dsirr48088 2015-10-17 16:31
浏览 54

这些标题是基于在CodeIgniter中使用AJAX的主要困惑

In my homepage AJAX is working properly, but after some action when I again return this same page, AJAX not working, until I again run CodeIgniter. I am using CodeIgniter 3.0.2

What should I do?

Model:

<?php
class tele extends CI_Model
{
    public function industry_fetch()
    {
        $qr=$this->db->query("select * from industry");
        if($qr)
        {
            return $qr->result();
        }
        else
        {
            return false;
        }
    }
    public function subindustry_fetch()
    {
        $qr=$this->db->query("select * from subindustry");
        if($qr)
        {
            return $qr->result();
        }
        else
        {
            return false;
        }
    }
    public function company_fetch()
    {
        $qr=$this->db->query("select * from company");
        if($qr)
        {
            return $qr->result();
        }
        else
        {
            return false;
        }
    }
    public function contact_fetch($v)
    {
        $qr=$this->db->query("select * from company where NAME like '$v%'");
        if($qr)
        {
            return $qr->result();
        }
        else
        {
            return false;
        }
    }
    public function company_registration()
    {
        $cname=mysql_real_escape_string($this->input->post('n'));
        $cemail=mysql_real_escape_string($this->input->post('e'));
        $cpassword=mysql_real_escape_string($this->input->post('p'));
        $cadress=mysql_real_escape_string($this->input->post('ad'));
        $cphone=mysql_real_escape_string($this->input->post('cn'));
        $ctime=mysql_real_escape_string($this->input->post('st'));
        $csi=$this->input->post('si');
        $qr=$this->db->query("select * from subindustry where NAME='$csi'");
        $r=$qr->row();
        //print_r($r);
        //exit;
        $csiid=$r->S_ID;
        $qry=$this->db->query("insert into company values('','$cname','$cemail','$cpassword','$cadress','$cphone','$ctime','$csiid',0)");
        if($qry)
        {
            return true;
        }
        else
        {
            return false;
        }
    }
    public function login_fetch()
    {
        $email=$this->input->post('e');
        $password=$this->input->post('p');
        $qr=$this->db->query("select * from company where EMAIL='$email' and PASSWORD='$password'");
        $r=$qr->num_rows();
        if($r==1)
        {
            $row=$qr->row();
            $id=$row->C_ID;

            $this->session->set_userdata('id',$id);
            return true;
        }
        else
        {
            return false;
        }
    }
    public function login_success()
    {
        $p=$this->session->userdata('id');
        $qr=$this->db->query("select * from company where C_ID='$p'");
        if($qr)
        {
            $st=$qr->row();//mysql_fetch_row()
            return $st;
        }
    }
}
?>
  • 写回答

1条回答 默认 最新

  • dongqian3198 2015-10-17 18:27
    关注

    As I understood you are running your ajax call onclik function like this

    $('.yourclass').click(function(){
    
      $.ajax({
       Your code
    
       })
    })
    

    change this to this

    $(document).on('click','.yourclass',function(){
    
          $.ajax({
           Your code
    
           })
        })
    

    Or you can run the ajax call on onclick function with java script it will give you any error. Read this tutorial for how to run ajax on javascript onclick function dynamically without page refresh. http://w3code.in/2015/10/how-to-edit-delete-and-update-data-without-refreshing-page-in-codeigniter/

    评论

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