javascript - thinkphp 异步操作执行到die()

Question

controller:
public function index(){

    $user = M('User');
    $userinfo = session('user');
    if(IS_POST){
        $info = I($data);
        //var_dump($info['nickname']); exit;
        if( !$info['nickname']){ 
            die(json_encode(['status'=>'0','info'=>'请输入用户名'])); 
        }
        return false;
        //未输入密码
        if( !$info['password']){
            die(json_encode(['status'=>'0','info'=>'请输入密码'])); 
        }
        $rs = $user->where(['nickname'=>$info['nickname']])->find();
        //var_dump($rs);
        if(!$rs){
            die(json_encode(['status'=>'5' , 'info'=>'不存在此用户，请注册!']));
        }else{
             if($rs['password'] == md5($info['password'])){
                $login['username'] = $info['nickname'];
                $login['logintime'] =  time();
                $login['loginip'] = get_client_ip();
                session('user', $login);
                $userinfo = session('user');
                die(json_encode(['status'=>'3' , 'info'=> '登录成功！']));
            }else{
                die(json_encode(['status'=>'4' , 'info'=>'密码错误!']));
            }
        }    
    }else{

        $this->assign('nickname', $userinfo['nickname']);
        $this->display();
    
    }              
  
}

页面：
  $('#button').click(function(){
      var url= "{:U('')}";
      var $data = $('#logininfo').serialize();
      $.ajax({
           type: 'post',
           url: url,
           data: $data,
           dataType: 'json',
           success:function(d){
                if(d.status == '0'){
                    alert(d.info);
                    return false;
                    window.location.href="{:U('')}";
                    
                }else if(d.status == '4'){
                    alert(d.info);  
                    $('#upass').val("");
                    $('#upass').focus();
                }else if(d.status == '5'){
                    alert(d.info);
                    $('#uname').val("");  
                    $('#upass').val("");
                    $('#uname').focus();
                }
                
           }

      });
});


点击登录之后就会跳转到die()

Answer 1

跳到哪个die

Answer 2

问题解决了，还是老问题，以前犯过一次，就是form表单提交时用的input标签，重复提交了