How do I enable another page to open if checkbox is checked

Oscar11 · January 4, 2014

I am designing a site where, if the student is under 19, they get transferred to the client page. Where am I going wrong as it's not working.

Should I use a checkbox or a radiobox?

Many thanks

This is my HTML code for the :

<form name ="contactform" action="send_form_email.php" method="post">

<br>
<table width="900" border="0" align="center" cellpadding="3" cellspacing="0">
<div align="center"><span class="style18">Student Enquiry Form</span></div>

<tr>
   <td></td> <td align="left"> <input type="radio" class="radio" rel="clientdetails.php">
   <?php if($("input[@name=under19]:checked").val()=="private")
{
attr("action"clientdetails.php");
?>php

<span class="style8"> The student is under 19</span></td>
</tr>

This is the PHP code:

<?php

if(isset($_POST['email'])) {
    // CHANGE THE TWO LINES BELOW
        $email_to = "ftnby@yahoo.com";
        $email_subject = "Tuition enquiry form";
   }
function died($error) {
   // your error code can go here
        echo "We're sorry, but there's errors found with the form you submitted.<br /><br />";
        echo $error."<br /><br />";
        echo "Please go back and fix these errors.<br /><br />";

    }

// your required code can go here
   if (empty($_POST["your_first_name"]))
    {$nameErr = "Name is required";}
else
    {$name = test_input($_POST["name"]);}

if (empty($_POST["your_surname"]))
    {$nameErr = "Name is required";}
else
    {$name = test_input($_POST["name"]);}

if (empty($_POST["email"]))
    {$emailErr = "Email is required";}
else
    {$email = test_input($_POST["email"]);}

if (empty($_POST["home_address"]))
    {$nameErr = "Address is required";}
else
    {$name = test_input($_POST["address"]);}

if (empty($_POST["post_code"]))
    {$nameErr = "Post code is required";}
else
    {$name = test_input($_POST["post_code"]);}

if (empty($_POST["landline_number"]))
    {$numberErr = "Please provide a Landline or Mobile number";}
else
    {$number = test_input($_POST["comment"]);}

if (empty($_POST["students_first_name"]))
    {$nameErr = "Name is required";}
else
    {$name = test_input($_POST["name"]);}

if (empty($_POST["students_surname"]))
    {$nameErr = "Name is required";}
else
    {$name = test_input($_POST["name"]);}

if (empty($_POST["subject_required"]))
    {$nameErr = "Subject is required";}
else
    {$name = test_input($_POST["name"]);}

if (empty($_POST["level_of_study"]))
    {$nameErr = "Level_of_Study is required";}
else
    {$name = test_input($_POST["name"]);}

Ch0cu3r · January 4, 2014

this will never work

<?php  if($("input[@name=under19]:checked").val()=="private")
{
attr("action"clientdetails.php");

You cannot mix javascript (JQuery) code together with PHP code. They are completely different languages and both run at completely different times. PHP is ran on the server, but. Javascript runs in the browser long after PHP had parsed the script.

You have two options when the checkbox is checked have JQuery submit the form to another page. Or

Redirect the user to the other page when PHP processes the form.

Edited January 4, 2014 by Ch0cu3r

eldan88 · January 4, 2014

Are you using trying to accomplish this with jquery? If so you can add a value to your radio option named "19" and add an if condition of that value has been selected by using PHP only and not jQuery.

hansford · January 6, 2014

Use a little jQuery to change the action attribute of the form upon submit.

$(document).ready(function() {

	if($('#age_form').length > 0) {

		$('#age_form').submit(function() {

			if($('#youngster:checked').val() == true) {

				$('#age_form').attr('action','http://127.0.0.1/test_page_3.php');
			}
			else {

				$('#age_form').attr('action','http://127.0.0.1/test_page_2.php');	
				
			}
		});

	}
});

Don't forget to add jQuery link in the head of your form:

<head>
<script src="//ajax.googleapis.com/ajax/libs/jquery/1.10.2/jquery.min.js"></script> 
</head>

Here is a sample form:

	<form id="age_form" method="post" action="">
		
		Under 19 years of age: <input type="radio" name="age_restriction" id="youngster" value="young" /><br />
		19 years of age or older: <input type="radio" name="age_restriction" value="old" checked /><br />
		<input type="submit" value="submit"/>
	</form>

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How do I enable another page to open if checkbox is checked

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