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This FAQ is split into two sections: common questions, and some specifics of implementation that are helpful to understand fully.
First, the common questions.
There are a few implementation details of the namespace implementations that are helpful to understand.
No. Namespaces do not affect any existing code in any way, or any as-yet-to-be-written code that does not contain namespaces. You can write this code if you wish:
Przykład #1 Accessing global classes outside a namespace
<?php
$a = new \stdClass;
This is functionally equivalent to:
Przykład #2 Accessing global classes outside a namespace
<?php
$a = new stdClass;
Przykład #3 Accessing internal classes in namespaces
<?php
namespace foo;
$a = new \stdClass;
function test(\ArrayObject $typehintexample = null) {}
$a = \DirectoryIterator::CURRENT_AS_FILEINFO;
// extending an internal or global class
class MyException extends \Exception {}
?>
Przykład #4 Accessing internal classes, functions or constants in namespaces
<?php
namespace foo;
class MyClass {}
// using a class from the current namespace as a type hint
function test(MyClass $typehintexample = null) {}
// another way to use a class from the current namespace as a type hint
function test(\foo\MyClass $typehintexample = null) {}
// extending a class from the current namespace
class Extended extends MyClass {}
// accessing a global function
$a = \globalfunc();
// accessing a global constant
$b = \INI_ALL;
?>
Names that begin with a \ always resolve to what they look like, so \my\name is in fact my\name, and \Exception is Exception.
Przykład #5 Fully Qualified names
<?php
namespace foo;
$a = new \my\name(); // instantiates "my\name" class
echo \strlen('hi'); // calls function "strlen"
$a = \INI_ALL; // $a is set to the value of constant "INI_ALL"
?>
Names that contain a backslash but do not begin with a backslash like my\name can be resolved in 2 different ways.
If there is an import statement that aliases another name to my, then the import alias is applied to the my in my\name.
Otherwise, the current namespace name is prepended to my\name.
Przykład #6 Qualified names
<?php
namespace foo;
use blah\blah as foo;
$a = new my\name(); // instantiates "foo\my\name" class
foo\bar::name(); // calls static method "name" in class "blah\blah\bar"
my\bar(); // calls function "foo\my\bar"
$a = my\BAR; // sets $a to the value of constant "foo\my\BAR"
?>
Class names that do not contain a backslash like name can be resolved in 2 different ways.
If there is an import statement that aliases another name to name, then the import alias is applied.
Otherwise, the current namespace name is prepended to name.
Przykład #7 Unqualified class names
<?php
namespace foo;
use blah\blah as foo;
$a = new name(); // instantiates "foo\name" class
foo::name(); // calls static method "name" in class "blah\blah"
?>
Function or constant names that do not contain a backslash like name can be resolved in 2 different ways.
First, the current namespace name is prepended to name.
Finally, if the constant or function name does not exist in the current namespace, a global constant or function name is used if it exists.
Przykład #8 Unqualified function or constant names
<?php
namespace foo;
use blah\blah as foo;
const FOO = 1;
function my() {}
function foo() {}
function sort(&$a)
{
sort($a);
$a = array_flip($a);
return $a;
}
my(); // calls "foo\my"
$a = strlen('hi'); // calls global function "strlen" because "foo\strlen" does not exist
$arr = array(1,3,2);
$b = sort($arr); // calls function "foo\sort"
$c = foo(); // calls function "foo\foo" - import is not applied
$a = FOO; // sets $a to value of constant "foo\FOO" - import is not applied
$b = INI_ALL; // sets $b to value of global constant "INI_ALL"
?>
The following script combinations are legal:
file1.php
<?php
namespace my\stuff;
class MyClass {}
?>
another.php
<?php
namespace another;
class thing {}
?>
file2.php
<?php
namespace my\stuff;
include 'file1.php';
include 'another.php';
use another\thing as MyClass;
$a = new MyClass; // instantiates class "thing" from namespace another
?>
There is no name conflict, even though the class MyClass exists within the my\stuff namespace, because the MyClass definition is in a separate file. However, the next example causes a fatal error on name conflict because MyClass is defined in the same file as the use statement.
<?php
namespace my\stuff;
use another\thing as MyClass;
class MyClass {} // fatal error: MyClass conflicts with import statement
$a = new MyClass;
?>
PHP does not allow nesting namespaces
<?php
namespace my\stuff {
namespace nested {
class foo {}
}
}
?>
However, it is easy to simulate nested namespaces like so:
<?php
namespace my\stuff\nested {
class foo {}
}
?>
The only elements that are affected by use statements are namespaces and class names. In order to shorten a long constant or function, import its containing namespace
<?php
namespace mine;
use ultra\long\ns\name;
$a = name\CONSTANT;
name\func();
?>
It is very important to realize that because the backslash is used as an escape character within strings, it should always be doubled when used inside a string. Otherwise there is a risk of unintended consequences:
Przykład #9 Dangers of using namespaced names inside a double-quoted string
<?php
$a = new "dangerous\name"; // \n is a newline inside double quoted strings!
$obj = new $a;
$a = new 'not\at\all\dangerous'; // no problems here.
$obj = new $a;
?>
Inside a single-quoted string, the backslash escape sequence is much safer to use, but it is still recommended practice to escape backslashes in all strings as a best practice.
Any undefined constant that is unqualified like FOO will produce a notice explaining that PHP assumed FOO was the value of the constant. Any constant, qualified or fully qualified, that contains a backslash will produce a fatal error if not found.
Przykład #10 Undefined constants
<?php
namespace bar;
$a = FOO; // produces notice - undefined constants "FOO" assumed "FOO";
$a = \FOO; // fatal error, undefined namespace constant FOO
$a = Bar\FOO; // fatal error, undefined namespace constant bar\Bar\FOO
$a = \Bar\FOO; // fatal error, undefined namespace constant Bar\FOO
?>
Any attempt to define a namespaced constant that is a special, built-in constant results in a fatal error
Przykład #11 Undefined constants
<?php
namespace bar;
const NULL = 0; // fatal error;
const true = 'stupid'; // also fatal error;
// etc.
?>
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