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How to prevent file declared in <form action="process-form.php"> from showing results to users when form is submitted?

I have a form with a form-handler 'process-form.php. This form works as planned, however, whenever the form is submitted, before redirecting the user to another URL, (say, thank-you.html) it is showing the contents received from the form in process-form.php. How can I prevent that from happening?
if ($mail->send())
{
?>
<script type="text/javascript">
alert('Message Sent!');
window.location.replace("thank.html");
</script>
<?php
}
else
{
echo "<script type='text/javascript'>alert('ERROR!');</script>";
}

You mean shown in the URL, or is it in the HTML of the page? If it is the former, you can use
<form method="POST" ...>
If you mean the latter, you should change your PHP / JavaScript to not show the data

Hey so the issue you're seeing is that not all form fields are filled in so that you get an undefined variable error and you don't want users to see that.
Usually in a production environment (like my shared server) errors are hidden anyway just not on my localhost. To hide the errors though you can just use this:
error_reporting(0);
ini_set('display_errors', 0);
However, You'd be much better off making sure there was no error in the first place. To do this check if the fields were actually filled in otherwise set your variable equal to nothing so technically it will still exist even if it is blank.
if(isset($_POST['my_var'])){
$my_var = $_POST['my_var'];
}else{
$my_var = "";
}
You can also use the same if(isset()) logic where you are actually using the variables rather than where you set them

how to access html element color in php

My html file makes a tag: Good car with like and dislike button. So that user can select like or unlike. Also there is submit button on screen to submit response. Code in html file is as below :
Good Car
Save Feedback
Also I do have PHP file. What I want is that when user clicks submit then my php file/code records the color of my like or dislike button (i applied these in java script code part ) and save in DB.
I don't know how to access the color of like or dislike button from html in PHP. So that if like is green then it means it has been clicked and I could increment the value in DB and similar for for dislike.

The only way i can see how you could do this with php is if you submit the color in a form.
PHP is a server-side language as a comment allready explained. Meaning that it cant directly access everything that happens on a client side to make it simple.
An idea how you could solve your problem tho is:
If a user clicks on a from a get request is submited and it appends an attribute to the url (example.com?color=green)
<form action="/action_page.php" target="_blank" method="GET">
//This is your button
<input type="submit" value="Submit">
//this is the color
<input type="hidden" name="color" value="green">
</form>
in php you can now check if the color get param is set and if it is you can get its value
<?php
var color = "";
if (isset($_GET['color'])) {
color = htmlspecialchars($_GET['color']);
//To check the color use always 3 = to evade type juggling
if(color === "green")
//do something
elseif(color === "red")
//do something else
}
?>
//make sure not to blindly handle user input tho.

Don't know how to solve (Between Php and Javascript)

I don't know how to solve the variable $cat by following script.
"text" variable from Form to javascript and pass to php function to be $Categories_name and to be $cat.
I already test the $cat variable, it is not "String", it is "object", I don't understand.
But I need $cat to be "String".
when Test = "This is Cat" (3 words),
I test $cat by php str_word_count and the output is 1 (I need to correct answer 3);
I test $cat by php var_dump and no output (I need to correct answer "String").
<p id="CaTable"></p>
<script>
function CaFunction()
{
var text = document.getElementById("CategorySelect").value;
document.getElementById("CaTable").innerHTML = "<?php php_catable('" + text + "'); ?>";
}
</script>
<!-- Generate Table by php and MySQL-->
<?php
function php_catable($Categories_name)
{
$cat = $Categories_name;
.................
.................
$sql = "select * from table where xyz = '" .$cat. "'";
}
?>

You are confusing server side code and client side code. Your php code live on the server and can only be executed on the server. And your javascript is on your client's browser and does not know about the server (remember, php generate a text file, and only that text file is sent to the browser). if you want to use the php_catable() function from your client, you will need to do an AJAX call or to redesign your page to do a form submit (just like what Steve is proposing).

Your First Page:
Assuming CategorySelect is a dropdown select box, create a script for its onChange event and create a method="post"post form with a hidden input that goes to "generate_table.php".
<input type="hidden" name="ca_table" id="ca_table" />
You make ca_table a hidden input so php will pick up the value from it when this page gets submitted to a second page where you can generate your table using the php function.
<script language="javascript" type=text/javascript>
function CaFunction(){
documentGetElementById('ca_table').value = documentGetElementById('CategorySelect').value;
submit();
}
</script>
add this to your select dropdown:
onChange="CaFunction();"
Your Receiving Page:
So your receiving page "generate_table.php" would have
<?php
function php_catable($Categories_name)
{
$cat = $Categories_name;
.................
.................
$sql = "select * from table where xyz = '" .$cat. "'";
}
$category_name = $_POST['ca_table']; // cleaned up at least with suitable preg_replace etc
// and call your catable function
php_catable($category_name);
?>
So that way your result will have been posted back to the server as per comments about client side/server side by #Fluinc and answer by #litelite. To get it to do something which performs looking like innerHTML which changes a part of the page without submitting the whole page you will need AJAX, again as per #litelite's answer.
Might get marked down for being dependant on JavaScript but intended mostly to help clarify client v server.
If you want to avoid the JavaScript dependency of this script you could leave out the onChange altogether and add a submit button, then collect $_POST['CategorySelect']; assuming that is its name - ensure it has name="CategorySelect" for php as well as its Id for your css/javascript. Php gets its variable from the item's name.
To get something a bit like the effect of AJAX visually (though the page is still submitted) you could submit the page to itself using action="<?php echo $_SERVER['PHP_SELF']; ?>" on the form and have all the code on the one page. You can put the table generating code in the div where you want the table to appear - it would need a default state set, of course.
#litelite's comment regarding not using posted data directly in an sql query is also vital to prevent attack - make sure you clean it up before you use it!

