如何根據另一個數組的鍵和值打印出一個數組的鍵和值？

Question

我有一個看起來像這樣的數組。

[DETROIT] => Array
(
    [NORTH] => 20.00%
    [SOUTH] => 30.00%
    [WEST] => 25.00%

)

[CHICAGO] => Array
(
    [NORTH] => 59.14%
    [SOUTH] => 12.94%
    [WEST] => 0.00%
    [EAST] => 34.60%
)

 [NEW YORK] => Array
(
    [WEST] => 38.00%
    [EAST] => 49.00%
)

[DALLAS] => Array
(
    [WEST] => 60.57%
)

在另一個用戶的幫助下，我得到了將其打印為該表的代碼。

          DETROIT     CHICAGO   NEW YORK   DALLAS
NORTH     20.00       59.14      N/A       N/A
SOUTH     30.00       12.94      N/A       N/A
WEST      25.00       0.00       38.00     60.57
EAST      N/A         34.60      49.00     N/A

這是我用於打印下表的代碼：

echo "<table>";
$heading = "<tr><td>&nbsp;</td>";
$stats_key = array("Stat 1","Stat 2","Stat 3");
$cities = array();
foreach ($stats as $city=>$city_stats){
$cities[] = $city;
$heading .= "<td>" . $city . "</td>";
}
$heading .= "</tr>";
foreach ($stats_key as $key){
$table .= "<tr><td>" . $key . "</td>";
foreach ($cities as $cit){
 $table .= "<td>" . $stats[$cit][$key] . "</td>";
}
$table .= "</tr>";
}

echo $heading;
echo $table;

我還有另一個陣列，關鍵是負責這些位置的人員。

[John Doe] => Array
(
    [0] => DETROIT
    [1] => DALLAS

)

[Sara Smith] => Array
(

    [1] => NEW YORK
)

 [Donald Duck] => Array
(
    [0] => CHICAGO
)

現在，我只想打印出同一張桌子，但是同一個人下面的位置將一起打印出來，如下所示：

           DETROIT     DALLAS   NEW YORK   CHICAGO
  NORTH     20.00       N/A      N/A       59.14
  SOUTH     30.00       N/A      N/A       12.94
  WEST      25.00       60.57    38.00     0.00
  EAST      N/A         N/A      49.00     34.60

任何幫助，將不勝感激，謝謝。

編輯：這是我到目前為止所做的：

echo "<table>";
$heading = "<tr><td>&nbsp;</td>";
$stats_key = array("Stat 1","Stat 2","Stat 3");
$cities = array();

foreach($PERSON as $per){
foreach ($stats as $city=>$city_stats){
$cities[] = $city;
$heading .= "<td>" . $city . "</td>";
}
$heading .= "</tr>";
foreach ($stats_key as $key){
$table .= "<tr><td>" . $key . "</td>";
foreach ($cities as $cit){
 $table .= "<td>" . $stats[$cit][$key] . "</td>";
}
$table .= "</tr>";
}

echo $heading;
echo $table;
}
}

但是那個額外的foreach循環讓我重復了很多。 同樣，沒有兩個人可以掌管一個地方。

Answer 1

代碼中的最后一個foreach

foreach ($cities as $cit){
    foreach($person_name as $charge){
        if($charge = $cit){
            $table .= "<td>" . $stats[$cit][$key] . "</td>";
            break;
        }else{
            $table .= "<td> N/A </td>";
        }
}

Answer 2

據我了解，您需要以與第二個數組中顯示的順序相同的順序顯示城市。

嘗試這個：

$array1 = array(
    'DETROIT' => array(
        'NORTH' => '20.00%',
        'SOUTH' => '30.00%',
        'WEST' => '25.00%',
    ),
    'CHICAGO' => array(
        'NORTH' => '59.14%',
        'SOUTH' => '12.94%',
        'WEST' => '0.00%',
        'EAST' => '34.60%',
    ),
    'NEW YORK' => array(
        'WEST' => '38.00%',
        'EAST' => '49.00%',
    ),
    'DALLAS' => array(
        'WEST' => '60.57%',
    ),
);
$array2 = array(
    'John Doe' => array(
        'DETROIT',
        'DALLAS',
    ),
    'Sara Smith' => array(
        'NEW YORK',
    ),
    'Donald Duck' => array(
        'CHICAGO',
    ),
);

function get2ndLevel($arr, $bKeys = false) {
    $vals = array();
    array_walk($arr, function($v, $k) use(&$vals, $bKeys){
        $vals = array_merge($vals, $bKeys? array_keys($v) : array_values($v));
    });
    return array_values(array_unique($vals));
}

$dirs = get2ndLevel($array1, true);
$cities = get2ndLevel($array2);
$cities = array_merge($cities,
    array_values(array_diff(array_keys($array1), $cities))
);

$table = "<tr><td>&nbsp;</td>";
foreach ($cities as $city){
    $table .= "<td>" . $city . "</td>";
}
$table .= "</tr>";
foreach ($dirs as $dir){
    $table .= "<tr><td>" . $dir . "</td>";
    foreach ($cities as $city){
        $table .= "<td>" . (isset($array1[$city][$dir])? $array1[$city][$dir] : "N/A") . "</td>";
    }
    $table .= "</tr>";
}

echo "<table>". $table. "</table>";

演示版