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Asked: 2026-06-17T11:04:58+00:00

Im just learning javascript the last days (started PHP some months ago). So, my

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Im just learning javascript the last days (started PHP some months ago).

So, my code make this:

e.g 

 http://controljuridico.com/video01/

I have three files.

  1. An Html file with the form.

  2. A javascript file with the functions after click on “enviar” (send) button.

  3. A php that process the data.

HTML file (index.html)

        <!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01//EN"
       "http://www.w3.org/TR/html4/strict.dtd">

       <html xmlns="http://www.w3.org/1999/xhtml" lang="en">
        <head>
  <meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>index</title>
<link href="css/css.css" type="text/css" rel="stylesheet" media="screen" />
<script src="js/jquery.js" type="text/javascript" language="JavaScript"></script>
<script src="js/js.js" type="text/javascript" language="JavaScript"></script>
    </head>
      <body>
   <div id="content">
    <h1>Test.</h1>
    <h1>Step 1) choose city</h1>
    <br />
    <br />
    <label>test(*)</label>
    <hr />
    <br />
    <form name="form_data" id="form_data">

        <label>(*) City</label>
        <br />
        Medellin
        <input type="radio" name="ciudad" value="Medellin" />
        Manizales
        <input type="radio" name="ciudad" value="Manizales"/>
        Cali
        <input type="radio" name="ciudad" value="Cali"/>
        <br />
        <label>(*) Especialidad</label>
        <br />
        <input type="button"  value="Enviar" id="btn_enviar" /> 
        <br />
        <br />
        <label id="mensaje"></label>
    </form>
    <div id="resultado" ></div>
        </div>
               </body>
            </html>

js.js File

       $(document).ready(function(){
   $('#btn_enviar').click(function(){

    if( validaRadio( 'ciudad','Ciudad' ) == false) return false;


    $.ajax({
        type:'POST',
        url :'upload.php',
        data: $('#form_data').serialize(),
        beforeSend : function(){
            $('#mensaje').html('Enviando datos...');
        },
        success: function (data){
                        $('#mensaje').html('Datos enviados  correctamente.');
                $('#resultado').html(data); 
        },
        complete: function(){
             $('#form_data').slideUp();
             $('#resultado').slideDown();

        }
    });

        }); 


          });

Php File (upload.php)

 <?php

  $sergu = $_POST['ciudad'];
   if ($_POST['ciudad'] == "Medellin") {
    ?>
    <label>Ciudad Ingresada:</label>
    <br />
    <label><?php echo $_POST['ciudad'] ?></label>
    <hr />
    <?php
    }else{
    echo "it is not medellin....";

   }
       ?>

So, this works very well but. What if I want this:

After click on “enviar” button also show another similar form at the left of this form.

I mean its just like choosing steps if you choose the step one I need another step and so on and I want that this another step just appear to the left of the previus form.

It is possible? how?

Thanks in advance for your help! I really appreciate it.

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1 Answer

  1. Editorial Team

    You could for example just hide the ‘second form’ when the page first loads via css like this:

    <form id="second_form" style="display: none">...

    and when your success function fires you remove the css like so:

    success: function(){
  ...
  $('#second_form').show();
}
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