英文:
How to convert WebClient to HttpClient C# WinForms
问题
我需要将已经过时的 webclient 代码转换为 httpclient。
using HttpClient Client = new();
Client.DefaultRequestHeaders.UserAgent.ParseAdd(CompanyName + "/" + ProductName);
HttpContent content = new StreamContent(new MemoryStream(fileimage));
HttpResponseMessage responseMessage = await Client.PostAsync("http://blahblah.com/upload.php", content);
string response = await responseMessage.Content.ReadAsStringAsync();
_ = MessageBox.Show(response, Text);
英文:
I need to convert the webclient code which is obsolte to httpclient.
using WebClient Client = new();
Client.Headers["User-Agent"] = CompanyName + "/" + ProductName;
byte[] responseBytes = Client.UploadFile("http://blahblah.com/upload.php", fileimage);
string response = Encoding.Default.GetString(responseBytes);
_ = MessageBox.Show(response, Text);
I need the exact same way to do this. I have seen other answers but none meet my needs.
答案1
得分: 1
1- 将文件的所有字节读取到upfilebytes数组中。
2- 创建新的HttpClient和MultipartFormDataContent。
3- 将字节数组添加到内容中。
4- 异步上传MultipartFormDataContent内容并将响应存储在response中。
var upfilebytes = File.ReadAllBytes("fileimage");
HttpClient client = new HttpClient();
client.DefaultRequestHeaders.Add("User-Agent", CompanyName + "/" + ProductName);
MultipartFormDataContent content = new MultipartFormDataContent();
ByteArrayContent baContent = new ByteArrayContent(upfilebytes);
content.Add(baContent, "File", "filename.png");
var response = await client.PostAsync("http://blahblah.com/upload.php", content);
var responsestr = await response.Content.ReadAsStringAsync();
MessageBox.Show(response, responsestr);
英文:
1- read AllByte of file into upfilebytes array
2-create new HttpClient and MultipartFormDataContent
3- add byte array to content
4-upload MultipartFormDataContent content async and store response in response.
var upfilebytes = File.ReadAllBytes("fileimage");
HttpClient client = new HttpClient();
client.DefaultRequestHeaders.Add("User-Agent", CompanyName + "/" + ProductName);
MultipartFormDataContent content = new MultipartFormDataContent();
ByteArrayContent baContent = new ByteArrayContent(upfilebytes);
content.Add(baContent, "File", "filename.png");
var response = await client.PostAsync("http://blahblah.com/upload.php", content);
var responsestr = await response.Content.ReadAsStringAsync();
MessageBox.Show(response, responsestr);
答案2
得分: 0
最简单的上传文件方法类似于这样。然而，最好使用HttpClientFactory来生成HttpClient实例。
var client = new HttpClient
{
BaseAddress = new Uri("http://blahblah.com/upload.php")
};
await using var stream = System.IO.File.OpenRead("filePathHere");
using var request = new HttpRequestMessage(HttpMethod.Post, "file");
request.Headers.Add("User-Agent", CompanyName + "/" + ProductName);
using var content = new MultipartFormDataContent
{
{ new StreamContent(stream), "file", "fileNameHere" }
};
request.Content = content;
var response = await client.SendAsync(request);
MessageBox.Show(response, Text);
英文:
The simplest way you can upload a file is something like here.However,
it's better to use HttpClientFactory to generate httpclient instance.
var client = new HttpClient
{
BaseAddress = new("http://blahblah.com/upload.php")
};
await using var stream = System.IO.File.OpenRead("filePathHere");
using var request = new HttpRequestMessage(HttpMethod.Post, "file");
request.Headers.Add("User-Agent",CompanyName + "/" + ProductName);
using var content = new MultipartFormDataContent
{
{ new StreamContent(stream), "file", "fileNameHere" }
};
request.Content = content;
var response = await client.SendAsync(request);
MessageBox.Show(response, Text);
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