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php - 试图获取非对象/Php codeigniter 的属性

转载 作者:行者123 更新时间:2023-11-29 03:35:19 24 4
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我似乎无法弄清楚这段代码有什么问题。谢谢你的帮助。 我收到错误消息:

A PHP Error was encountered

Severity: Notice

Message: Trying to get property of non-object

Filename: views/members.php

Line Number: 24

我的模型:

public function show_user()
{
$this->db->select('');
$this->db->from('users');
$this->db->where('email',$this->input->post('email'));

$q=$this->db->get('');

if($q->num_rows() > 0 )
{
$data = array();

foreach($q->result() as $row)
{
$data=$row;
}

return $data;
}
}

我的 Controller :

public function members()
{
if ($this->session->userdata('is_logged_in'))
{
$this->load->model('model_users');
$data['member'] = $this->model_users->show_user();
$this->load->view('members', $data);
}
else
{
redirect('main/restricted');
}
}

我的观点:

echo "<p> Congratulations you are logged in!!</p>";
foreach($member as $row)
{
echo $row->Name. "<br/>";
echo $row->email. "<br/>";
}

所以这是我的代码。我想显示数据库中的用户信息,但出现错误。请帮帮我..

最佳答案

只需要返回$q-result();

 if($q->num_rows() > 0 )
{
return $q-result();
}

关于php - 试图获取非对象/Php codeigniter 的属性,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/22608051/

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