PHP数组:将值与匹配日期相加

时间:2012-08-27 12:39:04

标签: php arrays

如果它们具有相似的日期,我试图弄清楚如何对多维数组的某些值求和。

这是我的数组:

<?$myArray=array(

array(
        'year' => 2011,
        'month ' => 5,
        'day' => 13,
        'value' => 2
    ),
array(
        'year' => 2011,
        'month '=> 5,
        'day' => 14,
        'value' => 5
    ),
array(
        'year' => 2011,
        'month ' => 5,
        'day' => 13,
        'value' => 1
    ),
array(
        'year' => 2011,
        'month ' => 5,
        'day' => 14,
        'value' => 9
    )
);?>

这是我想要输出的样子:

<?$output=array(

array(
        'year' => 2011,
        'month ' => 5,
        'day' => 13,
        'value' => 3 //the sum of 1+2
    ),
array(
        'year' => 2011,
        'month '=> 5,
        'day' => 14,
        'value' => 14 //the sum of 5+9
    )
);?>

注意year/month/day上如何匹配4个子阵列,然后只汇总value。我见过其他SO threads on this topic,但找不到只有value求和的值,而不是year/month/day值。

思想?

2 个答案:

答案 0 :(得分:4)

最初使用年/月/日的组合索引输出数组可能最简单:

注意:上面的示例数组的所有month键都带有尾随空格。我只是在month处使用 no 尾随空格。

// Initialize output array...
$out = array();

// Looping over each input array item
foreach ($myArray as $elem) {
  // Initialize a new element in the output keyed as yyyy-mm-dd if it doesn't already exist
  if (!isset($out[$elem['year'] . "-" . $elem['month '] . "-" . $elem['day']])) {
    $out[$elem['year'] . "-" . $elem['month '] . "-" . $elem['day']] = array(
      // Set the date keys...
      'year' => $elem['year'],
      'month' => $elem['month '],
      'day' => $elem['day'],
      // With the current value...
      'value' => $elem['value']
    );
  }
  // If it already exists, just add the current value onto it...
  else {
     $out[$elem['year'] . "-" . $elem['month '] . "-" . $elem['day']]['value'] += $elem['value'];
  }
}

// Now your output array is keyed by date.  Use array_values() to strip off those keys if it matters:
$out = array_values($out);

输出(在调用array_values()之前):

array(2) {
  '2011-5-13' =>
  array(4) {
    'year' =>
    int(2011)
    'month' =>
    int(5)
    'day' =>
    int(13)
    'value' =>
    int(3)
  }
  '2011-5-14' =>
  array(4) {
    'year' =>
    int(2011)
    'month' =>
    int(5)
    'day' =>
    int(14)
    'value' =>
    int(14)
  }
}

更新

使用单键日期(而不是3部分)做同样的事情,没有连接就更容易:

$myArray=array(
      array(
            'date' => '2011-05-13',
            'value' => 2
        ),
    array(
            'date' => '2011-05-14',
            'value' => 5
        ),
        array(
            'date' => '2011-05-13',
            'value' => 7
        ),
    array(
            'date' => '2011-05-14',
            'value' => 3
        ),

);

   foreach ($myArray as $elem) {
      // Initialize a new element in the output if it doesn't already exist
      if (!isset($out[$elem['date']])) {
        $out[$elem['date'] = array(
          // Set the date keys...
          'date' => $elem['date'],
          // With the current value...
          'value' => $elem['value']
        );
      }
      else {
        $out[$elem['date']]['value'] += $elem['value'];
      }
    }

答案 1 :(得分:2)

这是我怎么做的。结果将在$newArray中,日期时间对象作为键。如果你只是想把它作为一个索引数组,它应该很容易做到。

// Example array
$myArray = array(
    array(
    'date' => new DateTime('1993-08-11'),
    'value' => 3
    ),

    array(
    'date' => new DateTime('1993-08-11'),
    'value' => 5
    )
);

$newArray = array();

foreach($myArray as $element)
{
    $iterationValue = $element['value'];
    $iterationDate = $element['date'];
    $dateKey = $iterationDate->format('Y-m-d');

    if(array_key_exists($dateKey, $newArray))
    {
        // If we've already added this date to the new array, add the value
        $newArray[$dateKey]['value'] += $iterationValue;
    }
    else
    {
        // Otherwise create a new element with datetimeobject as key
        $newArray[$dateKey]['date'] = $iterationDate;
        $newArray[$dateKey]['value'] = $iterationValue;
    }
}

nl2br(print_r($newArray));

实际上最终与@MichaelBerkowski解决方案完全相同。尽管如此,如果您不想在应用程序中稍后处理日期,那么拥有DateTime对象总是更灵活。

编辑:现在对其进行测试并修复错误