I have a form with an input element in my view.

<input asp-for="Image" id="mainImageInput" type="file" onchange="readURL(this,'mainImage')" accept="image/jpeg, image/png" />

I am storing all my images in a database as byte[]. How can I convert byte[] to IFormFile and send it to my view in the input element? And no, I am not asking how to display the image, I want to insert it directly into the input element as if the user has selected it using "Browse".

I tried writing this, but it doesn't work:

var ad = _context.Advertisements
            .Select(x => new {
                Id = x.Id,
                Title = x.Title,
                Description = x.Description,
                Price = x.Price,
                Location = x.Location,
                CategoryId = x.CategoryId,
                ImageInBase64 = Convert.ToBase64String(x.Image),
                UserId = x.UserId,
                ImageInByteArray=x.Image
            })
            .FirstOrDefault(x=>x.Id==id);

IFormFile file = null;
using (var stream = new MemoryStream(ad.ImageInByteArray))
{
    file = new FormFile(stream, 0, ad.ImageInByteArray.Length, "name", "fileName");
}

var viewModel = new AdvertisementViewModel()
            {
                Id = ad.Id,
                Title = ad.Title,
                Description = ad.Description,
                Price = ad.Price,
                Location = ad.Location,
                CategoryId = ad.CategoryId,
                ImageInBase64 = ad.ImageInBase64,
                UserId = ad.UserId,
                Image = file
            };
return View("Edit",viewModel);

This is the action where I should receive my image after submiting my form, all other properties are there, but I am getting null for the Image.

        [HttpPost]
        [ValidateAntiForgeryToken]
        [Authorize]
        public async Task<IActionResult> Edit(AdvertisementViewModel viewModel) 
        {
            // viewmodel.Image is null, so my model state will always be invalid
            // need a workaround

            if (!ModelState.IsValid)
            {
                return View(viewModel);
            }
        }

CodePudding user response：

And no, I am not asking how to display the image, I want to insert it directly into the input element as if the user has selected it using "Browse".

Due to the security reason in Broswer, I think we can't set value to <input type='file' />.

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