Using PHP to create dynamic forms with MySQL

Using PHP to create dynamic forms with MySQL

By : user2955543
Date : November 22 2020, 10:40 AM
help you fix your problem I've searched the site and was unable to find an answer. , Below is an object-oriented, mysqli prepared statements version
code :
<form method="post" action="">
<select id="domain" name="domain">
$db = new mysqli($DBServer, $DBUser, $DBPass, $DBName, $DBPort);
$stmt = $db->prepare("select `domains` from `domains_subdomains_cop`");
$result = $stmt->get_result();
while($row = $result->fetch_object()){
    echo "<option>".$row->domains."</option>";

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How do I create and process dynamic forms in ASP.NET MVC 3?

How do I create and process dynamic forms in ASP.NET MVC 3?

By : Krishana Kumar
Date : March 29 2020, 07:55 AM
it fixes the issue I would imagine you would probably want to take advantage of Display / EditorTemplates. You would define an interface IQuestion or something, and then have a bunch of different form options that implement that interface. So your model would have a List, and then for each question in the list, Html.EditorFor(item) or so.
Then some kind of standardized way of storing the answers into a table (perhaps a unique save / load method on IQuestion. That's my take anyways. You could define the questions via DB and then your models could have varying counts (and elements) in the List. Just run a DB script (or some kind of admin page) and you could dynamically change the form displayed.
Create dynamic forms

Create dynamic forms

By : CrazyMod
Date : March 29 2020, 07:55 AM
hop of those help? Up to this point I've avoided javascript, mostly because I'm a newbie. But I think I can't avoid it anymore? , Basic javascript approach:
code :
<script type="text/javascript">
function toggle_visibility(id) {
    var e = document.getElementById(id);
    if(e.style.display === 'block')
        e.style.display = 'none';
        e.style.display = 'block';
<input type="button" value="Details" onClick="toggle_visibility('detailsWrapper')">
        function toggle_visibility(id) {
           var control = document.getElementById(controlId);
           if(control.style.visibility == "visible" || control.style.visibility == "")
              control.style.visibility = "hidden";
              control.style.visibility = "visible";
create and submit dynamic forms

create and submit dynamic forms

By : Steph J
Date : March 29 2020, 07:55 AM
I wish this help you You actually don't need to make a unique index or unique name. The HTTP protocol supports sending multiple data points with the same name.
For example, this is totally fine: &name=me&name=you&name=them&name=her. On the back end, depending on which framework and language you are using, you simply get an array.
code :
<label> Product 1 <input name="product_name" type="text" /></label>
<label> Product 2 <input name="product_name" type="text" /></label>
Create dynamic forms in ASP.NET MVC

Create dynamic forms in ASP.NET MVC

By : Alex The Programmer
Date : March 29 2020, 07:55 AM
hop of those help? I have an ASP.NET MVC application that tracks vendors for our company by purchase orders. Each PO rating is assigned a type. Each type can have multiple rating sections and each rating section can have multiple categories. The classes look like that: , You should change your foreach to for, and the binding will work!
How to create a dynamic admin panel with php forms and mysql?

How to create a dynamic admin panel with php forms and mysql?

By : Adityarso
Date : March 29 2020, 07:55 AM
will help you First off, if this question is duplicated, please let me know, as I am unsure as to how to search this particular question. , 1) Type Button
Change Line
code :
<a onclick='myFunction()' href='#'><span class='glyphicon glyphicon-trash'></span></a>
<input type='submit' value='del'><span class='glyphicon glyphicon-trash'></span>
<a onclick='myFunction()' href='#'><span class='glyphicon glyphicon-trash'></span></a>
<a href='#' onclick='formSubmit();'>Example to post<span class='glyphicon glyphicon-trash'></span></a>
function formSubmit()
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