如果找到鍵，則刪除數組對象

Question

如果$key=>value匹配，如何刪除整個數組對象？

例如，在我的數據中，我有["endDate"]=> string(10) "2017-06-24" ，如果該日期與今天的date('Ym-d')相匹配，我希望從中刪除整個對象$row

碼：

foreach ($json->data as $row)
{       
   $date = date('Y-m-d');

   if($row->endDate == $date){

    $search = array_search($date, array_column('endDate', $row));

    unset($row[$search]);

    if (!in_array($row->guestEmail, $emails) && date('Y-m-d', strtotime($row->startDate))== date('Y-m-d'))
    {
        $guests[] = array(
            'FirstName'      => $row->guestFirstName,
            'LastName'       => $row->guestLastName,
            'email'          => $row->guestEmail,
            'country'        => $row->guestCountry,
            'check-in_date'  => $row->startDate,
            'check-out_date' => $row->endDate,
        );
        $emails[] = $row->guestEmail;
    }
  }
}

JSON：

$response = $o->curl(sprintf($url, $propertyID, $pageSize, $pageNumber, $resultsFrom));
$json = json_decode($response);

Answer 1

如果在您的foreach中的$ row之前放置一個＆符號，則可以直接更改$ json-> data。 從您寫的內容來看，我想我理解您的問題，這可能就是您所需要的。

foreach ($json->data as &$row)
    {       
       $date = date('Y-m-d');

       if ($row->endDate == $date) {

        $search = array_search($date, array_column('endDate', get_object_vars($row)));

        unset($row[$search]);

        if (!in_array($row->guestEmail, $emails) && date('Y-m-d', strtotime($row->startDate))== date('Y-m-d'))
        {
            $guests[] = array(
                'FirstName'      => $row->guestFirstName,
                'LastName'       => $row->guestLastName,
                'email'          => $row->guestEmail,
                'country'        => $row->guestCountry,
                'check-in_date'  => $row->startDate,
                'check-out_date' => $row->endDate,
            );
            $emails[] = $row->guestEmail;
        }
      }
    }

如果沒有，請改一下您的問題，以便我提供您所需的答案。

Answer 2

$date = date('Y-m-d');
foreach ($json->data as $index=> $row){
    if($row->endDate == $date) unset($json->data{$index});
}

foreach ($json->data as $row){
    if (!in_array($row->guestEmail, $emails) && date('Y-m-d', strtotime($row->startDate))== date('Y-m-d'))
     {
        $guests[] = array(
            'FirstName'      => $row->guestFirstName,
            'LastName'       => $row->guestLastName,
            'email'          => $row->guestEmail,
            'country'        => $row->guestCountry,
            'check-in_date'  => $row->startDate,
            'check-out_date' => $row->endDate,
        );
        $emails[] = $row->guestEmail;
    }
  }

您可能還考慮跳到下一個迭代，因為看起來您正在構建訪客電子郵件列表，並且沒有嘗試修改json feed的來源。

$date = date('Y-m-d');
foreach ($json->data as $row){
    if($row->endDate == $date) {
       continue;
    }

     if (!in_array($row->guestEmail, $emails) && date('Y-m-d', strtotime($row->startDate))== date('Y-m-d'))
     {
        $guests[] = array(
            'FirstName'      => $row->guestFirstName,
            'LastName'       => $row->guestLastName,
            'email'          => $row->guestEmail,
            'country'        => $row->guestCountry,
            'check-in_date'  => $row->startDate,
            'check-out_date' => $row->endDate,
        );
        $emails[] = $row->guestEmail;
    }
}

在我提出的兩個選項之間，第二個選項需要較少的代碼，並且執行時間會稍短一些。