How to upload files in asp.net core?
C#File Uploadasp.net Core-MvcImage ManipulationC# Problem Overview
How to upload files or images using Asp.net MVC 6 with some model data? Example, I have a form like this;
<form>
<input type="file">
<input type="text" placeholder="Image name...">
<input type="text" placeholder="Image description...">
<input type="submit" value="Submit">
</form>
I read many Tutorials in how to upload but I don't see anything uploading with some data like the form above.
Also, is there any library for image manipulation for re-sizing and Image Watermarking same as Codeigniter image manipulation class? (https://codeigniter.com/user_guide/libraries/image_lib.html
C# Solutions
Solution 1 - C#
You can add a new property of type IFormFile to your view model
public class CreatePost
{
public string ImageCaption { set;get; }
public string ImageDescription { set;get; }
public IFormFile MyImage { set; get; }
}
and in your GET action method, we will create an object of this view model and send to the view.
public IActionResult Create()
{
return View(new CreatePost());
}
Now in your Create view which is strongly typed to our view model, have a form tag which has the enctype attribute set to "multipart/form-data"
@model CreatePost
<form asp-action="Create" enctype="multipart/form-data">
<input asp-for="ImageCaption"/>
<input asp-for="ImageDescription"/>
<input asp-for="MyImage"/>
<input type="submit"/>
</form>
And your HttpPost action to handle the form posting
[HttpPost]
public IActionResult Create(CreatePost model)
{
var img = model.MyImage;
var imgCaption = model.ImageCaption;
//Getting file meta data
var fileName = Path.GetFileName(model.MyImage.FileName);
var contentType = model.MyImage.ContentType;
// do something with the above data
// to do : return something
}
If you want to upload the file to some directory in your app, you should use IHostingEnvironment to get the webroot path. Here is a working sample.
public class HomeController : Controller
{
private readonly IHostingEnvironment hostingEnvironment;
public HomeController(IHostingEnvironment environment)
{
hostingEnvironment = environment;
}
[HttpPost]
public IActionResult Create(CreatePost model)
{
// do other validations on your model as needed
if (model.MyImage != null)
{
var uniqueFileName = GetUniqueFileName(model.MyImage.FileName);
var uploads = Path.Combine(hostingEnvironment.WebRootPath, "uploads");
var filePath = Path.Combine(uploads,uniqueFileName);
model.MyImage.CopyTo(new FileStream(filePath, FileMode.Create));
//to do : Save uniqueFileName to your db table
}
// to do : Return something
return RedirectToAction("Index","Home");
}
private string GetUniqueFileName(string fileName)
{
fileName = Path.GetFileName(fileName);
return Path.GetFileNameWithoutExtension(fileName)
+ "_"
+ Guid.NewGuid().ToString().Substring(0, 4)
+ Path.GetExtension(fileName);
}
}
This will save the file to uploads folder inside wwwwroot directory of your app with a random file name generated using Guids ( to prevent overwriting of files with same name)
Here we are using a very simple GetUniqueName method which will add 4 chars from a guid to the end of the file name to make it somewhat unique. You can update the method to make it more sophisticated as needed.
Should you be storing the full url to the uploaded image in the database ?
No. Do not store the full url to the image in the database. What if tomorrow your business decides to change your company/product name from www.thefacebook.com to www.facebook.com ? Now you have to fix all the urls in the table!
What should you store ?
You should store the unique filename which you generated above(the uniqueFileName varibale we used above) to store the file name. When you want to display the image back, you can use this value (the filename) and build the url to the image.
For example, you can do this in your view.
@{
var imgFileName = "cats_46df.png";
}
<img src="~/uploads/@imgFileName" alt="my img"/>
I just hardcoded an image name to imgFileName variable and used that. But you may read the stored file name from your database and set to your view model property and use that. Something like
<img src="~/uploads/@Model.FileName" alt="my img"/>
Storing the image to table
If you want to save the file as bytearray/varbinary to your database, you may convert the IFormFile object to byte array like this
private byte[] GetByteArrayFromImage(IFormFile file)
{
using (var target = new MemoryStream())
{
file.CopyTo(target);
return target.ToArray();
}
}
Now in your http post action method, you can call this method to generate the byte array from IFormFile and use that to save to your table. the below example is trying to save a Post entity object using entity framework.
[HttpPost]
public IActionResult Create(CreatePost model)
{
//Create an object of your entity class and map property values
var post=new Post() { ImageCaption = model.ImageCaption };
if (model.MyImage != null)
{
post.Image = GetByteArrayFromImage(model.MyImage);
}
_context.Posts.Add(post);
_context.SaveChanges();
return RedirectToAction("Index","Home");
}
Solution 2 - C#
<form class="col-xs-12" method="post" action="/News/AddNews" enctype="multipart/form-data">
<div class="form-group">
<input type="file" class="form-control" name="image" />
</div>
<div class="form-group">
<button type="submit" class="btn btn-primary col-xs-12">Add</button>
</div>
</form>
My Action Is
[HttpPost]
public IActionResult AddNews(IFormFile image)
{
Tbl_News tbl_News = new Tbl_News();
if (image!=null)
{
//Set Key Name
string ImageName= Guid.NewGuid().ToString() + Path.GetExtension(image.FileName);
//Get url To Save
string SavePath = Path.Combine(Directory.GetCurrentDirectory(),"wwwroot/img",ImageName);
using(var stream=new FileStream(SavePath, FileMode.Create))
{
image.CopyTo(stream);
}
}
return View();
}
Solution 3 - C#
Fileservice.cs:
public class FileService : IFileService
{
private readonly IWebHostEnvironment env;
public FileService(IWebHostEnvironment env)
{
this.env = env;
}
public string Upload(IFormFile file)
{
var uploadDirecotroy = "uploads/";
var uploadPath = Path.Combine(env.WebRootPath, uploadDirecotroy);
if (!Directory.Exists(uploadPath))
Directory.CreateDirectory(uploadPath);
var fileName = Guid.NewGuid() + Path.GetExtension(file.FileName);
var filePath = Path.Combine(uploadPath, fileName);
using (var strem = File.Create(filePath))
{
file.CopyTo(strem);
}
return fileName;
}
}
IFileService:
namespace studentapps.Services
{
public interface IFileService
{
string Upload(IFormFile file);
}
}
StudentController:
[HttpGet]
public IActionResult Create()
{
var student = new StudentCreateVM();
student.Colleges = dbContext.Colleges.ToList();
return View(student);
}
[HttpPost]
public IActionResult Create([FromForm] StudentCreateVM vm)
{
Student student = new Student()
{
DisplayImage = vm.DisplayImage.FileName,
Name = vm.Name,
Roll_no = vm.Roll_no,
CollegeId = vm.SelectedCollegeId,
};
if (ModelState.IsValid)
{
var fileName = fileService.Upload(vm.DisplayImage);
student.DisplayImage = fileName;
getpath = fileName;
dbContext.Add(student);
dbContext.SaveChanges();
TempData["message"] = "Successfully Added";
}
return RedirectToAction("Index");
}
Solution 4 - C#
you can try below code
1- model or view model
public class UploadImage
{
public string ImageFile { get; set; }
public string ImageName { get; set; }
public string ImageDescription { get; set; }
}
2- View page
<form class="form-horizontal" asp-controller="Image" asp-action="UploadImage" method="POST" enctype="multipart/form-data">
<input type="file" asp-for="ImageFile">
<input type="text" asp-for="ImageName">
<input type="text" asp-for="ImageDescription">
</form>
3- Controller
public class Image
{
private IHostingEnvironment _hostingEnv;
public Image(IHostingEnvironment hostingEnv)
{
_hostingEnv = hostingEnv;
}
[HttpPost]
public async Task<IActionResult> UploadImage(UploadImage model, IFormFile ImageFile)
{
if (ModelState.IsValid)
{
var filename = ContentDispositionHeaderValue.Parse(ImageFile.ContentDisposition).FileName.Trim('"');
var path = Path.Combine(Directory.GetCurrentDirectory(), "wwwroot", "images", ImageFile.FileName);
using (System.IO.Stream stream = new FileStream(path, FileMode.Create))
{
await ImageFile.CopyToAsync(stream);
}
model.ImageFile = filename;
_context.Add(model);
_context.SaveChanges();
}
return RedirectToAction("Index","Home");
}
Solution 5 - C#
Controller Code:-
public class HomeController : Controller
{
private readonly ILogger<HomeController> _logger;
private IHostingEnvironment _environment;
public HomeController(ILogger<HomeController> logger,IHostingEnvironment environment)
{
_logger = logger;
_environment = environment;
}
public IActionResult Index()
{
return View();
}
[HttpPost]
public IActionResult uploader()
{
var filelist = HttpContext.Request.Form.Files;
if(filelist.Count>0)
{
foreach(var file in filelist)
{
var uploads = Path.Combine(_environment.WebRootPath, "files");
string FileName = file.FileName;
using (var fileStream = new FileStream(Path.Combine(uploads, FileName), FileMode.Create))
{
file.CopyToAsync(fileStream);
}
}
}
return View("Index");
}
}
View Code:-
@using (Html.BeginForm("uploader", "Home", FormMethod.Post, new { enctype = "multipart/form-data" }))
{
<input type="file" multiple name="file" />
<input type="submit" value="Upload" />
}