Upload image using postasync

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Hi,

I would like to upload an image from my project to my web service.

How can I put the image into my PostAsync in order to upload to my PHP web service?

Thanks,
Jassim

What I have tried:

Here is my code:

C#

byte[] fileBytes;

using (var fileStream = await imageEditor.GetStream())
{
    var binaryReader = new BinaryReader(fileStream);

    fileBytes = binaryReader.ReadBytes((int)fileStream.Length);
}

var imageBinaryContent = new ByteArrayContent(fileBytes);

var multipartContent = new MultipartFormDataContent();
multipartContent.Add(imageBinaryContent, "image");

and my PostAsync is like this:

C#

var client = new HttpClient();
client.BaseAddress = new Uri("https://my.domain.com/api/save_data.php");

client.DefaultRequestHeaders.Accept.Clear();
client.DefaultRequestHeaders.Accept.Add(new MediaTypeWithQualityHeaderValue("application/json"));

var content = new FormUrlEncodedContent(new[]
{
    new KeyValuePair<string, string>("image", IMAGE_DATA_COMES_HERE)
});

var response = await client.PostAsync("https://my.domain.com/api/save.php", content);

Posted 11-Sep-19 13:36pm

Jassim Rahma

Updated 13-Sep-19 4:46am

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Richard Deeming · Accepted Answer · 2019-09-13T04:47:00

Solution 1

You can't use FormUrlEncodedContent to upload a file; you have to use MultipartFormDataContent.

C#

var content = new MultipartFormDataContent();
var imageContent = new StreamContent(await imageEditor.GetStream());
content.Add(imageContent, "image");

var response = await client.PostAsync("https://my.domain.com/api/save.php", content);

Posted 13-Sep-19 4:47am

Richard Deeming

Comments

Jassim Rahma 13-Sep-19 22:07pm

but I need to pass more than one parameter like this:

var content = new FormUrlEncodedContent(new[]
{
new KeyValuePair<string, string="">("student", Convert.ToString(student_number)),
new KeyValuePair<string, string="">("photo", STUDENT_PHOTO),
});

so I should pass it as a string?

Richard Deeming 16-Sep-19 5:18am

You cannot use FormUrlEncodedContent to upload a file.

var content = new MultipartFormDataContent();

// Add the image:
var imageContent = new StreamContent(await imageEditor.GetStream());
content.Add(imageContent, "photo");

// Add another parameter:
content.Add(new StringContent(Convert.ToString(student_number)), "\"student\"");

Jassim Rahma 22-Sep-19 5:26am

Yes that is Great. Thanks.

Jassim Rahma 22-Sep-19 8:59am

I have one more question Richard.

I am trying to pass more than one data in addition to the file but it;s not sending anything although my PHP backend works fine when I test

my PHP expect:


$photo = $_FILES["photo"]["tmp_name"];
$user = $_POST["user"];

but now I want to add:


$ad = $_POST["ad"];

so it will be:


$ad = $_POST["ad"];
$photo = $_FILES["photo"]["tmp_name"];
$user = $_POST["user"];

here is what I tried but not working:


var content = new MultipartFormDataContent();

// Add the image:
var imageContent = new StreamContent(await imageEditor.GetStream());
imageContent.Headers.ContentType = new MediaTypeHeaderValue("image/png");
content.Add(imageContent, "photo", "AdImage.png");

// Add another parameter:
content.Add(new StringContent(ad_id), "\"ad\"");
content.Add(new StringContent(App.getUserInfo("MyUserID").Result), "\"user\"");

Jassim Rahma 22-Sep-19 10:12am

sorry it's ok now, it was a problem with the backend PHP