Question

我正在对每个项目进行计数和求和，并且需要数组看起来像这样。

[Date] => 
         [device_type1] => [all the data]
         [device_type2] => [all the data]

我正在搜索每个，然后所有人都返回一个日期，我需要所有这些都在这个日期之下，并且在该数组内部将每个设备保持为关键并且在所有设备内部信息。这是我到目前为止所做的。

$all_device_types = array('pole','covert','firewatch','internet','trailer','slave','project');
foreach ($all_device_types as $adt) {
    $du->select("CONCAT(YEAR(uptime_date),'-',MONTH(uptime_date),'-',DAY(uptime_date)) as year_month_day,

                             SUM(uptime_router) as router,

                             COUNT(uptime_router) as router_count,

                             uptime_device_type as device_type,

                            date(uptime_date) as data_date","

                            WHERE uptime_device_type = '".$adt."'

                            GROUP BY CONCAT(YEAR(uptime_date),'-',MONTH(uptime_date),'-',DAY(uptime_date))

                            ORDER BY uptime_date ASC");


                            $all_results = array();
                            while( $all_row = $du->fetch() ){

                                if ($adt == $all_device_types[0]) {
                                    $dates[] = $all_row['data_date'];
                                    $adt_0 = $all_row;
                                } elseif ($adt == $all_device_types[1]) {
                                    $adt_1[] = $all_row;
                                } elseif ($adt == $all_device_types[2]) {
                                    $adt_2[] = $all_row;
                                } elseif ($adt == $all_device_types[3]) {
                                    $adt_3[] = $all_row;
                                } elseif ($adt == $all_device_types[4]) {
                                    $adt_4[] = $all_row;
                                } elseif ($adt == $all_device_types[5]) {
                                    $adt_5[] = $all_row;
                                } elseif ($adt == $all_device_types[6]) {
                                    $adt_6[] = $all_row;
                                }

                            }

                        }

其中一个结果就是这样。

 [pole] => Array
        (
            [year_month_day] => 2017-9-5
            [router] => 4408
            [router_count] => 4620
            [device_type] => pole
            [data_date] => 2017-09-05
        )

Answer 1

看起来它可以简化很多。您可以将设备类型添加到GROUP BY，并在单个查询中获取所有结果，而不是循环设备类型并为每个类型运行单独的查询。

我认为CONCAT(YEAR(uptime_date),'-',MONTH(uptime_date),'-',DAY(uptime_date))应输出与date(uptime_date)相同的内容，您应该能够在GROUP BY和ORDER BY子句中使用别名。

$du->select("DATE(uptime_date) as data_date,
             uptime_device_type as device_type,
             SUM(uptime_router) as router,
             COUNT(uptime_router) as router_count,
             GROUP BY data_date, device_type
             ORDER BY data_date, device_type");

如果我错了并且您确实需要CONCAT版本，请将其添加回来，但主要是将device_type添加到GROUP BY并消除WHERE子句。

然后，当您获取结果时，可以将它们直接插入到您想要的多维结构中。

while( $row = $du->fetch() ) {
    $all_results[$row['data_date']][$row['device_type']][] = $row;

    // to calculate totals for all the devices for each date
    if (!isset($all_results[$row['data_date']]['all_devices']) {
        // initialize if not set
        $all_results[$row['data_date']]['all_devices'] = [
            'router' => 0, 'router_count' => 0];
    }
    // increment values
    $all_results[$row['data_date']]['all_devices']['router'] += $row['router'];
    $all_results[$row['data_date']]['all_devices']['router_count'] += $row['router_count'];
}

来自查询的多个数组

1 个答案: