I recommend that you use a Custom ModelBinder, and that you send the form data file separately, I would do it this way.public Task CustomModelAsync(ModelBindingContext bindingContext) { if (bindingContext == null) { throw new ArgumentNullException(nameof(bindingContext)); } // Check the value sent in var valueProviderResult = bindingContext.ValueProvider.GetValue(bindingContext.ModelName); if (valueProviderResult != ValueProviderResult.None) { bindingContext.ModelState.SetModelValue(bindingContext.ModelName, valueProviderResult); // Attempt to convert the input value var valueAsString = valueProviderResult.FirstValue; var result = Newtonsoft.Json.JsonConvert.DeserializeObject(valueAsString, bindingContext.ModelType); if (result != null) { bindingContext.Result = ModelBindingResult.Success(result); return Task.CompletedTask; } } return Task.CompletedTask; } The controller is enough that you call the method to upload the file and perform your business logic, then save the route that returns the Upload method: [HttpPut("{id}")] public async Task<IActionResult> Put(int id, [ModelBinder(BinderType = typeof(JsonModelBinder))] Infraestructura model , IList<IFormFile> files) { if (model == null) return BadRequest(); //Traer la entidad que deseas modificar var filesPath = await _fileService.Upload(files,"NombreDelArchivo"); Entidad.RUTA_FOTO_PORTADA = filesPath; //Logia de negocio //Guardar datos return Ok(); } Method to upload a file, which receives a list of File, but you still do well with what you try to do:public async Task<string> Upload(IList<IFormFile> files, string name) { var filesPath = string.Empty; if (files.Count != 0) { foreach (var formFile in files) { if (formFile.Length > 0) { var filePath = Path.Combine(@$"{RootDirectory}{name}{MediaType}"); filesPath += filePath + ";"; await using var stream = File.Create(filePath); await formFile.CopyToAsync(stream); } } filesPath = filesPath.Remove((filesPath.Length-1)); } return filesPath; }

Hi try to put the image as Resource

You're asking a question of what's called the https://es.meta.stackoverflow.com/q/3360/21 I'll start answering what you're asking:You cannot write a relative route that includes the port, or any other element before the root folder of the web server to which your page belongs.In other words: You cannot use relative routes for what you want to achieve. (from there the reference to the XY problem).Let's talk about relative routesTo get any resources, a relative route means just sharing root of the resource route that refers to it.Speaking of https://es.wikipedia.org/wiki/Localizador_de_recursos_uniforme , the farthest you can get with a relative path is the root folder, which is beyond the domain and the possible specification of the port.If your page is on the route:https://www.midominio.com:9972/carpeta1/pagina2.html Any relative path you write within the HTML of that page will depart from https://www.midominio.com:9972/carpeta1/ if the route does not start with /or https://www.midominio.com:9972/ if the route starts with /.This makes sense, and is that any route that doesn't start with https://www.midominio.com:9972/it actually belongs to another server/service, and therefore it is not relative a https://www.midominio.com:9972/.Talking about your problemI understand that you:You have two web servers that have the same folder structure, and you want to refer, in the pages served by the first, resources located in the second.You are writing pages that either will be installed in different domains/servants and therefore you want a solution that is adapt to the address where the page was obtained, since, in any other case, it would be better to write absolute routes in each image.You can't do this with pure HTML, but if you put some javascript, you could do it (to a certain extent).For example, with pure javascript, you can put a id in every image and worth that id to put, already in time of cumshot, the direction of the image, also pulling of the attributes of https://www.w3schools.com/js/js_window_location.asp To get the URL data from where the page was loaded, a simple example would look like this: function ponerruta(imagen) { result = window.location.protocol + "//" + window.location.hostname + ":4450" + "/imagenes/" + imagen; console.log(result); return result; } window.onload = function() { document.getElementById("img1").src = ponerruta("imagen.jpeg"); };<img id="img1" src="/img/placeholder.jpeg"/>With this, you'd have to have a single image in /img/placeholder.jpeg to be shown while real images are loaded, process that would start once the full page has been loaded into the browser.

You tested with a Z-index ? the z-index is to accommodate visualization levels by layer, give a class to that image, and then to that class dale a.tuclase{ z-index: -1; } For example:But send me your codepen and check it out.

It's because it puts int contador=0; every time for the loop. I want it outside (better said, before) the loop.

After fighting for a while I found the solution, the problem was the PUBLIC_HTML folder solve it by adding in index.php $app->bind('path.public', function() { return __DIR__; }); This updates your public route to the current directory. If someone needs more info I got it from here: https://dcblog.dev/laravel-55-change-public-to-public-html

From the JSP page you can access the context of the Web directory. Example of page JSP:<% File f = new File(application.getRealPath("/")); String nombreArchivo = "archivo.jpg"; String ruta = f.getAbsolutePath() + File.separator + "Documentos" + File.separator + "Fotosperfil" + File.separator + nombreArchivo; %>

You don't have to make comparisons, just place the number on the image route.Images are not shown because you are using disk route and not URL, I think all browsers block access for security reasons.<html> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Tirada de dados</title> </head> <body> <?php $num = rand(1,6); // Ya tienes el número, solo muestra la imagen, sin comparaciones echo "<img src=\"dados/$num.jpg\">"; ?> </body> </html>

To redirect you can do it in several ways, either through htaccess with the sentence Redirect 301 / https://example.com/ And the .htaccess you save on the desktop of the image folder or with an index.php document in the image folder and redirect it to the index with a header('Location:https//miweb.com');

To avoid that exception you simply shouldn't start g with nullin fact, that object should not be created Graphics or call. crearImagenes in the builder Panel because you can perfectly redefine the method paintComponent of JPanel So:JPanel panel=new JPanel() { @Override public void paintComponent(Graphics g) { crearImagenes(g); } }; That way you already have the argument g of paintComponent and in turn deliver it to crearImagenes to do his job.Additionally the method crearImagenes I shouldn't have super.paintComponents

You can try clicking on the images and recover another specific image. Try this: function cambio1 () { document.getElementById('cambio1').src = "https://www.w3schools.com/images/lamp.jpg" } function cambio2 () { document.getElementById('cambio2').src = "https://www.w3schools.com/images/w3schools_green.jpg" }</code></pre> <!DOCTYPE html> <html> <body> <h4>Cambio imagenes</h4> <img id="cambio1" onclick="cambio1()" src="https://www.w3schools.com/images/w3schools_green.jpg" alt="Stickman" width="34" height="59"> <img id="cambio2" onclick="cambio2()" src="https://www.w3schools.com/images/lamp.jpg" alt="Lamp" width="32" height="32"> </body> </html>

Tkinter in if you do not support image formats (the most common), supports .ppm files. These generated through objects that are created with the class ImageTkavailable in the library https://en.wikipedia.org/wiki/Python_Imaging_Library (Python Imaging Library).In the case of Tkinter it is concerned to generate a type object Image and based on this to generate another type PhotoImagethrough the module ImageTk:imagen = ImageTk.PhotoImage(Image.open("australia.png")) Like any image that opens with PIL, it is used Image.open(). The difference here is that this object then serves to initialize a type object PhotoImagewhich is specific to Tkinter. Now let's see a practical example:from PIL import Image,ImageTk import tkinter as tk root = tk.Tk() imagen = ImageTk.PhotoImage(Image.open("australia.png")) fondo = tk.Label(root,image = imagen) fondo.pack() #Se ubica la imagen de fundo root.mainloop() The image australia.png I got it from Google, it works the same way you already have. In this case I used pack() for the element location, but can also be used place() or even grid() (the latter I do not recommend). The code results in this: I hope it served you something.

I think you need to open the route with opendir and modify the move_uploaded and the $directory $path= '../img/uploads/'; $imgPath=$path.$_FILES['imagen']['name']; $directorio = $_SERVER['DOCUMENT_ROOT'].$imgPath; $dir = opendir($directorio); move_uploaded_file($_FILES['imagen']['tmp_name'],$directorio); //recuerda que la ruta a colocar es la del servidor closedir($dir);

Here's the solution:In the view form you should go like this:<div class="abmformright"> <?php if(isset($_SESSION['img_prod'])): ?> <?php echo '<img src="data:image/jpeg;base64,'.base64_encode($_SESSION['img_prod']->i_content ).'" />'; ?> <?php endif; ?> </div> Thank you so much!

I work this: you do your folder within your project, but you must check the external image box

To start, do not use the data type image. According to the documentation https://docs.microsoft.com/es-es/sql/t-sql/data-types/ntext-text-and-image-transact-sql?view=sql-server-ver15 and will disappear from the SQL Server in future versions.The recommendation is to use the data type https://docs.microsoft.com/es-es/sql/t-sql/data-types/binary-and-varbinary-transact-sql?view=sql-server-ver15 which is suitable to store binary data, such as an image.You can set the default value using hexadecimal notation, which is 0x followed by a string of hexadecimal numbers that represent the bytes of the image.For example, for the default to be the 3 bytes sequence [00][FF][AA], you can create the table like this:create table Prueba ( numero int , foto varbinary(max) default 0x00FFAA ); go insert into prueba (numero) values (1); go select * from Prueba; go This returns: (1 row affected) numero foto 1 0x00FFAA (1 row affected) Since you are unlikely to write a few Kilobits of information in hexadecimal format, you can read the content of the image file in a local variable and validate the dynamic SQL for generate the creation sentence of the table, which you then direct.I have performed this test that works properly to put my own photograph, stored in c:\test\tono.jpg as default in a table:declare @ImgDefault varbinary(max); declare @SQL nvarchar(max); select @ImgDefault = BulkColumn from OpenRowSet(Bulk 'c:\test\tono.jpg', single_blob) as img; print 'Tamaño del archivo = ' + cast(datalength(@ImgDefault) as varchar); set @SQL = 'create table PruebaImgDefault ( idImg int primary key , foto varbinary(max) default ' + convert(nvarchar(max), @ImgDefault, 1) + ' );'; exec(@SQL); go

The first thing to achieve the purpose you seek is to increase the contrast of the original image so that the almost white is white. To do this we build the following function:private Bitmap AplicaFiltroContrasteBrillo(Image p_ImgSource, float p_Contraste, float p_Brillo) { float brilloAjustado = p_Brillo - 1.0f; // Creamos una matriz para aplicar contraste y brillo // a una imagen dada. float[][] matrizContraste = { new float[] {p_Contraste, 0, 0, 0, 0}, // aplica a RED new float[] {0, p_Contraste, 0, 0, 0}, // aplica a GREEN new float[] {0, 0, p_Contraste, 0, 0}, // aplica a BLUE new float[] {0, 0, 0, 1.0f, 0}, // canal alpha new float[] {brilloAjustado, brilloAjustado, brilloAjustado, 0, 1}}; // Creamos los atributos en función de nuestra matriz // para aplicar a la imagen ColorMatrix cmx = new ColorMatrix(matrizContraste); ImageAttributes imgAttr = new ImageAttributes(); imgAttr.SetColorMatrix(cmx); // Trasladamos la imagen a un bitmap ya con el efecto de contraste // y brillo aplicado para poder trabajar sobre eso Bitmap bmpCanvas = new Bitmap(p_ImgSource.Width, p_ImgSource.Height, p_ImgSource.PixelFormat); Graphics g = Graphics.FromImage(bmpCanvas); g.DrawImage(p_ImgSource, new Rectangle(Point.Empty, bmpCanvas.Size), 0, 0, p_ImgSource.Width, p_ImgSource.Height, GraphicsUnit.Pixel, imgAttr); // Liberamos recursos g.Dispose(); imgAttr.Dispose(); // Retornamos la imagen con el filtro aplicado return bmpCanvas; } Now what we need is an algorithm that captures the image obtained and calculates the rectangle of cuts that we are looking for. It could be optimized and improved, but for the matter I created it as follows:private Rectangle ObtenerBordes(Bitmap p_Bmp) { int bordeIZQ = -1; int bordeDER = -1; int bordeSUP = -1; int bordeINF = -1; // Buscamos el borde SUPERIOR e INFERIOR for (int y = 0; y < p_Bmp.Height; y++) { for (int x = 0; x < p_Bmp.Width; x++) { if (bordeSUP == -1) { Color pixel = p_Bmp.GetPixel(x, y); if (pixel.Name != "ffffffff") bordeSUP = y; } if (bordeINF == -1) { Color pixel = p_Bmp.GetPixel(x, p_Bmp.Height - y-1); if (pixel.Name !="ffffffff") bordeINF = p_Bmp.Height - y - 1; } } if (bordeSUP != -1 && bordeINF != -1) break; } // Buscamos el borde DERECHO e IZQUIERDO for (int x = 0; x < p_Bmp.Width; x++) { for (int y = 0; y < p_Bmp.Height; y++) { if (bordeIZQ == -1) { Color pixel = p_Bmp.GetPixel(x, y); if (pixel.Name != "ffffffff") bordeIZQ = x; } if (bordeDER == -1) { Color pixel = p_Bmp.GetPixel(p_Bmp.Width - x - 1, y); if (pixel.Name != "ffffffff") bordeDER = p_Bmp.Width - x - 1; } } if (bordeIZQ != -1 && bordeDER != -1) break; } return new Rectangle(bordeIZQ, bordeSUP, bordeDER - bordeIZQ, bordeINF - bordeSUP); } Note that the only thing I do is go around the image until you find a pixel that is different from BLANCO and then set the edge on that coordinate.We already have all the work done, just add it to a function that will be the one we will call directly to get the cut of the image:private Image RecortarImage(Image p_ImgSource) { // Aplicamos un filtro a la imagen y retornamos el // resultado a nuestro bitmap // He bajado el brillo a la mitad y aumentado a 6 el contraste // lo que me ha permitido obtener un resultado adecuado al menos // con tu imagen de prueba. Bitmap bmpCanvas = AplicaFiltroContrasteBrillo(p_ImgSource, 6f, 0.5f); // Llamamos a la siguiente función que le pasaremos el Bitmap que hemos // creado con la imagen y que ya tiene un contraste alto para hallar los bordes Rectangle imgRectRecorte = ObtenerBordes(bmpCanvas); // Una vez calculado el rectángulo de recorte creamos un Bitmap para copiar // desde la imagen original a dicho bitmap la porción de imagen correspondiente // según el rectángulo de recorte calculado. Bitmap bmpCROP = new Bitmap(imgRectRecorte.Width, imgRectRecorte.Height); Graphics gCrop = Graphics.FromImage(bmpCROP); gCrop.DrawImage(p_ImgSource, new Rectangle(Point.Empty, bmpCROP.Size), imgRectRecorte, GraphicsUnit.Pixel); // Nos aseguramos de liberar recursos gCrop.Dispose(); // Retornamos la imagen recortada return bmpCROP; }

Problem solved, what you had to do is instead of casting a string on the customer's oeticion, use the decode method('utf-8') in the encoded image and send it in this way, once on the server, use the encode('utf-8') method to take it back to bytes, and then decode it on base 64.

Add this tag and try the following:<div class="loader"> </div> CSSUse the following CSS to show the load image in the div loader. It makes the image sit at the center of the page and covers the entire screen with a semi-transparent white background..loader { position: fixed; left: 0px; top: 0px; width: 100%; height: 100%; z-index: 9999; background: url('https://i.gifer.com/74H8.gif') 50% 50% no-repeat rgb(249,249,249); opacity: .8; } Javascript:First of all, include the jQuery library <script src="https://ajax.googleapis.com/ajax/libs/jquery/3.5.1/jquery.min.js">/script> Now add the following code line to hide the upload image when the page is finished uploading<script type="text/javascript"> $(window).load(function() { $(".loader").fadeOut("slow"); }); </script> This would serve your loader.

By the code you're sharing I think you should include the form inside the body. On the other hand, you can style the Gender label with a block display. Try this:<body> <div> <form id="formulario"> <label for="Usuario">Nombre de usuario</label><br> <input type="text" name="Usuario" class="usuario" method="GET" /><br /> <label for="Correo Electrónico">Correo electrónico</label> <input type="email" name="Correo Electrónico" class="correo-electronico" method="GET" /><br /> <label for="Contraseña">Contraseña</label> <input type="password" name="Contraseña" class="contraseña" method="POST" /> <label for="Género" style="display: block">Género</label> <input type="radio" name="Género" value="masculino" class="masculino" />Masculino<br /> <input type="radio" name="Género" value="femenino" class="femenino" />Femenino<br /> <input type="radio" name="Género" value="otro" class="otro" />Otro<br /> <input type="submit" value="Enviar" class="enviar" action="https://medicine.cheeck.com/pagina.principal" /> </form> </div> <img src="https://upload.wikimedia.org/wikipedia/commons/thumb/3/35/Tux.svg/165px-Tux.svg.png" class="cuadro-de-fondo"/> </body>Note also that it would be interesting for the div to include the entire form.I hope it helped you.