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Telegram BOT Api: how to send a photo using C#? – Telegram API

The sendPhoto command require an argument photo defined as InputFile or String.

The API doc tells:

Photo to send. You can either pass a file_id as String to resend a photo
that is already on the Telegram servers, or upload a new photo using
multipart/form-data.
And

InputFile

This object represents the contents of a file to be uploaded. Must be
posted using multipart/form-data in the usual way that files are
uploaded via the browser.

Tags:

2

Answers


  1. 0

    I’m not a C# Developer but I generated this code using Postman, it uses RestSharp lib

    var client = new RestClient("https://api.telegram.org/bot%3Ctoken%3E/sendPhoto");
var request = new RestRequest(Method.POST);
request.AddHeader("postman-token", "7bb24813-8e63-0e5a-aa55-420a7d89a82c");
request.AddHeader("cache-control", "no-cache");
request.AddHeader("content-type", "multipart/form-data; boundary=---011000010111000001101001");
request.AddParameter("multipart/form-data; boundary=---011000010111000001101001", "-----011000010111000001101001rnContent-Disposition: form-data; name="photo"; filename="[object Object]"rnContent-Type: falsernrnrn-----011000010111000001101001rnContent-Disposition: form-data; name="chat_id"rnrn2314123rn-----011000010111000001101001--", ParameterType.RequestBody);
IRestResponse response = client.Execute(request);

    Just tweak it and it should work.

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  2. 0

    here is a working, parametrized code sample:

    using System.Linq;
using System.IO;
using System.Text;
using System.Net.Http;
using System.Threading.Tasks;

namespace ConsoleApplication
{
    public class Program
    {
        public static void Main(string[] args)
        {
            SendPhoto(args[0], args[1], args[2]).Wait();
        }

        public async static Task SendPhoto(string chatId, string filePath, string token)
        {
            var url = string.Format("https://api.telegram.org/bot{0}/sendPhoto", token);
            var fileName = filePath.Split('\').Last();

            using (var form = new MultipartFormDataContent())
            {
                form.Add(new StringContent(chatId.ToString(), Encoding.UTF8), "chat_id");

                using (FileStream fileStream = new FileStream(filePath, FileMode.Open, FileAccess.Read))
                {
                    form.Add(new StreamContent(fileStream), "photo", fileName);

                    using (var client = new HttpClient())
                    {
                        await client.PostAsync(url, form);
                    }
                }
            }
        }
    }
}
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