Question

in_array函数，由于某种原因，返回false;

这是我正在使用的代码：

$query = "SELECT users_invites.invite_user_id
          FROM users_invites
          JOIN users ON users.id = users_invites.user_id AND users_invites.user_id =". param('session_id')."
          ORDER BY users.id ASC";

$response = $this->_db->query($query)->result_array();

 foreach ($response as $key => $value)
 {
    if (!in_array($_REQUEST['invite_user_id'][$key],$response[$key]))
        $this->_db->insert("users_invites",array("user_id"=> param('session_id') , "invite_user_id"=>$_REQUEST['invite_user_id'][$key]));


 }

in_array函数不起作用，插入命令总是执行。有什么想法吗？

P.S： invite_user_id是一个数组。修改：响应值为：

Array
(
    [0] => Array
        (
            [invite_user_id] => 20
        )

    [1] => Array
        (
            [invite_user_id] => 19
        )

    [2] => Array
        (
            [invite_user_id] => 3
        )

)

$ _REQUEST ['invite_user_id']值：

Array
(
    [0] => 3
    [1] => 4
)

Answer 1

鉴于你得到的结构，我会对数据进行foreach将其分解为单个数组，并从表中返回所有id。

从那里我foreach $_REQUEST['invite_user_id']对阵新阵列$ids，这是一个示例：

$ids = array();
foreach ($response as $item)
{
   $ids[] = $item['invite_user_id'];
}

foreach ($_REQUEST['invite_user_id'] as $id) 
{
    if (!in_array($id, $ids))
    {
        echo "{$id} was not found...\n";
    }
    else
    {
        echo "{$id} was found...\n";
    }
}

<强> And here is a live DEMO.

Answer 2

您的请求应该是这样的

$query = "SELECT users_invites.invite_user_id
          FROM users_invites
          JOIN users ON users.id = users_invites.user_id AND users_invites.user_id =". param('session_id')."
          ORDER BY users.id ASC";

$response = $this->_db->query($query)->result_array();

$invite_user_id = array();


 foreach ($response as $key => $value)
 {
   $invite_user_id[] = $value;
 }

 foreach( $_REQUEST['invite_user_id'] AS $key => $value )
 {
      if ( !in_array( $value,$invite_user_id) )
      {
       $this->_db->insert("users_invites",array("user_id"=> param('session_id') , "invite_user_id"=> $value));
      }
 }

Answer 3

以下内容应该运作良好：

$query = 'SELECT users_invites.invite_user_id
            FROM users_invites
            JOIN users ON users.id = users_invites.user_id AND
                 users_invites.user_id = ' . param('session_id') . '
        ORDER BY users.id ASC';
$response = $this->_db->query($query)->result_array();

$storedInvites = array();
foreach ($response as $index => $data) {
    $storedInvites[] = $data['invite_user_id'];
}

foreach ($_REQUEST['invite_user_id'] as $index => $inviteUserId) {
    if (!in_array($inviteUserId, $storedInvites)) {
        $insertData =  array(
            'user_id' => param('session_id'),
            'invite_user_id' => $inviteUserId
        );

        $this->_db->insert('users_invites', $insertData);
    }
}

in_array命令不起作用

3 个答案: