与其他人发布的内容相比，您不能使用 max() 函数可解决此问题，因为这些函数无法理解传入的数据结构(数组).这些函数仅适用于标量数组元素.

In contrast to what others have posted, you cannot use the min()/max() functions for this problem as these functions do not understand the datastructure (array) which are passed in. These functions only work for scalar array elements.

开始编辑

使用 min()的原因 和 max() 似乎产生了正确的答案，这与将类型类型转换为整数的数组有关，这是一个

The reason why the use of min() and max() seem to yield the correct answer is related to type-casting arrays to integers which is an undefined behaviour:

转换为整数的行为对于其他类型未定义.不要依靠任何观察到的行为，因为它可以更改，恕不另行通知.

我上面关于类型转换的陈述是错误的.实际上 min() 和 max() 可以处理数组，但不能处理数组OP需要他们工作的方式.使用 min() 和 max() 带有多个数组或一个数组元素的数组从左到右逐个元素进行比较:

My statement above about the type-casting was wrong. Actually min() and max() do work with arrays but not in the way the OP needs them to work. When using min() and max() with multiple arrays or an array of arrays elements are compared element by element from left to right:

$val = min(array(2, 4, 8), array(2, 5, 1)); // array(2, 4, 8)
/*
 * first element compared to first element: 2 == 2
 * second element compared to second element: 4 < 5
 * first array is considered the min and is returned
 */

翻译成OP的问题，这表明了为什么直接使用 min() 和 max() 似乎产生了正确的结果.数组的第一个元素是 id 值，因此 min() 和 max() 首先将它们进行比较，因为正确的 id 是最小的 count ，而偶然的结果是正确的 id 最高的是 count 最高的一个.

Translated into the OP's problem this shows the reason why the direct use of min() and max() seems to yield the correct result. The arrays' first elements are the id-values, therefore min() and max() will compare them first, incidentally resulting in the correct result because the lowest id is the one with the lowest count and the highest id is the one with the highest count.

END EDIT

正确的方法是使用循环.

The correct way would be to use a loop.

$a = array(
        array('id' => 117, 'name' => 'Networking', 'count' => 16),
        array('id' => 188, 'name' => 'FTP', 'count' => 23),
        array('id' => 189, 'name' => 'Internet', 'count' => 48)
);
$min = PHP_INT_MAX;
$max = 0;
foreach ($a as $i) {
    $min = min($min, $i['count']);
    $max = max($max, $i['count']);
}