Send an HTTP POST request with C#

Question

I'm try to Send Data Using the WebRequest with POST But my problem is No data has be streamed to the server.

string user = textBox1.Text;
string password = textBox2.Text;  

ASCIIEncoding encoding = new ASCIIEncoding();
string postData = "username" + user + "&password" + password;
byte[] data = encoding.GetBytes(postData);

WebRequest request = WebRequest.Create("http://localhost/s/test3.php");
request.Method = "POST";
request.ContentType = "application/x-www-form-urlencoded";
request.ContentLength = data.Length;

Stream stream = request.GetRequestStream();
stream.Write(data, 0, data.Length);
stream.Close();

WebResponse response = request.GetResponse();
stream = response.GetResponseStream();

StreamReader sr99 = new StreamReader(stream);
MessageBox.Show(sr99.ReadToEnd());

sr99.Close();
stream.Close();

here the result

Solution

It's because you need to assign your posted parameters with the = equal sign:

byte[] data = Encoding.ASCII.GetBytes(
    $"username={user}&password={password}");

WebRequest request = WebRequest.Create("http://localhost/s/test3.php");
request.Method = "POST";
request.ContentType = "application/x-www-form-urlencoded";
request.ContentLength = data.Length;
using (Stream stream = request.GetRequestStream())
{
    stream.Write(data, 0, data.Length);
}

string responseContent = null;

using (WebResponse response = request.GetResponse())
{
    using (Stream stream = response.GetResponseStream())
    {
        using (StreamReader sr99 = new StreamReader(stream))
        {
            responseContent = sr99.ReadToEnd();
        }
    }
}

MessageBox.Show(responseContent);

See the username= and &password= in post data formatting.

You can test it on this fiddle.

Edit :

It seems that your PHP script has parameters named diffently than those used in your question.

Comments

[abdallah]

I try it But Still the same problem

[Fabien ESCOFFIER]

I updated my answer. I also added a fiddle so you can test yourself, it does work.

[Fabien ESCOFFIER]

It's because your PHP script accept parameters that are named differently thant you exposed in your question.

[Fabien ESCOFFIER]

What are the parameters name that your PHP script use ?

[abdallah]

How I Can make it without adding "=&Login=Login" ?

[Fabien ESCOFFIER]

It seems your issue come from your PHP script, as you can see the fiddle work.