如何獲得這種格式的二維數組？

Question

我得到了這種格式的數組。

   Date            Employee               Notes
   2013-03-08        ABC                 Notes of ABC on 08-03-2013
   2013-03-08        PQR                 Notes of PQR on 08-03-2013
   2013-03-08        XYZ                 Notes of XYZ on 08-03-2013
   2013-03-09        ABC                 Notes of ABC on 09-03-2013
   2013-03-09        PQR                 Notes of PQR on 09-03-2013
   2013-03-09        XYZ                 Notes of XYZ on 09-03-2013

我想要這種格式的結果數組

 Date            Employee               Notes
 2013-03-08        ABC                 Notes of ABC on 08-03-2013
                   PQR                 Notes of PQR on 08-03-2013
                   XYZ                 Notes of XYZ on 08-03-2013
 2013-03-09        ABC                 Notes of ABC on 09-03-2013
                   PQR                 Notes of PQR on 09-03-2013
                   XYZ                 Notes of XYZ on 09-03-2013

那么我怎么寫我的PHP代碼來得到這樣的結果呢？ 這是一個二維數組，我使用過foreach循環來顯示。

有人可以幫我嗎？

Answer 1

您只需要迭代初始數組，並將當前日期存儲在var中。 當下一個日期不變時，您將子數據存儲在同一子條目中。

假設你有

$initial = array(
    array('2013-03-08','ABC','Notes of ABC on 08-03-2013'),
    array('2013-03-08','PQR','Notes of PQR on 08-03-2013'),
    array('2013-03-08','XYZ','Notes of XYZ on 08-03-2013'),
    array('2013-03-09','ABC','Notes of ABC on 09-03-2013'),
    array('2013-03-09','PQR','Notes of PQR on 09-03-2013'),
    array('2013-03-09','XYZ','Notes of XYZ on 09-03-2013')
)

您可以使新數組像

$final = array();
$currentDate = false;
foreach($initial as $index => $subArray)
{
    if ($currentDate === false || $currentDate != $subArray[0])
    {
        $currentDate = $subArray[0];
        $final[$currentDate] = array();
    }

    $final[$currentDate][] = array($subArray[1], $subArray[2]);
}

Answer 2

這將起作用：

$entries = array(
    array("Date" => "2013-03-08", "Employee" => "ABC", "Notes" => "Notes of ABC on 08-03-2013"),
    array("Date" => "2013-03-08", "Employee" => "PQR", "Notes" => "Notes of PQR on 08-03-2013"),
    array("Date" => "2013-03-08", "Employee" => "XYZ", "Notes" => "Notes of XYZ on 08-03-2013"),
    array("Date" => "2013-03-09", "Employee" => "ABC", "Notes" => "Notes of ABC on 09-03-2013"),
    array("Date" => "2013-03-09", "Employee" => "PQR", "Notes" => "Notes of PQR on 09-03-2013"),
    array("Date" => "2013-03-09", "Employee" => "XYZ", "Notes" => "Notes of XYZ on 09-03-2013")
);

$y = array();
foreach ($entries as $entry) {
    $date = $entry["Date"];
    if (!isset($y[$date])) $y[$date] = array();
    unset($entry["Date"]);
    $y[$date][] = $entry;
}

print_r($y);