Php if Else in form action

You did not echo the variable to show the output:

<?php 
if($tickettype == 'VIP'){
    $action = "seats.php"; 
}else if($tickettype == 'VVIP'){
    $action ="seats1.php";
}
?>
<form action= "<?php echo $action; ?>" method="post">

Notice the <?php echo $action; ?>

Note: This is not the answer, and you should not accept this as answer, I am posting it here instead of posting in the comment section, because it's to long.

Note - You can solve your problem by your self, but for other it is impossible, as your code is just not making any sense at the moment, but I am pretty sure it will if you are able to read and understood what I am trying to say here in the solution.

First thing first,

Access Denied problem –

Note - you don’t have to fix this problem it will get fix by it self, but you should know why you have this problem, so in the future if you get the problem, you will know what to correct.

The reason it denying the access is this, that you are not allow to give special character for form action so after adding the echo in the form you are going to have this problem, because as soon as you add the echo in our form action you unknowingly giving special character to the action, because you never set the value for your variable $action so value for variable $action is equal to error message. So after adding the echo you are displaying that error message.

before doing any thing run your file in the browser and check the source code, and see the value in the form action, it will be some thing like this

<br /><b>Notice</b>: Undefined variable: action in <b>C:\in here you are going to have file path\form.php</b> on line <b>1</b><br />

notice how many angle brackets colon you got in the here, you are not even allowed one angle brackets, so browser is thinking you are trying to do some thing malicious and that's why its blocking your access.

If you don’t know how to check source code, just type how to see source code and give your browser name (don’t ask me it just one click away)

so your code is basically generating error message and displaying it for the action after adding the echo and that's why browser is blocking the access

so why it was not doing the same thing before adding the echo, because in php, if you don’t use echo you are not displaying any thing, so before adding the echo, your action was empty, just remove the echo from the form action and check the source code again, you will understand what I am trying to say.

Again you don’t have to fix it, it will get fix by it self, remember this.

Now the solution start from here

First I have to be right about my assumption, that what you are trying to do,

So these are my assumption, what you want to do,

1. People should be able to buy different different types of ticket, and they can choose the number of ticket from 1 to 30,
2. Than you want to run different file according to the TicketType, in one case you want to run seat.php and in one case you want to run index.php,
3. And to achieve your goal, you created a form, use the if statement to set your action value in the form, as I can see you are using variable $action in the form for action, and you set this value through your if statement.

If I am wrong about any thing up until now, don’t read further it will be waste of time.

basically to achieve your goal, you are using form, array and some logic,

in your case you seems to know array and logic, but you lack pre basic knowledge about the form, and that's why your code is not working.

To be honest. In my opinion you know nothing about the form. That's why it is impossible to give you solution(at least for me). But I am quite sure, if you read this fully you will be able to solve your problem by your self,

so give a quick read to the points, and you will be surprised how easy it is to solve your problem for you, and you will also realized why it is impossible for other to solve it for you, at the moment.

Now Start from here

I am dividing the problem in Four part

1. Some important Notes about the form.
2. $Ticket_Type is not equal $TicketType - an you can not even use $TiccketType here
3. Select option only passes the value from the value tag-- Two part solution
4. Your logic is at the wrong place,

1. Some important point about the form.

1. When you use form, you have to tell php, that you are submitting the form, and if you want to do it only with php, you have to use submit button for it(I already mention this in my comment other day),

2. second you need to know, what happened when you click on the submit button in the form, form will run the action file and also will pass the data from. opening form tag <form> to closing form tag </form> to that file

3. but any data which is out of these opening and closing form tag, form will not pass that data, so point is it only passes the data which is inside the opening form and closing form tag.

Look at the code below to understand it better, and please also run this code, so you can understand it better,

Few points about the code

1. in the below code, we have two input field location and name, location input field is out of the form tag and name input field is inside the form tag.
2. and our form action is seat.php, and name for form file is form.php for this example.
3. as our action is seat.php, so if you click on the submit button, it will pass the data to seat.php, and will run that file.
4. in seat php, we are just using echo to display both value,

see the code below for form.php and for seat.php,

form.php

Location<input type="text" name="location"/><br><br>//out side of the form tag

<form action="seat.php" method="post">
    Name <input type="text" name="name"/><br><br>
    Location <input type="text" name="location"/><br><br>
    <input type="submit" value="Hot Air">
</form>

seat.php

<?php
$name = $_POST['name'];

echo $name.'<br>';

$location = $_POST['location'];

echo $location;

Now go and run form.php and type some thing in the location and name box, and click on the submit button, and if you do that, it will only display you the value for name but for location value it will display you the undefined index location error.

And the reason is, because your form never passed that value for location to seat.php1, becauselocationis out ofform` opening and closing tag,

Now put this Location field inside the form tag

now put that location inside the opening and closing form tag,

after doing that our code should look like this

<form action="seat.php" method="post">
    Name <input type="text" name="name"/><br><br>
    Location <input type="text" name="location"/><br><br>//Now location is inside of the form
    <input type="submit" value="Hot Air">
</form>

Now again run your form.php type some thing for name and for location and click on submit button, this time it will display you both the value, as this time both values are inside the form tag so both value get passed.

Note – If you can understand the difference between the first result and second result, you will be able to understand, why you need to have your select tag inside the opening and closing form tag, you can not have them out side of,

2. $Ticket_Type is not equal to $TicketType – and you can not even use $TicketType in here as it is equal to array you want value from the select option

For Second mistake, pay attention to this portion of your code

echo ' <select name="Ticket_Type">';

    foreach ($TicketType  as $key => $value) {
        echo "<option value=\"$key\">$value</option>";
    }
    echo '</select>';


if($TicketType == 'VIP'){
    $action = "index.php";
}else if ($TicketType == 'VVIP') {
    $action = "seats.php";
}

in here if you read this line echo ' <select name="Ticket_Type">' you give name for select Ticket_Type and than in the if statement You use the variable $TicketType, I am assuming you are checking the value for $TicketType, This problem you can solve it by changing the variable name $TicketType to $Ticket_Type In the if statement.

you have to correct this as well at the same place

So even after apply the above fix, your code wont work, cause you have second problem in this code

Now comes to the second problem, which you made in your if statement , in your if statement in here you checking $TicketType value, but you don’t have value for variable $TicketType because when you use the form php don’t set the variable it set the index so first you have to set the value for $TicketType than only you can use the if statement,

After applying both fixes your code should look like this

Note – (I am changing the name for select to TicketType )

echo ' <select name="Ticket_Type">'; 
    foreach ($TicketType  as $key => $value) {
        echo "<option value=\"$key\">$value</option>";
    }
    echo '</select>';

$Ticket_Type = $_POST['Ticket_Type']; // Now you set the value for variable 
// as your value is set, now you can type your if statement below.

if($Ticket_Type == 'VIP'){
    $action = "index.php";
}else if ($Ticket_Type == 'VVIP') {
    $action = "seats.php";
}

3. Select option, pass value only from the value tag

Third mistake you make and this one you will not notice with this code, but if you don’t learn, you will likely to make that mistake in the future, so correct this mistake as well,

Just pay attention to this code

$TicketType = array('VIP' =>'VIP', 'VVIP' =>'VVIP', 'Upper Box A'=>'Upper Box A', 'Upper Box B'=>'Upper Box B', 'Lower Box A'=>'Lower Box A', 'Lower Box B'=>'Lower Box B', 'General Admission'=>'General Admission') ;

    echo ' <select name="Ticket_Type">';
    foreach ($TicketType  as $key => $value) {
        echo "<option value=\"$key\">$value</option>";
    }
    echo '</select>';

In here you have array, and than you are creating select option from your array which is fine, but you are making one fundamental mistakes in here, form will only passes the value form the value tag so come to this line in your code

echo "<option value=\"$key\">$value</option>";

now change variable to $key to variable $value as well, because when you click on the submit button, again form only passes the value from value.

echo "<option value=\"$key\">$value</option>"; to echo "<option value=\"$value\">$value</option>";

it wont change the out come, for this code, as in your array key and values are same, but you should know what value you are passing for the future.

To Understand what I am trying to say here, run these two code from below, you will understand what I am trying to say.

form.php

<form action="seat.php" method="post">
    <select name="Ticket_Type">
<option value="key1">First</option>
        <option value="key2">Second</option>

    </select>
    <input type="submit" value="Hot Air">
</form>

seat.php

<?php

$value = $_POST['Ticket_Type'];
echo $value;

run this file in the browser and than select the option, and click on the submit button, it will display you key1 or key2 depending on the option you choose, not the first or second, as form only submit the value from value tag now change value tag value to One and two, and run your code again,

<form action="seat.php" method="post">
    <select name="Ticket_Type">;

        <option value="One">First</option>
        <option value="Two">Second</option>

    </select>
    <input type="submit" value="Hot Air">
</form>

Now again run the file in the browser, and chose the option and click on the submit, and see the outcome, and try to understand what is happening.

4 Your logic is at the wrong place

see this code

if($Ticket_Type == 'VIP'){
    $action = "index.php";
}else if ($Ticket_Type == 'VVIP') {
    $action = "seats.php";
}

What your are trying to do here, in your head it make sense, that you will run the different different file, according the ticket type, but in reality it does not make any sense, because when you click on the submit button, than only form will pass the value, to the action file,

this one is hard to explain

ok listen you have this variable $TicketType in this code, it only get set after you click on the submit button, and the action file it never going get set in this files, so you can not have these lines in this file. (and that's why it was showing to access forbidden after adding the echo)

Ok let's say you move this line to the different file, but still what you are trying to do with these lines is still silly, just don’t use these silly lines, and don’t try to run different different file, run single file, means give single action, and lets say action in the form is book.php

so our file name is book.php,

so starting shell for your book.php should look like this

if(!empty($_POST['Ticket_Type'])){

    // what you are basically saying that if TicketType is not empty run this code, 
// if it is empty run the else statement so the user will know that he did not select any ticket, 
//now we are going to add some more code In here

}else 'Hey you did not select any ticket';

After adding the code in the middle our code will look like this

if(!empty($_POST['Ticket_Type'])){

    $Ticket_Type = $_POST['Ticket_Type']; // set the value

    if($Ticket_Type == 'VIP'){ // 
//    Copy and paste the index.php file code here
}else if ($Ticket_Type == 'VVIP') {
    //copy and paste the seat.php file code here
    echo $action;
}// if you have another Ticket_Type just add another else if here, 
//Note: you can add as many else if you like, 

}else 'Hey you did not select any ticket';

so just run single file action

i am still alive, after typing all this, if you have any confusion let me know.

Your Proble is here:

$TicketType = array('VIP' =>'VIP', 'VVIP' =>'VVIP', 'Upper Box A'=>'Upper Box A', 'Upper Box B'=>'Upper Box B', 'Lower Box A'=>'Lower Box A', 'Lower Box B'=>'Lower Box B', 'General Admission'=>'General Admission') ;

And your condition for if statment is

if($TicketType == 'VIP'){
    $action = "index.php";
}else if ($TicketType == 'VVIP') {
    $action = "seats.php";
}

$TicketType is an array and you can't check it with a string, so your $action variable is not setting. and you have to provide index to check.

I had a similar problem where my error read "Object not found! The requested URL was not found on this server. The link on the referr..."

Going off of aria_suhail_123's answer, you simply need to remove any php code surrounding your $action variable in your html like so:

<?php 
    if($tickettype == 'VIP'){
        $action = "seats.php"; 
    }else if($tickettype == 'VVIP'){
        $action ="seats1.php";
    }
?>

<form action= $action method="post">

The reason why is explained by aria_suhail_123, but since his/her response is very long... maybe this will help someone else!

Consider this :

<?php
if(!empty($_POST['tickettype'])){
  
  if($tickettype == "VIP"){
    ?>
<script type="text/javascript">
window.location = "./seats.php";
</script>      
<?php  
}else if ($tickettype == "VVIP") {
?>
<script type="text/javascript">
window.location = "./seats1.php";
</script>      
<?php
        
}   
}else 'Hey you did not select any ticket';

?>