Drop down menu selection ditactes what form will d

Question

I'm currently working on a project that saves details of different types of objects to a database e.g. book, webpage and journal article. To save the different attributes of these objects I am trying to get different forms to display that depend on the selection in a drop down menu.

Here's the dropdown menu:

<div class="dropdown">
        <button class="btn btn-default dropdown-toggle" type="button" id="dropdownMenu1" data-toggle="dropdown">
            Select Reference Type...
            <span class="caret"></span>
        </button>
        <ul class="dropdown-menu" role="menu" aria-labelledby="dropdownMenu1">
            <li role="presentation"><a role="menuitem" tabindex="-1" href="book.php">Book</a></li>
            <li role="presentation"><a role="menuitem" tabindex="-1" href="journal.php">Journal</a></li>
            <li role="presentation"><a role="menuitem" tabindex="-1" href="webpage.php">Webpage</a></li>
        </ul>
    </div>

How to I get a different form to load on screen without redirecting to a different page. I've been trying to do this in php but I get the feeling that php isn't the right way of going about doing this. Also, apologies in advance as I have no previous experience in Javascript, AJAX or jQuery.

Answer 1

you can also keep all the 3 forms on the same page and can hide or show them on the basis of the dropdown selection.

Lets say there are 3 forms: 1. book 2. webpage 3. journal

Create forms for all three in html, and by default keep all of them hidden(for that we will use a class "display-none"), after that detect change in the dropdown on basis of whom the form populates and do action as required. Please look at the following peace of code, it might help:

<style type="text/css">
    .display-none {
        display: none;
    }
</style>

<select id="dropdown" onchange="myFunction()">
    <option value="-1">-- Select --</option>
    <option value="book">Book</option>
    <option value="webpage">Webpage</option>
    <option value="journal">Journal</option>
</select>
<form id="book" class="display-none" method="post" action="where-you-want-to-post/book">
    <input type="submit" value="Submit Book" />
</form>
<form id="webpage" class="display-none" method="post" action="where-you-want-to-post/webpage">
    <input type="submit" value="Submit Webpage" />
</form>
<form id="journal" class="display-none" method="post" action="where-you-want-to-post/journal">
    <input type="submit" value="Submit Journal" />
</form>

<script type="text/javascript">

    function myFunction() {
        var allForms = document.getElementsByTagName('form');
        var dropdown = document.getElementById("dropdown");
        if (dropdown.value != "-1") {
            var form = document.getElementById(dropdown.value);
            for (var i = 0; i < allForms.length; i++) {
                allForms[i].setAttribute("class", "display-none");
            }
            form.setAttribute("class", "");
        }
    }
</script>

Demo: http://jsfiddle.net/oynjj3jn/

Answer 2

In order to accomplish your goal, you would need the following

A menu in your HTML page like this one. Let's say that the values of the select-list will be your PHP files, just as it's shown below

<!-- HTML -->
<p>Select reference type...</p>
<!-- This is the menu -->
<select id="menu">
  <option value="book.php">Book</option>
  <option value="jurnal.php">Jurnal</option>
  <option value="webpage.php">Webpage</option>  
</select>

<!-- This is the container for your HTML content -->
<div id="content"></div> 

You'll need this snippet to make AJAX requests to your server

// JavaScript
var s = document.getElementById('menu');     // reference to the <select> menu
var c = document.getElementById('content');  // reference to the content <div> 
s.onchange = function(){                     // hook the change event on <select>
  xhr = new XMLHttpRequest();
  var url = 'your-domain/' + this.value;     // here we're passing the selected value
  xhr.open("GET", url, true);
  xhr.onreadystatechange = function () { 
    if (xhr.readyState == 4 && xhr.status == 200) {  // wait for the response
      c.innerHTML = xhr.responseText;                // here it comes; populate the <div> 
    }
  };
  xhr.send(); 
};

And these are your PHP files on the server-side; they may contain any valid PHP / HTML code

// PHP
// book.php or jurnal.php or webpage.php
echo 'anything';

Answer 3

Okay, so without knowing the rest of your code, I would suggest that the best option would be to have the different forms in separate documents. For example

HTML

<div class="dropdown">
    <button class="btn btn-default dropdown-toggle" type="button" id="dropdownMenu1" data-toggle="dropdown">
        Select Reference Type...
        <span class="caret"></span>
    </button>
    <ul class="dropdown-menu" role="menu" aria-labelledby="dropdownMenu1">
        <li role="presentation"><a onclick="bookinclude()" role="menuitem" tabindex="-1" href="book.php">Book</a></li>
        <li role="presentation"><a role="menuitem" tabindex="-1" href="journal.php">Journal</a></li>
        <li role="presentation"><a role="menuitem" tabindex="-1" href="webpage.php">Webpage</a></li>
    </ul>
</div>
<div id="contentwrapper">
    Some Initial Content Here
</div>

Javascript

function bookinclude(){
  $("#contentwrapper").fadeOut(400);
  setTimeout(function(){$("#contentwrapper").load("book.php").fadeIn();}, 400); 
};

And remember to include Jquery!In the head of your HTML:

<script src="//ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

Basically, on the click of the book link, it will load the book.php into the contentwrapper div.

I think this is what you want? All you need to do is replicate the function, and the link, but replace the book with journal, and with webpage.

Hope this helps!