.ReadAllBytes(filePath);
var form = new MultipartFormDataContent();
var content = new StreamContent(new MemoryStream(bytes));
form.Add(content, "csvFile");
content.Headers.ContentDisposition = new ContentDispositionHeaderValue("form-data")
{
Name = "csvFile",
FileName = fileName
};
content.Headers.Remove("Content-Type");
content.Headers.Add("Content-Type", "application/octet-stream; boundary=----WebKitFormBoundaryMRxYYlVt8KWT8TU3");
form.Add(content);
//Act
var postResponse = await _sellerClient.PostAsync("items/upload", form);
答案 4 :(得分:0)
将附件作为 MultipartFormDataContent
发布var type = typeof(Startup);
var stream = type.Assembly.GetManifestResourceStream(type, "Resources.New.docx");
var fileContent = new StreamContent(stream);
var data = new MultipartFormDataContent
{
{ fileContent, "file", "New.docx" }
};
var response = await _client.PostAsync("upload", multipartContent);
来源:https://medium.com/@woeterman_94/c-webapi-upload-files-to-a-controller-e5ccf288b0ca