Share files easily
You can also upload files from the command line using
curl -F "file=@yourfile.ext" https://file.xeon.kr/upload
Invoke-RestMethod -Uri "https://file.xeon.kr/upload" -Method Post -Form @{
file = Get-Item -Path "yourfile.ext"
}
import requests
with open('yourfile.ext', 'rb') as f:
files = {'file': f}
response = requests.post('https://file.xeon.kr/upload', files=files)
print(response.json())
package main
import (
"bytes"
"fmt"
"io"
"mime/multipart"
"net/http"
"os"
)
func main() {
file, _ := os.Open("yourfile.ext")
defer file.Close()
body := &bytes.Buffer{}
writer := multipart.NewWriter(body)
part, _ := writer.CreateFormFile("file", "yourfile.ext")
io.Copy(part, file)
writer.Close()
req, _ := http.NewRequest("POST", "https://file.xeon.kr/upload", body)
req.Header.Set("Content-Type", writer.FormDataContentType())
client := &http.Client{}
resp, _ := client.Do(req)
defer resp.Body.Close()
respBody, _ := io.ReadAll(resp.Body)
fmt.Println(string(respBody))
}
use reqwest::blocking::multipart;
fn main() -> Result<(), Box<dyn std::error::Error>> {
let form = multipart::Form::new()
.file("file", "yourfile.ext")?;
let client = reqwest::blocking::Client::new();
let res = client.post("https://file.xeon.kr/upload")
.multipart(form)
.send()?;
println!("{}", res.text()?);
Ok(())
}
import okhttp3.*;
import java.io.File;
import java.io.IOException;
public class Upload {
public static void main(String[] args) throws IOException {
OkHttpClient client = new OkHttpClient();
File file = new File("yourfile.ext");
RequestBody requestBody = new MultipartBody.Builder()
.setType(MultipartBody.FORM)
.addFormDataPart("file", file.getName(),
RequestBody.create(MediaType.parse("application/octet-stream"), file))
.build();
Request request = new Request.Builder()
.url("https://file.xeon.kr/upload")
.post(requestBody)
.build();
try (Response response = client.newCall(request).execute()) {
System.out.println(response.body().string());
}
}
}