引用 PHP 中的静态方法？
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/192111/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Reference to static method in PHP?
提问by Eric
In PHP, I am able to use a normal function as a variable without problem, but I haven't figured out how to use a static method. Am I just missing the right syntax, or is this not possible?
在 PHP 中，我可以毫无问题地使用普通函数作为变量，但我还没有弄清楚如何使用静态方法。我只是缺少正确的语法，还是不可能？
(EDIT: the first suggested answer does not seem to work. I've extended my example to show the errors returned.)
（编辑：第一个建议的答案似乎不起作用。我扩展了我的示例以显示返回的错误。）
function foo1($a,$b) { return $a/$b; }
class Bar
{
static function foo2($a,$b) { return $a/$b; }
public function UseReferences()
{
// WORKS FINE:
$fn = foo1;
print $fn(1,1);
// WORKS FINE:
print self::foo2(2,1);
print Bar::foo2(3,1);
// DOES NOT WORK ... error: Undefined class constant 'foo2'
//$fn = self::foo2;
//print $fn(4,1);
// DOES NOT WORK ... error: Call to undefined function self::foo2()
//$fn = 'self::foo2';
//print $fn(5,1);
// DOES NOT WORK ... error: Call to undefined function Bar::foo2()
//$fn = 'Bar::foo2';
//print $fn(5,1);
}
}
$x = new Bar();
$x->UseReferences();
(I am using PHP v5.2.6 -- does the answer change depending on version too?)
（我使用的是 PHP v5.2.6 - 答案是否也会因版本而异？）
回答by Peter Bailey
PHP handles callbacks as strings, not function pointers. The reason your first test works is because the PHP interpreter assumes foo1as a string. If you have E_NOTICE level error enabled, you should see proof of that.
PHP 将回调作为字符串处理，而不是函数指针。您的第一个测试有效的原因是 PHP 解释器将foo1假定为字符串。如果您启用了 E_NOTICE 级别错误，您应该会看到证明。
"Use of undefined constant foo1 - assumed 'foo1'"
“使用未定义的常量 foo1 - 假定为 'foo1'”
You can't call static methods this way, unfortunately. The scope (class) is relevant so you need to use call_user_func instead.
不幸的是，您不能以这种方式调用静态方法。范围（类）是相关的，因此您需要改用 call_user_func。
<?php
function foo1($a,$b) { return $a/$b; }
class Bar
{
public static function foo2($a,$b) { return $a/$b; }
public function UseReferences()
{
$fn = 'foo1';
echo $fn(6,3);
$fn = array( 'self', 'foo2' );
print call_user_func( $fn, 6, 2 );
}
}
$b = new Bar;
$b->UseReferences();
回答by rewbs
In php 5.2, you can use a variable as the method name in a static call, but to use a variable as the class name, you'll have to use callbacks as described by BaileyP.
在 php 5.2 中，您可以在静态调用中使用变量作为方法名，但要使用变量作为类名，您必须使用 BaileyP 描述的回调。
However, from php 5.3, you canuse a variable as the class name in a static call. So:
但是，从 php 5.3 开始，您可以在静态调用中使用变量作为类名。所以：
class Bar
{
public static function foo2($a,$b) { return $a/$b; }
public function UseReferences()
{
$method = 'foo2';
print Bar::$method(6,2); // works in php 5.2.6
$class = 'Bar';
print $class::$method(6,2); // works in php 5.3
}
}
$b = new Bar;
$b->UseReferences();
?>
回答by Jiangge Zhang
You could use the full name of static method, including the namespace.
您可以使用静态方法的全名，包括命名空间。
<?php
function foo($method)
{
return $method('argument');
}
foo('YourClass::staticMethod');
foo('Namespace\YourClass::staticMethod');
The name array array('YourClass', 'staticMethod')is equal to it. But I think the string may be more clear for reading.
name 数组array('YourClass', 'staticMethod')等于它。但我认为字符串可能更易于阅读。
回答by dwallace
In PHP 5.3.0, you could also do the following:
在 PHP 5.3.0 中，您还可以执行以下操作：
<?php
class Foo {
static function Bar($a, $b) {
if ($a == $b)
return 0;
return ($a < $b) ? -1 : 1;
}
function RBar($a, $b) {
if ($a == $b)
return 0;
return ($a < $b) ? 1 : -1;
}
}
$vals = array(3,2,6,4,1);
$cmpFunc = array('Foo', 'Bar');
usort($vals, $cmpFunc);
// This would also work:
$fooInstance = new Foo();
$cmpFunc = array('fooInstance', 'RBar');
// Or
// $cmpFunc = array('fooInstance', 'Bar');
usort($vals, $cmpFunc);
?>
回答by hek2mgl
In addition to what was said you can also use PHP's reflection capabilities:
除了上面所说的，您还可以使用 PHP 的反射功能：
class Bar {
public static function foo($foo, $bar) {
return $foo . ' ' . $bar;
}
public function useReferences () {
$method = new ReflectionMethod($this, 'foo');
// Note NULL as the first argument for a static call
$result = $method->invoke(NULL, '123', 'xyz');
}
}
回答by Camilo Martin
Coming from a javascript background and being spoiled by it, I just coded this:
来自 javascript 背景并被它宠坏了，我只是这样编码：
function staticFunctionReference($name)
{
return function() use ($name)
{
$className = strstr($name, '::', true);
if (class_exists(__NAMESPACE__."\$className")) $name = __NAMESPACE__."\$name";
return call_user_func_array($name, func_get_args());
};
}
To use it:
要使用它：
$foo = staticFunctionReference('Foo::bar');
$foo('some', 'parameters');
It's a function that returns a function that calls the function you wanted to call. Sounds fancy but as you can see in practice it's piece of cake.
它是一个函数，它返回一个调用您想要调用的函数的函数。听起来很花哨，但正如您在实践中看到的那样，这是小菜一碟。
Works with namespaces and the returned function should work just like the static method - parameters work the same.
使用命名空间和返回的函数应该像静态方法一样工作 - 参数工作相同。
回答by Camilo Martin
This seems to work for me:
这似乎对我有用：
<?php
class Foo{
static function Calc($x,$y){
return $x + $y;
}
public function Test(){
$z = self::Calc(3,4);
echo("z = ".$z);
}
}
$foo = new Foo();
$foo->Test();
?>
回答by Joe
"A member or method declared with static can not be accessed with a variable that is an instance of the object and cannot be re-defined in an extending class"
“用静态声明的成员或方法不能用作为对象实例的变量访问，不能在扩展类中重新定义”
(http://theserverpages.com/php/manual/en/language.oop5.static.php)
( http://theserverpages.com/php/manual/en/language.oop5.static.php)
