php - PHP的:array_intersect没有给出预期的结果

Question

我正在尝试计算PHP数组中expected和actual之间的匹配，我有这个...

$array = array(
    "item" => array(
        'expected' => array(
            '1' => 25,
            '2' => 4,
            '3' => 4,
        ),
        'color' => 'red',
        'actual' => array(
            '1' => 25,
            '2' => 4,
            '3' => 3,
        ),
    ),
);

foreach ($array as $key => $arrayItem) {

    $matches = array (
        'matches'  => count ( array_intersect ( $arrayItem['expected'], $arrayItem['actual'] ) ),
    );

}

echo "Matches = " . $matches['matches'];

我期望它返回 2，但实际上它正在返回 3。如果我像下面的示例中那样更改值，则它确实可以工作...

$array = array(
    "item" => array(
        'expected' => array(
            '1' => 25,
            '2' => 84,
            '3' => 4,
        ),
        'color' => 'red',
        'actual' => array(
            '1' => 25,
            '2' => 84,
            '3' => 3,
        ),
    ),
);

foreach ($array as $key => $arrayItem) {

    $matches = array (
        'matches'  => count ( array_intersect ( $arrayItem['expected'], $arrayItem['actual'] ) ),
    );

}

echo "Matches = " . $matches['matches'];

任何人都知道为什么最高版本没有给我预期的结果吗？

Answer 1

因为它返回一个包含array1中所有值的数组，其值存在于所有参数中。

array_intersect(array $array1, array $array2[, array $... ]): array

https://www.php.net/manual/en/function.array-intersect.php

也许您可以从以下角度清楚地看到它：

var_dump(array_intersect([25, 4, 4, 4], [25, 4, 3])); // [25, 4, 4, 4] 
// because the number `4` is in the second array!

var_dump(array_intersect([25, 4, 3], [25, 4, 4, 4])); // [25, 4]