具有命名空间的函数中的PHP函数

Question

I can't really seem to call my function within a function with namespaces. I know that

\foo\bar::foo_bar();

Will call this function

function foo_bar() {
    return('This is foo_bar');
}

But what if I have a function like so

static function foo_bar() {
    function foo() {
        return('This is foo');
    }
}

I have many other functions within foo_bar like

static function foo_bar() {
    function foo() {
        return('This is foo');
    }
    function bar() {
        return('This is bar');
    }
}

So I was thinking \foo(); but it just says Fatal error: Call to undefined function foo() in. Then if I do something like \foo\bar\foo_bar::foo();, it'll give me this error Fatal error: Class '\foo\bar\foo_bar' not found in.

How would I go about to call my function within a function?

Answer 1

There are no inner/nested functions/methods/classes in php (yet?).
Once foo_bar() gets executed the function foo() gets declared/defined "outside" any function/method/class, i.e. in your case as a function directly under the namespace \foo

<?php
namespace foo {
    class bar {
        static function foo_bar() {
            // even though function baz() only is declared once foo_bar()
            // is invoked
            function baz() {
                return('This is baz');
            }
            // the scope where it gets declared is outside
            // the function/method and class
        }
    }
    // <--- so that would be here - figuratively speaking
}

namespace lalala {
    echo 'baz() exists: ', function_exists('\foo\baz') ? 'yes':'no', "
";

    \foo\bar::foo_bar(); // now \foo\baz gets declared
    echo 'baz() exists: ', function_exists('\foo\baz') ? 'yes':'no', "
";
    echo \foo\baz();

    \foo\bar::foo_bar(); // will raise an error
}

prints

baz() exists: no
baz() exists: yes
This is baz
Fatal error: Cannot redeclare foo\baz() (previously declared in ...

It's most likely not what you want ...or if you want it, you shouldn't ;-)