Question

我有两个数组：$bluescan和$maclistResults。 $bluescan遗失了一些['company']个值，我需要通过搜索$maclistResults来更新。

$bluescan有30个子阵列，$maclistResults是每个制造商的列表，这些制造商曾向IEEE注册了一个MAC地址块，总共20285个子阵列。

$bluescan数组：

$maclistResults数组的示例：

Array
(
    [0] => Array
        (
            [mac] => E8:5B:5B
            [company] => LG ELECTRONICS INC
        )

    [1] => Array
        (
            [mac] => E8:5B:F0
            [company] => Imaging Diagnostics
        )

    [2] => Array
        (
            [mac] => E8:5D:6B
            [company] => Luminate Wireless
        )

    [3] => Array
        (
            [mac] => E8:5E:53
            [company] => Infratec Datentechnik GmbH
        )

    [4] => Array
        (
            [mac] => E8:61:1F
            [company] => Dawning Information Industry Co.,Ltd
        )

    [5] => Array
        (
            [mac] => E8:61:7E
            [company] => Liteon Technology Corporation
        )

    [6] => Array
        (
            [mac] => E8:61:83
            [company] => Black Diamond Advanced Technology, LLC
        )

    [7] => Array
        (
            [mac] => E8:61:BE
            [company] => Melec Inc.
        )

    [8] => Array
        (
            [mac] => E8:66:C4
            [company] => Datawise Systems
        )

    [9] => Array
        (
            [mac] => E8:6D:52
            [company] => ARRIS Group, Inc.
        )

    [10] => Array
        (
            [mac] => E8:6D:54
            [company] => Digit Mobile Inc
        )
)

$maclistResult['mac']字段小于$bluescan['mac']，但前三对在两个数组上匹配，使用substr()。

以下代码正在执行我需要的操作，但执行任务需要63秒。

$maclist = 'maclist';
$maclistResults = $db->$maclist->find(array('mac'=>substr($value['mac'], 0, 8)), array('mac'=>1, 'company'=>1));
$maclistResults = iterator_to_array($maclistResults);
foreach ($bluescan as $key => $value)
{
    foreach ($maclistResults as $value2) 
    {
        if (empty($value['company']))
        {
            $bluescan[$key]['company'] = $value2['company'];
        }
    }
}

我想知道如何仅使用一个['company']来更新$bluescan的{{1}}值，因为另一个foreach不是最佳值。

Answer 1

$ bluescan [$ key] ['company'] = $ value2 ['company'];您应该将$maclistResults数组转换为以下格式：

Array
(
    'E8:5B:5B' => LG ELECTRONICS INC,
    'E8:5B:5A' => SAMSUNG
    ...
)

然后查看它会很快，比如

$bluescan[$key]['company'] = $macResults[$mac];

Answer 2

因为你只需要与bluescan中的元素一样多的mac开始。

//prepare all macs from bluescan and keep reference to their elements
$macs = array();
foreach ($bluescan as $blueIndex => $blueElement) {
    $macs[substr($blueElement['mac'], 0, 8)][] = $blueIndex;
}

现在$macs是地图mac =＆gt; [bluescan指数]

//query for maclist
$maclist = query("SELECT mac, company FROM maclist_table WHERE mac IN (" . implode(',', array_keys($macs)) . ")");

//finally assign companies to bluescans
foreach ($maclist as $macElement) {
    foreach ($macs[$macElement['mac']] as $blueIndex) {
        $bluescan[$blueIndex]['company'] = $macElement['company'];
    }
}

根据另一个数组设置数组值

2 个答案: