nested functions in php throws an exception when the outer is called more than once

Question

lest assume that i have the following

function a(){
  function b(){}
}
a(); //pass
a(); //error

why in the second call an exception is thr

Answer 1

It's exactly what is says, when you call a() again it tries to redeclare b(), declare b() outside of a() and call b() from within a() like so:

function a() {
  b();
}

function b() {}

a();
a();

Answer 2

Declaring a function within another function like that would be considered bad practice in php. If you really need a function inside a() you should create a closure.

function a() {
  $b = function() {

  };
}

Answer 3

This is because when you execute function a it declares function b. Executing it again it re-declares it. You can fix this by using function_exists function.

function a(){
  if(!function_exists('b')){
    function b(){}
  }
}

But what I suggest is, you should declare the function outside. NOT inside.

Answer 4

It's because you're calling a() in a global scope. Add a function_exists call to make the above code work, but really there are few scenarios where you should really do something like this.

Answer 5

Named functions are always global in PHP. You will therefore need to check if function B has already been created:

function A() {
    if (!function_exists('B')) {
        function B() {}
    }
    B();
}

A different solution is to use an anonymous function (this will more likely fit your needs, as the function is now stored in a variable and therefore local to the function scope of A):

function A() {
    $B = function() {};
    $B();
}