尝试获取非对象/Php codeigniter的属性
我似乎不知道这段代码有什么问题。谢谢你的帮助。 我收到错误消息:尝试获取非对象/Php codeigniter的属性,php,mysql,codeigniter,Php,Mysql,Codeigniter,我似乎不知道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 我的型号: public function show_user() { $this->db->select(''); $thi
A PHP Error was encountered
Severity: Notice
Message: Trying to get property of non-object
Filename: views/members.php
Line Number: 24
我的型号:
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 $data;
}
}
public function members()
{
if ($this->session->userdata('is_logged_in'))
{
$this->load->model('model_users');
$data['member'] = $this->model_users->show_user();
$this->load->view('members', $data);
}
else
{
redirect('main/restricted');
}
}
echo "<p> Congratulations you are logged in!!</p>";
foreach($member as $row)
{
echo $row->Name. "<br/>";
echo $row->email. "<br/>";
}
我的控制器:
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 $data;
}
}
public function members()
{
if ($this->session->userdata('is_logged_in'))
{
$this->load->model('model_users');
$data['member'] = $this->model_users->show_user();
$this->load->view('members', $data);
}
else
{
redirect('main/restricted');
}
}
echo "<p> Congratulations you are logged in!!</p>";
foreach($member as $row)
{
echo $row->Name. "<br/>";
echo $row->email. "<br/>";
}
我的观点:
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 $data;
}
}
public function members()
{
if ($this->session->userdata('is_logged_in'))
{
$this->load->model('model_users');
$data['member'] = $this->model_users->show_user();
$this->load->view('members', $data);
}
else
{
redirect('main/restricted');
}
}
echo "<p> Congratulations you are logged in!!</p>";
foreach($member as $row)
{
echo $row->Name. "<br/>";
echo $row->email. "<br/>";
}
echo“祝贺您登录!!”;
foreach($row成员)
{
echo$row->Name.“
”;
echo$row->email。“
”;
}
这就是我的代码。我想显示数据库中的用户信息,但出现错误。请帮帮我 只需返回
$q-result()代码>
我不确定您的数据库看起来像什么,但我怀疑这是因为名称
或电子邮件
属性在$row
对象上不存在。您确定users
表中的字段名为Name
,还是Name
(小写)?如果是小写,则代码应为
foreach($member as $row)
{
echo $row->name. "<br/>"; //Changed this line
echo $row->email. "<br/>";
}
谢谢你,我的错误已经解决了,但是我没有得到预期的结果。我只想要那个刚刚登录的用户的电子邮件和姓名,而不是数据库用户的所有姓名和电子邮件的输出。。。。你能帮我吗???@Sushmita,从你的查询中,你会得到所有拥有你提供的电子邮件地址的用户的回复