Asp.Net Web Api multiple files with additional data for each

Question

I am trying to send multiple files along with some data for every file. This is my model:

public class FileDTO
{
    [Required]
    public IFormFile File { get; set; }
    [Required]
    public string CategoryName { get; set; }
    [Required]
    public string CategoryDescription { get; set; }
    public string Detail { get; set; }
}

This is my controller:

[HttpPost("Upload/{id:int}")]
public async Task<IActionResult> Upload(int id, IEnumerable<FileDTO> appFileDTOs)
{
    ...
}

Is this even a correct way to do so? How do I send such a request in Postman to simulate it?

Thanks in advance!

Edit

I tried it like this in Postman:

Everything submits correctly besides the image. For some reason the image is always null...

When you it your controller, what is the value of 'var files = HttpContext.Request.Form.Files;' ? — Daboul
– Daboul, Commented Dec 2, 2019 at 9:15

itminus · Accepted Answer · 2019-12-02 09:52:16Z

1

[] represents collection/dictionary index while dot(.) represents there's a property.

So you should rename all the field names with the dot representation.

For example, change

appFileDTOs[0][File]

to

appFileDTOs[0].File

Demo

answered Dec 2, 2019 at 9:52

itminus

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1 Comment

Tom el Safadi Over a year ago

Thanks a lot!! Would be good if there would be a better GUI from Postman for that kind of scenario. The way you have to send the data seems a little bit like a hack to me.

Ashutosh Kushawaha · Accepted Answer · 2019-12-02 09:18:17Z

0

try this it may help you,

send from formData.

in model key send value as

[
{
"CategoryName":"Category1",
"CategoryDescription ":"Category1 Description",
"Detail":"Details "
},
{
"CategoryName":"Category2",
"CategoryDescription ":"Category2 Description",
"Detail":"Details2"
}

]

and for file send first file as file1 and second file as file2;

In server side , remove IEnumerable of FileDTO appFileDTOs from method name. get value of model as

var data = JsonConvert.DeserializeObject<List<FileDTO>>(Request.Form["model"]);

simillary for file

   var fileUpload1 = Request.Form.Files["file1"];
   var fileUpload2 = Request.Form.Files["file2"];

answered Dec 2, 2019 at 9:18

Ashutosh Kushawaha

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1 Comment

Tom el Safadi Over a year ago

Thanks, I will try it later!

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