按日期时间列的日期部分对 2d 数组进行分组，并对另一列求和 - Group 2d array by date portion of datetime column and sum another column

我有一个带有日期时间列和小计列的 2d 数组。我需要按日期时间值的日期部分对行进行分组，并对每个组中的小计值求和。

示例输入数组：

[
    ['id' => 81, 'placed' => '2013-09-19 16:32:53', 'sub_total' => 786],
    ['id' => 80, 'placed' => '2013-09-19 16:32:06', 'sub_total' => 780],
    ['id' => 79, 'placed' => '2013-09-18 17:06:48', 'sub_total' => 786],
    ['id' => 78, 'placed' => '2013-09-18 17:05:02', 'sub_total' => 756],
    ['id' => 77, 'placed' => '2013-09-17 17:02:53', 'sub_total' => 786],
    ['id' => 76, 'placed' => '2013-09-16 17:02:53', 'sub_total' => 756],
]

期望输出：

[
    ['placed' => '2013-09-19', 'sub_total' => 1566],
    ['placed' => '2013-09-18', 'sub_total' => 1542],
    ['placed' => '2013-09-17', 'sub_total' => 786],
    ['placed' => '2013-09-16', 'sub_total' => 756],
]

我也测试了这个：http://phpfiddle.org/main/code/rzv-ngp

$newarray = array();
foreach ($array as $value){
    $temp = explode(" ", $value['placed']);
    $date = $temp[0];
    $total = (isset($newarray[$date]['sub_total']) ? $newarray[$date]['sub_total'] + $value['sub_total']: $value['sub_total']);
    $newarray[$date] = array('placed' => $date, 'sub_total' => $total);
}
print_r($newarray);

$output=array();
foreach($yourArray as $values)
{
 $d=date("Y-m-d",strtotime($values["placed"]));
 $output[$d]["sub_total"]+=$values["sub_total"];
}
print_r($output);

输出：

Warning: Undefined array key "2013-09-19" in /in/SO65c on line 17
Warning: Undefined array key "sub_total" in /in/SO65c on line 17
Warning: Undefined array key "2013-09-18" in /in/SO65c on line 17
Warning: Undefined array key "sub_total" in /in/SO65c on line 17
Warning: Undefined array key "2013-09-17" in /in/SO65c on line 17
Warning: Undefined array key "sub_total" in /in/SO65c on line 17
Warning: Undefined array key "2013-09-16" in /in/SO65c on line 17
Warning: Undefined array key "sub_total" in /in/SO65c on line 17
Array
(
    [2013-09-19] => Array
        (
            [sub_total] => 1566
        )
    [2013-09-18] => Array
        (
            [sub_total] => 1542
        )
    [2013-09-17] => Array
        (
            [sub_total] => 786
        )
    [2013-09-16] => Array
        (
            [sub_total] => 756
        )
)

小提琴

您可以在查询时拥有这种数组。类似的东西

select date_field_name, other_field
from table_name
group by Day(date_field_name)

之后，您可以使用使用foreach()来处理每天的数据！

尝试使用子查询

select field_name,
       DAY(date_field) as date,
       (
        select sum(sub_total)
        where DAY(date_field)=date
       ) as sub_total
from table_name
group by day(date_field)

由于 datetime 表达式的格式始终首先使用其日期子字符串，因此只需隔离第一个空格之前出现的字符。使用此子字符串进行分组。

检查之前是否遇到过日期。如果没有，请创建一个引用变量并将其推送到结果数组中。如果/当再次遇到相同的日期时，将新的小计添加到相关引用中的缓存sub_total中。对引用变量的所有更改都将反映在结果数组中。

代码：（演示）

$result = [];
foreach ($array as $row) {
    $date = strtok($row['placed'], ' ');
    if (!isset($ref[$date])) {
        $ref[$date] = ['placed' => $date, 'sub_total' => $row['sub_total']];
        $result[] = &$ref[$date];
    } else {
        $ref[$date]['sub_total'] += $row['sub_total'];
    }
}
var_export($result);

试试这段代码

$result = array();
foreach($shop as $value)
{
    if (!isset($result[$value['placed']]))
    {
        $result[$value['placed']] = array('placed' => $value['placed'], 'sub_total' => 0);
    }
    $result[$value['placed']]['sub_total'] += $value['sub_total'];
}
print_r($result);