使用PHP变量实例化一个类 - 命名空间的问题

Question

All examples below are based on the guarantee that all files exist in their correct location. I've treble checked this.

(1) This works when NOT using namespaces:

$a = "ClassName";
$b = new $a();

This doesn't work:

// 'class not found' error, even though file is there

namespace path\to\here;
$a = "ClassName";
$b = new $a();

This DOES work:

namespace path\to\here;
$a = "path\to\here\ClassName";
$b = new $a();

So it seems the namespace declaration is ignored when using a variable to instantiate a class.

Is there a better way (than my last example) so I don't need to go through some code and change every variable to include the namespace?

Answer 1

The namespace is always part of the complete class name. With some use statements you'll only create an alias for a class during runtime.

<?php

use Name\Space\Class;

// actually reads like

use Name\Space\Class as Class;

?>

The namespace declaration before a class only tells the PHP parser that this class belongs to that namespace, for instantiation you still need to reference the complete class name (which includes the namespace as explained before).

To answer your specific question, no, there isn't any better way than the last example included in your question. Although I'd escape those bad backslashes in a double quoted string.*

<?php

$foo = "Name\\Space\\Class";
new $foo();

// Of course we can mimic PHP's alias behaviour.

$namespace = "Name\\Space\\";

$foo = "{$namespace}Foo";
$bar = "{$namespace}Bar";

new $foo();
new $bar();

?>

*) No need for escaping if you use single quoted strings.