Php Codeigniter:集成离子认证屏幕和引导模式
若用户单击页面上的任何链接,我将检查用户是否以控制器方法登录。如果他未登录,则我会在屏幕上显示登录屏幕模式。Php Codeigniter:集成离子认证屏幕和引导模式,php,codeigniter,twitter-bootstrap,ion-auth,Php,Codeigniter,Twitter Bootstrap,Ion Auth,若用户单击页面上的任何链接,我将检查用户是否以控制器方法登录。如果他未登录,则我会在屏幕上显示登录屏幕模式。 到目前为止一切正常。 如果用户输入了错误的用户名和密码。下一页显示为没有任何CSS的新页。 如何将结果返回到模式登录屏幕? view/index.php $('.index-product img').click(function() { var product = $('<div id="modal-productDetail" class="modal hi
到目前为止一切正常。
如果用户输入了错误的用户名和密码。下一页显示为没有任何CSS的新页。
如何将结果返回到模式登录屏幕?
view/index.php
$('.index-product img').click(function() {
var product = $('<div id="modal-productDetail" class="modal hide fade" tabindex="-1" role="dialog" aria-labelledby="myModalLabel" aria-hidden="true"></div>');
$.ajax({
url: "<?php echo site_url('gallery/openProductDetail');?>",
type: 'POST',
success: function(msg) {
product.html(msg);
product.modal('show');
}
});
});
<div class="modal-header">
<h4 id="myModalLabel">Login To Your Account</h4>
</div>
<div id="loginModal" class="modal-body">
<div class="row">
<?php echo form_open("auth/login");?>
<div class="span3">
<ul class="unstyled" id="emailLogin">
<li><small class="muted">Login with your email</small></li>
<?php echo form_open("auth/login");?>
<li>
<input name="identity" value="" id="identity" type="text" placeholder="Email Address"/>
<span class="help-block" id="emailError" style="display:none;"></span>
</li>
<li>
<input name="password" value="" id="password" type="password" placeholder="Password"/>
<span class="help-block" id="passwordError" style="display:none;"></span>
</li>
<li>
<input style="display:none;" name="remember" value="1" id="remember" type="checkbox">
</li>
<li>
<a class="text-info" id="accountProblem" href="auth/forgot_password">Account problem?</a>
<input class="loginButton" type="submit" name="submit" value="Login">
<?php echo form_close();?>
</li>
<li>
<div id="infoMessage"><?php echo $message;?></div>
</li>
</ul>
</div>
</div>
</div>
查看/login.php
$('.index-product img').click(function() {
var product = $('<div id="modal-productDetail" class="modal hide fade" tabindex="-1" role="dialog" aria-labelledby="myModalLabel" aria-hidden="true"></div>');
$.ajax({
url: "<?php echo site_url('gallery/openProductDetail');?>",
type: 'POST',
success: function(msg) {
product.html(msg);
product.modal('show');
}
});
});
<div class="modal-header">
<h4 id="myModalLabel">Login To Your Account</h4>
</div>
<div id="loginModal" class="modal-body">
<div class="row">
<?php echo form_open("auth/login");?>
<div class="span3">
<ul class="unstyled" id="emailLogin">
<li><small class="muted">Login with your email</small></li>
<?php echo form_open("auth/login");?>
<li>
<input name="identity" value="" id="identity" type="text" placeholder="Email Address"/>
<span class="help-block" id="emailError" style="display:none;"></span>
</li>
<li>
<input name="password" value="" id="password" type="password" placeholder="Password"/>
<span class="help-block" id="passwordError" style="display:none;"></span>
</li>
<li>
<input style="display:none;" name="remember" value="1" id="remember" type="checkbox">
</li>
<li>
<a class="text-info" id="accountProblem" href="auth/forgot_password">Account problem?</a>
<input class="loginButton" type="submit" name="submit" value="Login">
<?php echo form_close();?>
</li>
<li>
<div id="infoMessage"><?php echo $message;?></div>
</li>
</ul>
</div>
</div>
</div>
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您需要更改application/controllers/auth.php
中的login()
方法。
我想您需要在重定向调用之前添加到login()
方法,如下所示:
$location = '';
if($this->form_validation->run() == true)
{
//check to see if the user is logging in
//check for "remember me"
$remember = (bool) $this->input->post('remember');
if ($this->ion_auth->login($this->input->post('identity'), $this->input->post('password'), $remember))
{
//if the login is successful
//redirect them back to the home page
$this->session->set_flashdata('message', $this->ion_auth->messages());
$data = $this->ion_auth->messages();
$location = '/';
}
else
{
$this->session->set_flashdata('message', $this->ion_auth->errors());
$data = $this->ion_auth->errors();
$location = 'auth/login';
}
}
else
{
$this->data['message'] = (validation_errors()) ? validation_errors() : $this->session->flashdata('message');
$this->data['identity'] = array('name' => 'identity',
'id' => 'identity',
'type' => 'text',
'value' => $this->form_validation->set_value('identity'),
);
$this->data['password'] = array('name' => 'password',
'id' => 'password',
'type' => 'password',
);
$data = $this->data;
}
if($this->input->is_ajax_request())
{
echo json_encode(array('data' => $data));
}
else
{
($location) ? redirect($location, 'refresh') : $this->_render_page('auth/login', $data);
}
并添加js代码来处理来自
login()
方法的响应 对于任何ajax请求,它不会返回错误吗?您可以在这里找到login()
方法抱歉,但我不理解您的评论。我应该在哪里将其添加到login()
方法?至于我,我会扩展Auth controller并更改login()
方法,但此解决方案需要更多时间。更简单的解决方案-将login()
中的所有重定向移动到一个位置,并添加$this->input->is\u ajax\u request()
检查,但您需要更改login()的逻辑。请查看我的更新代码。