我有登录页面,但我不知道在哪里可以收到此警报
我将采用这样的脚本
echo ("<SCRIPT LANGUAGE='JavaScript'> window.alert('Your Login Succesfully ,'); window.location.href='home'; </SCRIPT>");
但是我不知道如何在模型和控制器中查询条件。我使用num_rows但它没有用。
我的控制器
public function login()
{
$u = $this->input->post('username');
$p = $this->input->post('password');
$data = $this->app_model->getlogindata($u,$p);
return $data;
}
我的模特
public function getlogindata($username,$password)
{
$u = $username;
$p = md5($password);
$cek_login = $this->db->get_where('login', array('username' => $u,'password'=>$p));
if(count($cek_login->result())>0)
{
foreach ($cek_login->result() as $qck)
{
if($qck->level=='puskesmas')
{
// $ambil_data = $this->db->get_where('puskesmas',array('id_puskesmas' => $u));
$this->db->select('*');
$this->db->from('puskesmas');
$this->db->join('login', 'puskesmas.id_puskesmas=login.id_puskesmas');
$this->db->where('username', $u);
$ambil_data = $this->db->get();
foreach ($ambil_data->result() as $qad)
{
$sess_data['logged_in'] = 'yes';
$sess_data['id_puskesmas'] = $qad->id_puskesmas; //
$sess_data['nama_puskesmas'] = $qad->nama_puskesmas;
$sess_data['alamat_puskesmas'] = $qad->alamat_puskesmas;
$sess_data['nama_petugas'] = $qad->nama_petugas;
$sess_data['nomor'] = $qad->nomor;
$sess_data['email'] = $qad->email;
$sess_data['level'] = $qad->level;
$this->session->set_userdata($sess_data);
}
header('location:'.base_url().'puskesmas');
} //xammp mu aktif?
elseif($qck->level=='dinas')
{
//$ambil_data = $this->db->get_where('dinas',array('id_dinas' => $u));
$this->db->select('*');
$this->db->from('dinas');
$this->db->join('login', 'dinas.id_dinas=login.id_dinas');
$this->db->where('username', $u);
$ambil_data = $this->db->get();
foreach ($ambil_data->result() as $qad)
{
$sess_data['logged_in'] = 'yes';
$sess_data['id_dinas'] = $qad->id_dinas;
$sess_data['nama_dinas'] = $qad->nama_dinas;
$sess_data['alamat_dinas'] = $qad->alamat_dinas;
$sess_data['kode_pos'] = $qad->kode_pos;
$sess_data['level'] = $qad->level;
$this->session->set_userdata($sess_data);
}
header('location:'.base_url().'dinas');
}
elseif($qck->level=='admin')
{
//$ambil_data = $this->db->get_where('admin',array('id_admin' => $u));
$this->db->select('*');
$this->db->from('admin');
$this->db->join('login', 'admin.id_admin=login.id_admin');
$this->db->where('username', $u);
$ambil_data = $this->db->get();
foreach ($ambil_data->result() as $qad)
{
$sess_data['logged_in'] = 'yes';
$sess_data['id_admin'] = $qad->id_admin;
$sess_data['nama_admin'] = $qad->nama_admin;
$sess_data['alamat_admin'] = $qad->alamat_admin;
$sess_data['status'] = $qad->status;
$sess_data['level'] = $qad->level;
$this->session->set_userdata($sess_data);
}
header('location:'.base_url().'admin');
}
else{
echo ("<SCRIPT LANGUAGE='JavaScript'> window.alert('Record Updated Successfully'); window.location.href='web'; </SCRIPT>");// i add this and still didnt work
}
}
}
}
在此脚本中,我只是不知道可以在哪里接收此警报以及可以使用什么参数来添加此警报。就像num_rows = 1一样,echo = blablabla 谢谢你先生
答案 0 :(得分:1)
您必须在代码点火器中使用Flash数据,或者您可以成功使用ajax回调,从而可以警告您想要的内容并重定向到所需的内容。
控制器
$this->session->set_flashdata('error_message', 'Incorrect Username or Password ! Please try again.');
redirect(URL.'backend/login');
查看
<?php
if ($this->session->flashdata('error_message'))
{
?>
<div class="alert alert-danger" style="color: #fff;">
<!-- <button class="close" data-close="alert"></button> -->
<span><?php echo $this->session->flashdata('error_message'); ?></span>
</div>
<?php
}
if ($this->session->flashdata('ok_message'))
{
?>
<div class="alert alert-success">
<!-- <button class="close" data-close="alert"></button> -->
<span><?php echo $this->session->flashdata('ok_message'); ?></span>
</div>
<?php
}
?>
答案 1 :(得分:0)
尝试这个。希望对您有所帮助。
else{
echo '<script language="javascript">';
echo 'alert("Record Updated Successfully")';
window.location.href = 'web';
echo '</script>';
}