调用在另一个命名空间中定义的没有前缀的 PHP 函数
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Calling a PHP function defined in another namespace without the prefix
提问by David Harkness
When you define a function in a namespace,
在命名空间中定义函数时，
namespace foo {
function bar() { echo "foo!\n"; }
class MyClass { }
}
you must specify the namespace when calling it from another (or global) namespace:
从另一个（或全局）命名空间调用它时，您必须指定命名空间：
bar(); // call to undefined function \bar()
foo\bar(); // ok
With classes you can employ the "use" statement to effectively import a class into the current namespace [Edit: I thought you could "use foo" to get the classes, but apparently not.]
对于类，您可以使用“use”语句将类有效地导入当前命名空间 [编辑：我认为您可以“使用 foo”来获取类，但显然不是。]
use foo\MyClass as MyClass;
new MyClass(); // ok, instantiates foo\MyClass
but this doesn't work with functions [and would be unwieldy given how many there are]:
但这不适用于函数[并且考虑到有多少会很笨拙]：
use foo\bar as bar;
bar(); // call to undefined function \bar()
You can alias the namespace to make the prefix shorter to type,
您可以为命名空间设置别名以使前缀更短以便键入，
use foo as f; // more useful if "foo" were much longer or nested
f\bar(); // ok
but is there any way to remove the prefix entirely?
但是有没有办法完全删除前缀？
Background: I'm working on the Hamcrest matching library which defines a lot of factory functions, and many of them are designed to be nested. Having the namespace prefix really kills the readability of the expressions. Compare
背景：我正在研究 Hamcrest 匹配库，它定义了许多工厂函数，其中许多被设计为嵌套。拥有命名空间前缀真的会扼杀表达式的可读性。相比
assertThat($names,
is(anArray(
equalTo('Alice'),
startsWith('Bob'),
anything(),
hasLength(atLeast(12))
)));
to
到
use Hamcrest as h;
h\assertThat($names,
h\is(h\anArray(
h\equalTo('Alice'),
h\startsWith('Bob'),
h\anything(),
h\hasLength(h\atLeast(12))
)));
回答by Matthieu Napoli
PHP 5.6 will allow to import functions with the usekeyword:
PHP 5.6 将允许使用use关键字导入函数：
namespace foo\bar {
function baz() {
echo 'foo.bar.baz';
}
}
namespace {
use function foo\bar\baz;
baz();
}
See the RFC for more information: https://wiki.php.net/rfc/use_function
有关更多信息，请参阅 RFC：https: //wiki.php.net/rfc/use_function
回答by luka8088
By adding the helper hacks mentioned below, you can import everything from Hamcrest namespace to current namespace by calling:
通过添加下面提到的帮助程序，您可以通过调用将 Hamcrest 命名空间中的所有内容导入到当前命名空间：
import_namespace('Hamcrest', __NAMESPACE__);
Here are the hacks, function_alias works like http://www.php.net/manual/en/function.class-alias.phpexcept if works on functions:
以下是技巧，function_alias 的工作方式类似于http://www.php.net/manual/en/function.class-alias.php，除非适用于函数：
function function_alias ($original, $alias) {
$args = func_get_args();
assert('count($args) == 2', 'function_alias(): requires exactly two arguments');
assert('is_string($original) && is_string($alias)', 'function_alias(): requires string arguments');
// valid function name - http://php.net/manual/en/functions.user-defined.php
assert('preg_match(\'/^[a-zA-Z_\x7f-\xff][\\\\a-zA-Z0-9_\x7f-\xff]*$/\', $original) > 0',
"function_alias(): '$original' is not a valid function name");
assert('preg_match(\'/^[a-zA-Z_\x7f-\xff][\\\\a-zA-Z0-9_\x7f-\xff]*$/\', $alias) > 0',
"function_alias(): '$alias' is not a valid function name");
$aliasNamespace = substr($alias, 0, strrpos($alias, '\') !== false ? strrpos($alias, '\') : 0);
$aliasName = substr($alias, strrpos($alias, '\') !== false ? strrpos($alias, '\') + 1 : 0);
$serializedOriginal = var_export($original, true);
eval("
namespace $aliasNamespace {
function $aliasName () {
return call_user_func_array($serializedOriginal, func_get_args());
}
}
");
}
In combination with namespace importer:
结合命名空间导入器：
function import_namespace ($source, $destination) {
$args = func_get_args();
assert('count($args) == 2', 'import_namespace(): requires exactly two arguments');
assert('is_string($source) && is_string($destination)', 'import_namespace(): requires string arguments');
// valid function name - http://php.net/manual/en/functions.user-defined.php
assert('preg_match(\'/^([a-zA-Z_\x7f-\xff][\\\\a-zA-Z0-9_\x7f-\xff]*)?$/\', $source) > 0',
"import_namespace(): '$destination' is not a valid namespace name");
assert('preg_match(\'/^([a-zA-Z_\x7f-\xff][\\\\a-zA-Z0-9_\x7f-\xff]*)?$/\', $destination) > 0',
"import_namespace(): '$source' is not a valid namespace name");
foreach(get_declared_classes() as $class)
if (strpos($class, $source . '\') === 0)
class_alias($class, $destination . ($destination ? '\' : '') . substr($class, strlen($source . '\')));
$functions = get_defined_functions();
foreach(array_merge($functions['internal'], $functions['user']) as $function)
if (strpos($function, $source . '\') === 0)
function_alias($function, $destination . ($destination ? '\' : '') . substr($function, strlen($source . '\')));
}
回答by Borealid
I don't know an elegantsolution, but...
我不知道一个优雅的解决方案，但是......
You can create wrapper functions that encapsulate the functions in the external namespace. This will let you keep your code readability...
您可以创建将函数封装在外部命名空间中的包装函数。这将使您保持代码的可读性......
function assertThat($x, $y) { return h\assertThat($x, $y); }
function assertThat($x, $y) { return h\assertThat($x, $y); }
