Question

如何获取值为[existence] == 1

的对象

如果[existence] == 1我希望获得具有[existence] == 1

的所有对象

并删除所有[existence] == 0

这是我的阵列。

Array
(

    [0] => Array
        (
            [id] => 3
            [accountcode_naisen] => 
            [extentype] => 0
            [extenrealname] => 
            [name] => 0090000270
            [extenktaiemail] => 
            [secret] => Myojyo42_f
            [username] => 0090000270
            [guestIpAddr] => 192.168.236.15
            [participantSetting] => Array
                (
                    [id] => 13
                    [existence] => 1
                    [leader] => 1
                    [simultaneous] => 
                )

        )

)

这是我目前的代码

 foreach ($participants as $participant=>$c) {

            if ($c['existence'] != 1) {
              unset($participants[$participant]);
            }

          }

我收到此错误消息**********

[30-Sep-2015 15:42:38] PHP Notice: Use of undefined constant participantSetting - assumed 'participantSetting' in index.php on line 253

Answer 1

你可以使用比循环更有效的东西。您可以将array_filter()与匿名函数一起用作回调。像这样的东西

$participants = array_filter(
    $participants,
    function($element) { return $element['participantSetting']['existence'] == 1; }
);

Answer 2

我认为这是有效的

foreach ($participants as $participant=>$c) {

   if ($c['participantSetting']['existence'] != 1) {
          unset($participants[$participant]);
        }
}

Answer 3

以下代码可能对您有帮助..

foreach ($participants as $key => &$participant)
{    
     if ($participant['existence'] != 1)
     {
          unset($participant[$key]);
     }    
}

在上述迭代后打印数组时，您将获得预期的输出。

如何通过检查其值来获取数组的值

3 个答案: