Mengira tatasusunan berdasarkan syarat: panduan langkah demi langkah

Question

Saya mempunyai senarai hos dengan antara muka. Kod mesti mengira antara muka pendua untuk setiap hos. Akhir sekali, kod juga mesti memaparkan antara muka pendua X kali untuk setiap hos.

Saya bertanya soalan ini kerana saya ingin menghantar makluman bahawa hos X telah gagal dengan satu atau lebih antara muka yang gagal.

$data = array(
    array("192.168.0.1","eth1"),
    array("192.168.0.2","eth2"),
    array("192.168.0.3","eth3"),
    array("192.168.0.1","eth1"),
    array("192.168.0.4","eth1"),
    array("192.168.0.2","eth5")
);

Saya telah mengikuti contoh lain di sini, tetapi kebanyakannya adalah tatasusunan mudah, atau jika ia contoh berbilang dimensi, contohnya tidak serupa.

Saya sudah mencubanya...

<?php
$data = array(
    array("192.168.0.1","eth1"),
    array("192.168.0.2","eth2"),
    array("192.168.0.3","eth3"),
    array("192.168.0.1","eth1"),
    array("192.168.0.4","eth1"),
    array("192.168.0.2","eth5")
);


$counter_data = count($data);

$duplicated_host = array_filter(array_count_values(array_column($data, 0)), function($v) { return $v > 1; });
print_r($duplicated_host);
print ("<br>");

$duplicated_host_keys = (array_keys($duplicated_host));

for ($row_num = 0; $row_num < $counter_data; $row_num++)
{
    $host = $data[$row_num][0];
    $interface = $data[$row_num][1];
    if (in_array($host,$duplicated_host_keys))
    {
        print($host . " " . $interface . "<br>");
    }
    
}

Kod di atas adalah salah, ia agak berfungsi tetapi ia tidak seperti yang saya jangkakan... Adakah terdapat cara mudah untuk melakukan ini?

Keluaran akhir sepatutnya kelihatan seperti ini:

Host 192.168.0.1 has eth1 repeated 2 times. --> For current data only
Host 192.168.0.1 has eth9 repeated 5 times.
Host 192.168.0.4 has eth1 repeated 9 times.

Answer 1

Ini mungkin yang anda cari:

<?php
$input = array(
    array("192.168.0.1","eth1"),
    array("192.168.0.2","eth2"),
    array("192.168.0.3","eth3"),
    array("192.168.0.1","eth1"),
    array("192.168.0.4","eth1"),
    array("192.168.0.2","eth5"),
);
$output = [];
array_walk($input, function($entry) use (&$output) {
    [$host, $interface] = $entry;
    if (isset($host, $output) && isset($interface, $output[$host])) {
        $output[$host][$interface]++;
    } else {
        $output[$host][$interface] = 1;
    }
});
print_r($output);

Keluarannya ialah:

Array
(
    [192.168.0.1] => Array
        (
            [eth1] => 2
        )
    [192.168.0.2] => Array
        (
            [eth2] => 1
            [eth5] => 1
        )
    [192.168.0.3] => Array
        (
            [eth3] => 1
        )
    [192.168.0.4] => Array
        (
            [eth1] => 1
        )
)

Answer 2

Anda perlu mengumpulkan dua kumpulan, pertama hos dan kemudian antara muka.

Anda kemudian boleh mengulangi tatasusunan berkumpulan ini untuk memaparkan/menghantar output:

<?php

$data = array(
    array("192.168.0.1","eth1"),
    array("192.168.0.2","eth2"),
    array("192.168.0.3","eth3"),
    array("192.168.0.1","eth1"),
    array("192.168.0.4","eth1"),
    array("192.168.0.2","eth5")
);


$result = [];

foreach ($data as $arr) {
    [ $host, $nic ] = $arr;
    if (!isset($result[$host])) {
        $result[$host] = [];
    }
    if (!isset($result[$host][$nic])) {
        $result[$host][$nic] = 0;
    }

    $result[$host][$nic]++;
}


foreach ($result as $host => $nics) {
    foreach ($nics as $nic => $count) {
        echo "${host} has his '${nic}' interface fail ${count} time(s)" . PHP_EOL;
    }
}

192.168.0.1 has his 'eth1' interface fail 2 time(s)
192.168.0.2 has his 'eth2' interface fail 1 time(s)
192.168.0.2 has his 'eth5' interface fail 1 time(s)
192.168.0.3 has his 'eth3' interface fail 1 time(s)
192.168.0.4 has his 'eth1' interface fail 1 time(s)

Cuba dalam talian!

---

^{NIC --> "Kad Antara Muka Rangkaian"}