接受变量并将其用作函数

So, I'm trying to turn a variable into a function.

$go_menu=$_POST['menu'];
$go_menu();

Function potoato_chip()
{
    echo "I like Doritos!";
}

Function apples()
{
    echo "I like Fuji";
}

I am getting the data from the $go_menu=$_POST['menu'] but everything below $go_menu(); comes up blank.

The full code is here in a pastebin. Appreciate any help! I'm still learning!

You should not do this: it's highly insecure to execute a variable as a function when you just got that value from a POST! In particular because you're creating an admin interface. Any malicious person could execute other functions in your code if needed.

You should use if statements or a switch statement to control what needs to happen.

Note that Function should be written as function in PHP.

1. Make sure it's callable before calling it:

   $go_menu = $_POST['menu'];
   
   if(is_callable($go_menu))
       $go_menu();
   

2. Keep an array of valid values; this will fix both your "no content" problem and your security problem:

   $valid_actions = array('potato_chip', 'apples');
   
   if(isset($_POST['menu']) && in_array($go_menu = $_POST['menu'], $valid_actions)) {
       $go_menu();
   }
   

First off, your code is incredibly insecure. Let's say I edit your form and submit a POST request setting $_POST['menu'] to, let's say die. This crashes your program. You can just imagine what else I can do with this. A correct way to do this would be with switch:

switch $_POST['menu'] {
    case "menu_1":
        menu_1();
        break;
    case "menu_2":
        menu_1();
        break;
    [...]
}

Generally speaking, I don't like to use variable variables or variable functions, because it's hard to keep track of the possible values of a variable. However, they can still be used if there are properly validated, like in minitech's answer. In your case though, since you seem to have a limited selection of functions that can be called, I would use switch.