Question

我已经和他坐了一会儿，而且已经迟到了。我试图从上个月获得大多数销售额的前3名，我需要计算上个月数组1的id等于数组2的次数（6 =最后一个atm。），如id 4 = 2，id 7 = 3

这可能不是一个完美的解决方案，但我只是想通过我的自我来打破它，所以以后可能＆＃39;也许＆＃39;问题，当我撞墙时我会照顾，所以，如果有人能在这里帮助我，那就太好了。

更新 - 我将在这里添加我想要的结果:(对不起我之前没有，它让它变得更容易:-) - 下面的结果，是因为我希望计算从2014-06-01到那个月的最后一天monly，在数组[0] [1]下的数组，只有6-7-8不是来自2014 -06

希望现在更有意义^^

Array
(
[0] => Array
    (
        [0] => Array
            (
                [4] => 2
                [7] => 3
                [1] => 2
                [3] => 2
                [9] => 1
                [12] => 1
                [2] => 1
                [13] => 1
            )

    )

)




Array
(
[0] => Array
    (
        [0] => Array
            (
                [0] => 4
                [1] => 4
                [2] => 7
                [3] => 1
                [4] => 7
                [5] => 7
                [6] => 3
                [7] => 3
                [8] => 4
                [9] => 9
                [10] => 12
                [11] => 2
                [12] => 13
                [13] => 1
            )

        [1] => Array
            (
                [0] => 2014-06-18
                [1] => 2014-06-10
                [2] => 2014-06-05
                [3] => 2014-06-05
                [4] => 2014-06-12
                [5] => 2014-06-11
                [6] => 2013-12-12
                [7] => 2014-07-23
                [8] => 2014-05-13
                [9] => 2014-06-01
                [10] => 2014-06-12
                [11] => 2014-06-04
                [12] => 2014-06-04
                [13] => 2014-06-11
            )

    )

)

Answer 1

我希望我所理解的是你真正要求的。假设你的数组定义是：

$arr = Array
(
0 => Array
    (
        0 => Array
            (
                0 => 4,
                1 => 4,
                2 => 7,
                3 => 1,
                4 => 7,
                5 => 7,
                6 => 3,
                7 => 3,
                8 => 4,
                9 => 9,
                10 => 12,
                11 => 2,
                12 => 13,
                13 => 1
            ),

        1 => Array
            (
                0 => "2014-06-18",
                1 => "2014-06-10",
                2 => "2014-06-05",
                3 => "2014-06-05",
                4 => "2014-06-12",
                5 => "2014-06-11",
                6 => "2013-12-12",
                7 => "2014-07-23",
                8 => "2014-05-13",
                9 => "2014-06-01",
                10 => "2014-06-12",
                11 => "2014-06-04",
                12 => "2014-06-04",
                13 => "2014-06-11"
            )

    )

);

如果您需要测试日期是否低于6个月，然后放置他们的id，sales和date，则需要使用此代码

   $result = [];
   $index=0;
   foreach ($arr[0][0] as $key => $value) 
   {
      $date1 = new DateTime($arr[0][1][$key]);
      $date2 = new DateTime();
      $diff = $date1->diff($date2);
      $diff = ($diff->format('%y') * 12) + $diff->format('%m');
      if($diff<=6)
      {
         $result[$index]['id'] = $key;
         $result[$index]['sales'] = $value;
         $result[$index]['date'] = $arr[0][1][$key];
         $index++;
      }
   }

   var_dump($result);

Answer 2

array_count_values()将为您提供每个值在数组1中显示的次数。

$count = array_count_values($array[0][0]);

结果：

Array
(
    [4] => 3
    [7] => 3
    [1] => 2
    [3] => 2
    [9] => 1
    [12] => 1
    [2] => 1
    [13] => 1
)

然后你可以使用循环与数组2结合：

$result = array();
foreach($count as $key=>$val) {
    $result[$array[0][1][$key]] = $val;
}

结果：

Array
(
    [2014-06-12] => 3
    [2014-07-23] => 3
    [2014-06-10] => 2
    [2014-06-05] => 1
    [2014-06-01] => 1
    [2014-06-04] => 1
    [2014-06-11] => 1
)

See demo

2个数组计算id = date的次数

2 个答案: