Closed. This question needs details or clarity. It is not currently accepting answers. Want to improve this question? Add details and clarify the problem by editing this post. Closed 5 months ago. Improve this question I need to show/hide three forms depending on dropdown selected option without using conditional statements. Is it possible? Answer You can do it like so,
Tag: forms
Update MySQL Database With Data From Multiple Input Elements Inside A While Loop, And A Single Submit Button Outside of The While Loop
I have a page that fetches images from a database after an initial upload, that allows the user to add titles and tags to the images. I have set the while loop to output the individual images inside one form. Each image has two related input fields. What I would like to do is have it so when the form
Rendering conditional HTML elements with PHP
I’m trying to learn PHP to solve an Assesment challenge, where I have to build a Product List & Add Product pages. Reading here and there, watching some tutorials I was able to develop this code so far: Index.php: Thing is, inputs for DVD Size, BOOK Weight and Forniture dimensions ( H, W & L ) should be render depending
display the same form with new updated data after click on save button in PHP
I have a form that show the user information and when I load userpage.php all info appear in each field and user can edit his information all is good until the user click on save button, it have to show the same form with updated information, data updated in database but the new data doesn’t appear when user click save.
How to scale the number of form fields based on user input?
I’m new to PHP and making a website to add an arbitrary number of values in a given base. How would I generate several fields based on a user’s input in a previous field? Answer The simple code without any validation will be like this:
Input from Model not included in form POST data
I have a simple HTML POST form with a model contained within like so: https://jsfiddle.net/pilotman/rn9gspz8/6/ note: the JS fiddle is basic and just for demo and is not perfect. So when I submit the form and let the input inside the model has some text, my server script does not see ‘input2’ Does the data inside the model get sent
Form Processing php
I am new here so please point me in the right direction if I am misunderstanding how to submit a question here. 1. I am getting warning errors stating that I have “Undefined array keys”. I get this before and after the form has been submitted. I have defined the arrays at the top so I am unsure why they
required attribute not working in form in codeigniter php
i have a shopping website in codeigniter, where have a product details page, in which user should select the color and size which is mandotory before adding to cart, so i did the following code: however, the required attriubute is not working, user is able to click the add to cart button, can anyone please tell me how to fix
How to display $name Input that a user submitted in a form within a $alert when user pushes the submit button?
I am stuck and hoping someone can help. I am trying to get the user’s input (their $name) to show in the $alert” when user submit form. Example: User types in to the form the following: ** Name: Joe , Tel: XXX.XXX.XXXX , Email: Joe@example.com *** and pushes submit. — page refreshes and says. “Message sent Joe, thank you for
PHP form validation not functioning having copied the tutorial code
I am hoping the community can give me a little insight into what is not working with my code, I am following a Udemy course. I have followed the accompanying video which developed an undefined variable error, which after doing some research I believe I have fixed by declaring variables as empty strings being able to be over-ridden by the