在codeigniter中如何使用ajax登录后重定向
我有一个登录弹出窗口Modal。我正在通过ajax登录
I have a login popup Modal. and i am logging through ajax
模态
<div class="modal-body">
<form action="<?php echo base_url('Login');?>" method="POST">
<div class="form-group">
<input type="text" placeholder="Email or Mobile*" value="" id="loginEmail" name="email" class="form-control input-feild">
</div>
<div class="form-group">
<input type="password" placeholder="Password*" value="" id="loginPassword" name="password" class="form-control input-feild">
</div>
<div class="form-group">
<input type="button" id="l_submit" name="l_submit" value="Login" class="btn btn-primary input-feild">
</div>
</form>
<p id="error-msg"></p>
</div>
我正在尝试使用ajax成功登录后重定向。如果电子邮件和密码正确，则重定向到任何页面。如果没有，那么它将显示错误。
I am trying to redirect after successful login using ajax. If email and password is correct then redirect to any page. If not then it will show the Error.
控制器
function index() {
$this->form_validation->set_rules('email', 'Email', 'trim|required');
$this->form_validation->set_rules('password', 'Password', 'trim|required');
if ($this->form_validation->run() == false) {
echo validation_errors();
}else {
$email = $this->input->post("email");
$password = $this->input->post("password");
$user = $this->Perfect_mdl->check_user($email, $password);
if ($user) {
$logged_in_data = array();
foreach ($user as $logged_in_data) {
$logged_in_data = array(
'id' => $user[0]->id,
'email' => $user[0]->email
);
}
$this->session->set_userdata($logged_in_data);
$id = $this->session->userdata('email');
$data['details'] = $this->Perfect_mdl->get_login_user_detail($id);
echo "Yes";
}else {
echo "No";
}
}
}
这是我的控制器，其中我正在检查登录用户的电子邮件和密码是否正确/不正确。
This is my controller in which i am checking login user email and password correct/incorrect.
这是我的脚本
<script type="text/javascript">
$("#l_submit").click(function(){
var email = $("#loginEmail").val();
var password = $("#loginPassword").val();
$.ajax({
url : "<?php echo base_url('Login');?>",
type: 'POST',
data : {'email':email,'password':password},
success: function(msg) {
if (msg == "Yes")
window.location.href = "<?php echo current_url(); ?>";
else if (msg == "No")
$('#error-msg').html('<div class="alert alert-danger text-center">Incorrect Email & Password. Please try again ...</div>');
else
$('#error-msg').html('<div class="alert alert-danger">' + msg + '</div>');
}
});
return false;
});
</script>
正如您在成功部分中看到的那样，当我输入正确的电子邮件/密码时。它显示是。但是我想重定向另一个页面，为什么在正确的电子邮件/密码上显示 YES 。并且在错误的电子邮件/密码上显示否。
As you can see in the success part, when i entered the correct email/password. It is showing Yes. But i want to redirect another page, Why this is showing YES on correct email/password.and On incorrect email/password This is showing NO.
我做错了???
你需要更改几行代码在jQuery和Controller函数中。这里我附上你的代码的更新版本。请参阅以下内容：
You need to change few lines of codes in jQuery and Controller function. Here I am attaching updated version of your code. Please refer below:
查看（Bootstrap模态）
<div class="modal-body">
<form action="<?php echo base_url('Login');?>" method="POST">
<div class="form-group">
<input type="text" placeholder="Email or Mobile*" value="" id="loginEmail" name="email" class="form-control input-feild">
</div>
<div class="form-group">
<input type="password" placeholder="Password*" value="" id="loginPassword" name="password" class="form-control input-feild">
</div>
<div class="form-group">
<input type="button" id="l_submit" name="l_submit" value="Login" class="btn btn-primary input-feild">
</div>
</form>
<p id="error-msg"></p>
</div>
这是您的查看文件。它将保持不变。在单击按钮时，您已编写了需要修改的脚本。附加控制器功能后会发生：
This is your view file. It will remain same. On clicking on button you have written a script which need to be modified. Will attact after attaching controller's function:
控制器
function index() {
if (!$this->input->is_ajax_request()) {
echo 'No direct script is allowed';
die;
}
$this->form_validation->set_rules('email', 'Email', 'trim|required');
$this->form_validation->set_rules('password', 'Password', 'trim|required');
if ($this->form_validation->run() == false) {
$result['status'] = 'error';
$result['message'] = validation_errors();
}else {
$email = $this->input->post("email");
$password = $this->input->post("password");
$user = $this->Perfect_mdl->check_user($email, $password);
if ($user) {
$logged_in_data = array();
foreach ($user as $logged_in_data) {
$logged_in_data = array(
'id' => $user[0]->id,
'email' => $user[0]->email
);
}
$this->session->set_userdata($logged_in_data);
$id = $this->session->userdata('email');
$data['details'] = $this->Perfect_mdl->get_login_user_detail($id);
$result['status'] = 'success';
$result['message'] = 'Yeah! You have successfully logged in.';
$result['redirect_url'] = base_url();
}else {
$result['status'] = 'error';
$result['message'] = 'Whoops! Incorrect Email & Password. Please try again';
}
}
$this->output->set_content_type('application/json');
$this->output->set_output(json_encode($result));
$string = $this->output->get_output();
echo $string;
exit();
}
脚本
<script type="text/javascript">
$("#l_submit").click(function(){
var email = $("#loginEmail").val();
var password = $("#loginPassword").val();
$.ajax({
url : "<?php echo base_url('Login');?>",
type: 'POST',
data : {'email':email,'password':password},
success: function(resp) {
if (resp.status == "success")
window.location.href = resp.redirect_url;
else
$('#error-msg').html('<div class="alert alert-danger">' + resp.message + '</div>');
}
});
return false;
});
这是正确答案。如果您遇到任何问题，请告诉我。
This is the correct answer. Let me know if you face any issue.