Question

嗨我有结构数组，如

array:2 [
  0 => array:3 [
    "title" => "samba"
    "diff" => "06:30:00"
    "user" => "Janek"
  ]
  1 => array:3 [
    "title" => "Ramba"
    "diff" => "03:30:00"
    "user" => "Janek"
  ]
  2 => array:3 [
    "title" => "fit"
    "diff" => "03:30:00"
    "user" => "Pawel"
  ]
]

输出应该是

array:2 [
  0 => array:3 [
    "title" => "samba"
    "sum" => "10"
    "user" => "Janek"
  ]
  1 => array:3 [
    "title" => "Ramba"
    "sum" => "03:30:00"
    "user" => "Pawel"
  ]

]

我需要对所有diff键求和，但对于用户。例如user =＆gt; janek sum diff =＆gt; 10 user =＆gt; Pawel sum diff =＆gt;＆gt; 3:30

我试过

$values=array_sum(array_column($newArray,'diff))

但是这个总和所有数组。我希望你理解。

Answer 1

我找到了你的答案。那么这里的代码很乱，但我认为你可以自己管理休息......任何编辑建议也欢迎...

<?php
$sumArray = array();
$myArray = array(
  0 => array(
    "title" => "samba",
    "diff" => "06:30:00",
    "user" => "Janek"
  ),
  1 => array(
    "title" => "Ramba",
    "diff" => "03:30:00",
    "user" => "Janek",
  ),
  2 => array(
    "title" => "fit",
    "diff" => "03:30:00",
    "user" => "Pawel"
  )
  );
$i = 0;
$xarray = array();
foreach ($myArray as $k=>$subArray) 
{   
    $parts = explode(':', $subArray["diff"]);
    $seconds = ($parts[0] * 60 * 60) + ($parts[1] * 60) + $parts[2];    
     if(in_array($subArray["user"],$xarray)):        
          $key = myfunction( $sumArray,"user",$subArray["user"]);
          $sumArray[$key]["sum"] = gmdate("H:i:s", $sumArray[$key]["sum1"]+$seconds);
     else:       
         $sumArray[$i] = $subArray;
         $sumArray[$i]["sum1"] =  $seconds;
         $sumArray[$i]["sum"] = gmdate("H:i:s",$seconds);
         $xarray[] = $subArray["user"];
         $i++;
     endif;     
}
foreach($sumArray as $key=>$value)
{
    unset($sumArray[$key]["diff"]);
    unset($sumArray[$key]["sum1"]);
}
function myfunction($arrays, $field, $value)
{
   foreach($arrays as $key => $product)
   {
      if ( $product[$field] === $value )
         return $key;
   }
}
echo "<pre>";
print_r($sumArray);
echo "</pre>";

输出： -

Array
(
    [0] => Array
        (
            [title] => samba
            [user] => Janek
            [sum] => 10:00:00
        )

    [1] => Array
        (
            [title] => fit
            [user] => Pawel
            [sum] => 03:30:00
        )

)

Answer 2

要总结分组用户的小时数，请使用DateTime::add()，preg_replace和array_values函数使用以下方法：

// supposing $arr is your initial array
$result = [];
foreach ($arr as $el) {
    if (!isset($result[$el['user']])) {
        $result[$el['user']] = $el;
    } else {
        $interval = new \DateInterval(preg_replace("/^(\d{2}):(\d{2}):\d{2}$/", 'PT$1H$2M', $el['diff']));
        $result[$el['user']]["diff"] = (new \DateTime($result[$el['user']]["diff"]))->add($interval)->format('H:i:s');
    }
}
foreach ($result as &$v) {  // renaming keys 'diff'/'sum'
    $v["sum"] = $v["diff"];
    unset($v["diff"]);
}
print_r(array_values($result));

输出：

Array
(
    [0] => Array
        (
            [title] => samba
            [user] => Janek
            [sum] => 10:00:00
        )

    [1] => Array
        (
            [title] => fit
            [user] => Pawel
            [sum] => 03:30:00
        )
)

键的Sum数组

2 个答案: