PHP namespaces and "use"

04/12/2020 18:30:02

I am having a little trouble with namespaces and the use statements.

I have three files: ShapeInterface.php, Shape.php and Circle.php.

I am trying to do this using relative paths so I have put this in all of the classes:

namespace Shape; 

In my circle class I have the following:

namespace Shape;
//use Shape;
//use ShapeInterface;

include 'Shape.php';
include 'ShapeInterface.php';    

class Circle extends Shape implements ShapeInterface{ ....

If I use the include statements I get no errors. If I try the use statements I get:

Fatal error: Class 'Shape\Shape' not found in /Users/shawn/Documents/work/sites/workspace/shape/Circle.php on line 8

Could someone please give me a little guidance on the issue?

Verified Answer (164 Votes)

05/11/2012 02:17:29

The use operator is for giving aliases to names of classes, interfaces or other namespaces. Most use statements refer to a namespace or class that you'd like to shorten:

use My\Full\Namespace;

is equivalent to:

use My\Full\Namespace as Namespace;
// Namespace\Foo is now shorthand for My\Full\Namespace\Foo

If the use operator is used with a class or interface name, it has the following uses:

// after this, "new DifferentName();" would instantiate a My\Full\Classname
use My\Full\Classname as DifferentName;

// global class - making "new ArrayObject()" and "new \ArrayObject()" equivalent
use ArrayObject;

The use operator is not to be confused with autoloading. A class is autoloaded (negating the need for include) by registering an autoloader (e.g. with spl_autoload_register). You might want to read PSR-4 to see a suitable autoloader implementation.


Answer #2 (11 Votes)

02/24/2017 22:16:18

If you need to order your code into namespaces, just use the keyword namespace:


namespace foo\bar;

In file2.php

$obj = new \foo\bar\myObj();

You can also use use. If in file2 you put

use foo\bar as mypath;

you need to use mypath instead of bar anywhere in the file:

$obj  = new mypath\myObj();

Using use foo\bar; is equal to use foo\bar as bar;.

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