如何在回调函数中包含变量? [英] How can I include a variable inside a callback function?

问题描述

我正在尝试获取大于n的数组值.

I'm trying to get counts of array values greater than n.

我正在像这样使用array_reduce():

$arr = range(1,10);
echo array_reduce($arr, function ($a, $b) { return ($b > 5) ? ++$a : $a; });

这会打印出比硬编码的5还要大的数组中的元素数.

This prints the number of elements in the array greater than the hard-coded 5 just fine.

但是如何使5像$n这样的变量?

But how can I make 5 a variable like $n?

我尝试过引入这样的第三个参数:

I've tried introducing a third argument like this:

array_reduce($arr, function ($a, $b, $n) { return ($b > $n) ? ++$a : $a; });
//                                    ^                  ^

甚至

array_reduce($arr, function ($a, $b, $n) { return ($b > $n) ? ++$a : $a; }, $n);
//                                    ^                  ^                   ^

这些工作都没有.您能告诉我如何在此处添加变量吗?

None of these work. Can you tell me how I can include a variable here?

推荐答案

捕获父值的语法可以在

The syntax to capture parent values can be found in the function .. use documentation under "Example #3 Inheriting variables from the parent scope".

..从父级作用域继承变量[需要'use'形式，并且不与使用全局变量相同..闭包的父级作用域是闭包所在的函数被声明(不一定是调用它的函数).

.. Inheriting variables from the parent scope [requires the 'use' form and] is not the same as using global variables .. The parent scope of a closure is the function in which the closure was declared (not necessarily the function it was called from).

然后在use的帮助下转换原始代码:

Converting the original code, with the assistance of use, is then:

$n = 5;
array_reduce($arr, function ($a, $b) use ($n) {
    return ($b > $n) ? ++$a : $a;
});

从外部词汇范围使用" $n的地方.

Where $n is "used" from the outer lexical scope.

注意:在上面的示例中，提供了值的副本，并且变量本身未绑定.请参阅有关使用变量引用(例如&$n)以能够在父上下文中重新分配变量的文档.

NOTE: In the above example, a copy of the value is supplied and the variable itself is not bound. See the documentation about using a reference-to-a-variable (eg. &$n) to be able and re-assign to variables in the parent context.

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