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Php versions – PHP : array_intersect not giving expected result

I am trying to count the matches between expected and actual in a PHP array, I have this…

$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'];

I am expecting this to return 2 but it is actually returning 3. If I change the values like in the example below then it does work…

$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'];

Anyone any ideas why the top version is not giving me the expected result?

Tags:

3

Answers


  1. 0

    The count is actually correct.

    It doesn’t happen in your second example because you use the numbers 84 and 4, but essentially here are the matches:

    $arrayItem['expected'][1] matches with $arrayItem['actual'][1] (25 and 25)

    $arrayItem['expected'][2] matches with $arrayItem['actual'][2] (4 and 4)

    $arrayItem['expected'][3] matches with $arrayItem['actual'][2] (4 and 4)

    The count of 3 is correct.

    You can test this by changing your code to the following:

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

var_dump($matches);

    Here you’ll see this output:

    array(1) {
    ["matches"] => array(3) {
        [1]=> int(25) 
        [2]=> int(4) 
        [3]=> int(4)
    }
}
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  2. 0

    Because it returns an array containing all values in array1 whose values exist in all of the parameters.

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

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

    Maybe you can see it clearly from this perspective:

    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]
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  3. 0

    it returns 2

    <?php

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

echo count(array_intersect($array['item']['expected'],$array['item']['actual']));
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