在数组中的PHP阵列有条件的计数值

Question

我有阵列的一个这样的数组：

array('0'=> 
      array(
      'fromDate' =>2007, 
      'toDate' => 2008, 
      'value' => 7) 
     '1'=> 
      array(
      'fromDate' =>2010, 
      'toDate' => , 
      'value' => 3) 
     '2'=> 
      array(
      'fromDate' => 2007, 
      'toDate' => , 
      'value' => 4)) 

我想算“价值”，把它们加起来通过这样的“FROM日期”分组：

array('0' => 
      array(
      'fromDate' => 2007, 
      'sum' => 2) 
     '1' => 
      array(
      'fromDate' => 2010, 
      'sum' => 1)) 

任何帮助非常感谢！

Answer 1

$myData = array('0'=> 
      array(
      'fromDate' =>2007, 
      'toDate' => 2008, 
      'value' => 7), 
     '1'=> 
      array(
      'fromDate' =>2010, 
      'toDate' => NULL, 
      'value' => 3), 
     '2'=> 
      array(
      'fromDate' => 2007, 
      'toDate' => NULL, 
      'value' => 4) 
); 

$counts = array_reduce(
    $myData, 
    function ($counter, $entry) { 
     if (!isset($counter[$entry['fromDate']])) { 
      $counter[$entry['fromDate']] = array(
       'fromDate' => $entry['fromDate'], 
       'sum' => 1 
      ); 
     } else { 
      $counter[$entry['fromDate']]['sum']++; 
     } 
     return $counter; 
    }, 
    array() 
); 

var_dump($counts); 

Answer 2

如果您想要计算具有“fromDate”的阵列数量，例如2007你有：

function countByFromDate($array) { 
    $result = array(); 
    foreach($array as $item) { 
    $fromDate = $item['fromDate']; 
    if (!isset($result[$fromDate])) { 
     $result[$fromDate] = array('fromDate' => $fromDate, 'sum' => 0); 
    } 
    $result[$fromDate]['sum']++; 
    } 

    return $result; 
}