Set the default value of an input field
This is one way of doing it: document.getElementById(“nameofid”).value = “My value”;
This is one way of doing it: document.getElementById(“nameofid”).value = “My value”;
How about adding a space in the pattern attribute like pattern=”[a-zA-Z0-9 ]+”. If you want to support any kind of space try pattern=”[a-zA-Z0-9\s]+”
This gives you the root folder: System.AppDomain.CurrentDomain.BaseDirectory You can navigate from here using .. or ./ etc.. , Appending .. takes you to folder where .sln file can be found For .NET framework (thanks to Adiono comment) Path.GetFullPath(Path.Combine(AppDomain.CurrentDomain.BaseDirectory,”..\\..\\”)) For .NET core here is a way to do it (thanks to nopara73 comment) Path.GetFullPath(Path.Combine(AppContext.BaseDirectory, “..\\..\\..\\”)) ;
Use [ngValue] instead of “value”: <select [(ngModel)]=”selected.isConnected” id=”etat”> <option [ngValue]=”0″>Not connected</option> <option [ngValue]=”1″>Connected</option> </select>
You can do this from the twig template: {{ form_widget(form.birthdate, { ‘attr’: {‘class’: ‘calendar’} }) }} From http://symfony.com/doc/current/book/forms.html#rendering-each-field-by-hand
There is no meaning to say that WPF is better than windows forms or vice versa. It depends on many factors: What kind of UI you are building. Obviously, the complexity of the views you are designing will factor in to performance on both platforms. They have different layout and rendering pipelines. How effectively you …
Best practice: one form per product is definitely the way to go. Benefits: It will save you the hassle of having to parse the data to figure out which product was clicked It will reduce the size of data being posted In your specific situation If you only ever intend to have one form element, …
.nodeName gives the html tag used so you have to use .type to get the type of the node there. Try this one: console.log(event.target.type); Demo
Here is the complete example for previewing image before it gets upload. HTML : <html> <head> <link class=”jsbin” href=”http://ajax.googleapis.com/ajax/libs/jqueryui/1/themes/base/jquery-ui.css” rel=”stylesheet” type=”text/css” /> <script class=”jsbin” src=”http://ajax.googleapis.com/ajax/libs/jquery/1/jquery.min.js”></script> <script class=”jsbin” src=”http://ajax.googleapis.com/ajax/libs/jqueryui/1.8.0/jquery-ui.min.js”></script> <meta charset=utf-8 /> <title>JS Bin</title> <!–[if IE]> <script src=”http://goo.gl/r57ze”></script> <![endif]–> </head> <body> <input type=”file” onchange=”readURL(this);” /> <img id=”blah” src=”#” alt=”your image” /> </body> </html> JavaScript : function …
You can use the following: return Redirect::back()->withInput(Input::all()); If you’re using Form Request Validation, this is exactly how Laravel will redirect you back with errors and the given input. Excerpt from \Illuminate\Foundation\Validation\ValidatesRequests: return redirect()->to($this->getRedirectUrl()) ->withInput($request->input()) ->withErrors($errors, $this->errorBag());