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" class="form-control"/>
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);
}
}
