Question

如果$key=>value匹配，如何删除整个数组对象？

例如，在我的数据中，我有["endDate"]=> string(10) "2017-06-24"，如果该日期与今天date('Y-m-d')匹配，我希望从$row

中删除整个对象

代码：

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

如果您在foreach中放置＆符号前的＆符号，则可以直接更改$ json-＆gt;数据。根据你所写的内容，我认为我理解你的问题，这可能就是你所需要的。

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;
        }
      }
    }

如果没有，请重新解释您的问题，以便我提供您需要的答案。

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;
    }
  }

您也可以考虑跳到下一次迭代，因为看起来您正在构建来宾电子邮件列表，而不是尝试修改json供稿源。

$date = date('Y-m-d');
foreach ($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;
    }
}

在我提出的两个选项之间，第二个选项需要的代码更少，执行时间会稍微快一些。

如果找到密钥，则删除数组对象

2 个答案: