按日期列对二维数组的行进行分组，并覆盖每组中无值的关联元素

Question

我有一个二维数组。如何按日期值分组并合并每个日期的主动值和被动值？

[
    ["status" => "ONE",   "active" => "103347.02", "passive" => " ",    "date" => "2016-08-01"],
    ["status" => "TWO",   "active" => "65.32",     "passive" => " ",    "date" => "2016-08-01"],
    ["status" => "THREE", "active" => "3258.39",   "passive" => " ",    "date" => "2016-08-01"],
    ["status" => "FOUR",  "active" => "35470.01",  "passive" => " ",    "date" => "2016-08-01"],
    ["status" => "ONE",   "active" => "152685.68", "passive" => " ",    "date" => "2016-08-02"],
    ["status" => "TWO",   "active" => "1739.33",   "passive" => " ",    "date" => "2016-08-02"],
    ["status" => "THREE", "active" => "1300.00",   "passive" => " ",    "date" => "2016-08-02"],
    ["status" => "FOUR",  "active" => "31461.06",  "passive" => " ",    "date" => "2016-08-02"],
    ["status" => "ONE",   "active" => " ",         "passive" => "2236", "date" => "2016-08-01"],
    ["status" => "THREE", "active" => " ",         "passive" => "4496", "date" => "2016-08-02"]
]

预期结果：

[
    ["status" => "ONE",   "active" => "103347.02", "passive" => "2236", "date" => "2016-08-01"],
    ["status" => "TWO",   "active" => "65.32",     "passive" => " ",    "date" => "2016-08-01"],
    ["status" => "THREE", "active" => "3258.39",   "passive" => " ",    "date" => "2016-08-01"],
    ["status" => "FOUR",  "active" => "35470.01",  "passive" => " ",    "date" => "2016-08-01"],
    ["status" => "ONE",   "active" => "152685.68", "passive" => " ",    "date" => "2016-08-02"],
    ["status" => "TWO",   "active" => "1739.33",   "passive" => " ",    "date" => "2016-08-02"],
    ["status" => "THREE", "active" => "1300.00",   "passive" => "4496", "date" => "2016-08-02"],
    ["status" => "FOUR",  "active" => "31461.06",  "passive" => " ",    "date" => "2016-08-02"]
]

Answer 1

最简单的解决方案是按日期重新索引，并将数字加在一起。 您可以在 php 中输入杂耍字符串。

$output = [];
foreach($array as $item){
    if(!isset($output[$item['date']])){
        $output[$item['date']] = $item;
    } else {
        $output[$item['date']]['active'] += $item['active'];
        $output[$item['date']]['passive'] += $item['passive'];
    }
}

//You can optionally remove the date keys again with:
$output = array_values($output);

Answer 2

这是获得输出的另一种方式。

  $input=array( 
    0=> array( "status"=> "ONE" ,"active"=>  "103347.02","passive"=>" " ,"date"=> "2016-08-01" ),
    1=> array( "status"=> "TWO" ,"active"=> "65.32" ,"passive"=> " " ,"date"=> "2016-08-01" ),
    2=> array("status"=> "THREE" ,"active"=> "3258.39" ,"passive"=> " " ,"date"=>"2016-08-01" ),
    3=> array( "status"=> "FOUR" ,"active"=> "35470.01" ,"passive"=> " " ,"date"=>"2016-08-01" ),
    4=> array( "status"=> "ONE" ,"active"=> "152685.68" ,"passive"=> " " ,"date"=> "2016-08-02" ),
    5=> array( "status"=> "TWO" ,"active"=> "1739.33" ,"passive"=> " " ,"date"=> "2016-08-02" ),
    6=> array( "status"=> "THREE" ,"active"=> "1300.00" ,"passive"=> " " ,"date"=> "2016-08-02" ),
    7=> array( "status"=> "FOUR" ,"active"=> "31461.06" ,"passive"=>  " " ,"date"=>  "2016-08-02" ),
    8=> array( "status"=> "ONE" ,"active"=>  " " ,"passive"=> "2236" ,"date"=> "2016-08-01"),   
    9=> array( "status"=> "THREE" ,"active"=> " " ,"passive"=> "4496" ,"date"=>  "2016-08-02" )
);

foreach($input as $index_1 => $value_1){
    if(ctype_space($value_1['active']) && !ctype_space($value_1['passive'])) {
        foreach($input as $index_2 => $value_2){
            if($value_2['status']==$value_1['status'] && ctype_space($value_2['passive'])){
                $input[$index_2]['passive']=$value_1['passive'];
                unset($input[$index_1]);
                break; 
            }
        }
    }
}
var_dump($input);