Php Codeigniter如何在响应后将值传递给ajax?
我是codeigniter的新手,现在在响应后将值传递给ajax时陷入了困境。我尝试在php检查后回显值,但当ajax响应显示所有html元素时,我只希望回显值仅来自php,请提供任何想法。这是我在控制器中的代码Php Codeigniter如何在响应后将值传递给ajax?,php,ajax,Php,Ajax,我是codeigniter的新手,现在在响应后将值传递给ajax时陷入了困境。我尝试在php检查后回显值,但当ajax响应显示所有html元素时,我只希望回显值仅来自php,请提供任何想法。这是我在控制器中的代码 public function index(){ //Fetch page template if(isset($_POST['email'])){ $this->data['email'] = $this->sign_up_m->ge
public function index(){
//Fetch page template
if(isset($_POST['email'])){
$this->data['email'] = $this->sign_up_m->get_by(array('email' => $_POST['email']), TRUE );
if(empty($this->data['email'])){
$this->sign_up_m->save(array('email' => $_POST['email'], 'date' => date('Y-m-d H:i:s')));
}else{
$this->data['email'] = $_POST['email'];
echo'exist';
}
}
$page = $this->uri->segment(1);
if( is_null($page)){
$page = 'home';
}
$this->data['home'] = $this->page_m->get_by(array('nice_name' => $page),TRUE);
count($this->data['home']) || show_404(current_url());
//Fetch page data
$method = '_'. $this->data['home']->template;
if(method_exists($this, $method))
{
$this->$method();
}
else
{
log_message('errors','Could not load template'. $method. 'in file'. __FILE__. 'at line'. __LINE__);
show_error('Could not load template', $method);
}
$this->data['subview'] = $this->data['home']->template;
$this->load->view('_layout_main', $this->data);
}
这是我的ajax
$('.sign_up').click(function(){
var email = $('.email_signup').val();
var validate = isValidEmailAddress(email);
if(!isValidEmailAddress(email)){
$('.invalide_email').css('display','inline');
return;
}
$.ajax({
type: 'post',
url: 'http://localhost:8080/project/www.example.com/',
data: {
email: email,
},
beforeSend: function(){
$('#load').css('display','inline');
},
success: function( response ) {
$('#load').css('display','none');
console.log(response);
}
});
});
放置
exit()
或die()代码>在函数末尾,您将从该函数中回显此url的值。。。。r你确定这是正确的吗?Naincy,我工作正常,我只想在它响应时回显我想要的字符串,urfusion,我试过了,但不起作用urfusion,现在我解决了这个问题,感谢urfusion的想法,它在函数末尾不起作用,但在传导isset()中起作用