Perform function but stop page from submitting

I am using the php code to empty my cart on my website if javascript is disabled. Only problem is that it empties the cart but submit the page too. I want it to remain on the same page and empty the cart because the page is supposed to submit for checking out not emptying the cart. Also I can't use javascript for this part because this is meant for user's who don't have javascript enabled.
I have tried using:
return false;
in the post function but it just made the page blank.
Here is my code:
<form method='post' action='checkout.php'>
// some input
<input type='submit' id='cart-empty' name='cartEmpty' value='Empty' />
</form>
if($_POST['cartEmpty']) {
$this->shippingfee = 0;
$this->empty_cart();
}

You mentioned in one of the comments that you weren't able to modify the headers (i.e. header('Location: form.php); because the header information has already been sent. To get around this, you should move your check for the posted cartEmpty value to the top of your submit page.
Headers are delivered any time content is echoed out to the browser (even white space such as blank lines), but if you do this processing prior to any output being sent, you have an opportunity to modify those headers still:
<?php
if($_POST['cartEmpty']) {
$this->shippingfee = 0;
$this->empty_cart();
header("Location: cartEmptied.php");
exit();
}

You can use this php code for validation in checkout.php and return user to page(if user doesn't fill the form):
if(empty($_POST['cartEmpty'])){
header("Location: ./form.php");
die;
}

JavaScript variable passed to PHP

I'm working on a form that adds up the totals selected (via checkboxes). In my JavaScript file, build.js, the totals are added together. On my PHP page, the code takes the items selected on the previous form/HTML page and passes them to what is shown on the PHP page. I want to be able to take the total that was added up via JavaScript on the form page and bring it over to be listed as a total underneath all the options that were selected.
My knowledge of PHP and JavaScript are very rudimentary. This is the first real form I have created in either of these languages. I have poured over this site and the internet in general and have not been able to get any of the options I've found to work. I think I just lucked out on getting the form this far, so I apologize if my code isn't very clean!
Any help would be amazing, as specific as possible please. Here is my code:
The JavaScript that adds the total:
$(document).ready(function() {
$("input[type=checkbox]:checked").attr("checked", false);
function recalculate() {
var sum = 0;
$("input[type=checkbox]:checked").each(function() {
sum += parseInt($(this).attr("rel"));
});
$("#output").html(sum);
}
$("input[type=checkbox]").change(function() {
recalculate();
});
});
Code written on the form itself that shows the total:
<span id="output" class="total"></span><BR><BR>
Code written on the PHP page:
<b>Estimate:</b>
<?php
$aTruck = $_POST['formSelected'];
if(empty($aTruck))
{
echo("You didn't select a truck.<BR><BR>");
}
else
{
$N = count($aTruck);
echo("<h3>Truck Type: ");
for($i=0; $i < $N; $i++)
{
echo($aTruck[$i] . " ");
}}
$aAddons = $_POST['formAddons'];
if(empty($aAddons))
{
echo("You didn't select any options.");
}
else
foreach ($aAddons as $v)
{
echo "<h3> $v </h3>";
}
?>
If I'm not mistaken, the reason I can't currently pass the total is because of something I read on here: the PHP is run on the server while the JavaScript runs on the user's end. My options are thus to send the total in the form (possibly as a hidden variable, which I can't figure out either), pass it along in Ajax (I don't know if the server I'm on is capable of this- possibly so and it's all use error!), or use an XMLHttpRequest. I've tried anything I could find on any of those and either do not have the right variable listed inside, am placing it in the wrong spot, or it's just plain wrong.
As I mentioned, I've poured over the forums for everything I can that's related to this and nothing I've found is specific enough for the tiny bit of understanding I have. Among other things I've tried: Pass a javascript variable value into input type hidden value and Pass Javascript Variable to PHP POST along with using an XMLHttpRequest, using Ajax, passing it as a hidden variable (which I'm leaning towards but don't think I'm implementing correctly) and a ton more- it's pretty much all I did all day at work yesterday so I'm not trying to be redundant with my question- I just can't figure out where I'm going wrong.

It looks like you hit upon it right here:
send the total in the form (possibly as a hidden variable)
Since you're talking about one page posting to another page, and that other page showing the results, then there's no need for AJAX here. You can just use a form value like any other. The "hidden variable" in this case is actually an input element:
<input type="hidden" name="sum" />
In your JavaScript where you're displaying the sum on the first page:
$("#output").html(sum);
You can also set that sum to the form element's value:
$("#output").html(sum);
$("input[name=sum]").val(sum);
As long as that input is inside the same form as the other input elements (like formSelected and formAddons) then when the first page posts to the second page, the code in the second page can access the sum value the same way:
$_POST["sum"]

In your form you should add a hidden input like this :
<input type="hidden" name="sum" value="">
Then in your recalculate() (javasript) function, you should change the value of this input once you calculated everything :
function recalculate() {
var sum = 0;
$("input[type=checkbox]:checked").each(function() {
sum += parseInt($(this).attr("rel"));
});
$("#output").html(sum);
// Change the hidden input value
$("input[name='sum']").val(sum);
}
Now, when your form is submitted, you should access the sum value, server side (PHP), with a simple :
$sum = $_POST['sum'];