public function post(Request $request){ $validator = Validator::make($request->all(),[ ‘title’=>’required’, ‘job_type_id’=>’required|exists:job_type,id’, ‘work_level_id’=>’required|exists:work_level,id’, ‘no_of_candidate’=>’required’, ]); if ( $validator->fails()) { return $this->validationErrors($validator->errors()); } } this is my validation code which i have written in my controller.I just want to remove this validation from controller. So that i can make same Form Request Validation what we do in the laravel ..
Category : validation
When I upgraded the VS Code from version 1.55 to version 1.56 the inline PHP validator stopped working (onType). I downgrade TO 1.55 and everything works. Has anyone had this problem and how can I activate PHP validation in version 1.56? I tried HTML and JavaScript, this two validators works fine Sourc..
i’m trying to make validation update while user types, with laravel but there is no effect how to make validation run dynamically while user typing i used this form it shows what i need should i use ajax or what ? it’s good if some one help me with simple example form: <form action="{{route(‘about.store’)}}" method="POST"> ..
I’m making input validation in my controllers using RespectValidation library. I have two parameters that are optional but dependent, if the "a" parameter is present then I also need the "b" to be here and vice-versa and it is possible that there is none of them. For now i have the following but it allow ..
I have a simple form that sends a value to controller: <form action="{{route(‘mollie.payment’)}}" method="post" > @csrf {{– <input name="text" type="text">–}} <button type="submit" name="test" value="23.00">Send</button> </form> This value will be static and this value cannot be modified. But of course it can be easily changed in Chrome dev tools. How can I better implement this? I ..
New to all this so forgive my ignorance. I am trying to figure out how to change a "Change Your Password" field to my form. Using PHP and MySQL. I work on this a little bit and I want to share for you people that you help me to resolve my error. So first of ..
New to all this so forgive my ignorance. I am trying to figure out how to change a "Change Your Password" field to my form. Using PHP and MySQL. I work on this a little bit and I want to share for you people that you help me to resolve my error. So first of ..
I have a query to check username, email and voucher-code already exists and if voucher-code status = USED in php mysql, I have check every input, code works fine, but the problem is when a user enter a random voucher-code that do not exist in the database, the form will process his input and register ..
i am working on a project where i want to insert only one record in a specific time. For Example: in the project client rate a contractor only once.(when the construction is in working) once he rated he cannot rate second time whatsoever. I checked online but didn’t find what i am looking for. Now ..
Tenho um campo de data chamado ‘inicio’ no formato date (Y-m-d), porém na hora de salvar, quando eu tento colocar um DIA maior que o número 12, por exemplo 30/04/2021, aparece esse erro: CarbonExceptionsInvalidFormatException Could not parse ’30/04/2021′: DateTime::__construct(): Failed to parse time string (30/04/2021) at position 0 (3): Unexpected character Esse é o meu ..
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