Is it possible to use namespace without a class

Question

Is it possible to use namespace without a class?

For example:

namespace Foo;
// rest of the procedural code and functions

Asking because I have notion that namespace are only used where there are classes.

Thanks for your help.

Answer 1

Yes, you can use namespace without a class, like:

//demo.php file
namespace FooDemo;
function first() { return "First"; }
function second() { return "Second"; }
function third() { return "Third"; }

//test.php file
require_once 'demo.php';
foreach (array("first","second","third") as $funcs) {
    echo call_user_func('FooDemo\\'.$funcs);
}

Did you mean something like this

Answer 2

Yes, it is. See the documentation .

Extract:

<?php
namespace Foo\Bar;
include 'file1.php';

const FOO = 2;
function foo() {}
class foo
{
    static function staticmethod() {}
}

/* Unqualified name */
foo(); // resolves to function Foo\Bar\foo
foo::staticmethod(); // resolves to class Foo\Bar\foo, method staticmethod
echo FOO; // resolves to constant Foo\Bar\FOO

/* Qualified name */
subnamespace\foo(); // resolves to function Foo\Bar\subnamespace\foo
subnamespace\foo::staticmethod(); // resolves to class Foo\Bar\subnamespace\foo,
                              // method staticmethod
echo subnamespace\FOO; // resolves to constant Foo\Bar\subnamespace\FOO

/* Fully qualified name */
\Foo\Bar\foo(); // resolves to function Foo\Bar\foo
\Foo\Bar\foo::staticmethod(); // resolves to class Foo\Bar\foo, method staticmethod
echo \Foo\Bar\FOO; // resolves to constant Foo\Bar\FOO
?>

Answer 3

use function My\Full\functionName; //or
//use function My\Full\functionName as func;

functionName();