2017-07-29 104 views
0

Codeigniter与数据库连接但没有来自数据库表的查询响应。我的数据库已经从PostgreSQL的转换到MySQL,我们的表类型是InnoDB的,但我在创建服务器的Godaddy的样本表的表型是MyISAM.I不知道什么是我面临Codeigniter在Godaddy上查询空响应,但在本地主机上工作正常

.htaccess中的问题

RewriteEngine on 
RewriteCond $1 !^(index\.php|resources|robots\.txt) 
RewriteCond %{REQUEST_FILENAME} !-f 
RewriteCond %{REQUEST_FILENAME} !-d 
RewriteRule ^(.*)$ index.php?/$1 [L,QSA] 

login_view.php

var name = { 
ajax: 1, 
name: $('#login-username').val() 
} 
$.ajax({ 
url: base_url+"login/username_validation", 
type: 'POST', 
data: name, 
dataType: 'json', 
success: function(data) 
{  
$.each(data.response.userid, function (a, b) 
{ 
if((b.count) == 1) 
{ 
$("#login").removeAttr("style"); 
$(".alertbox").css("display", "none");  
$("#login-password") .focus(); 
} 
});  

的login.php

public function username_validation() 
{ 
    if ($this->input->post('ajax')) 
    { 
     $username = $this->input->post('name'); 
     $result['userid'] = $this->common_model->chk_username($username); 
$this->output->set_content_type('application/json')->set_output(json_encode(array('response'=>$result))); 
    } 
} 

common_model.php

public function chk_username($username) 
{ 
    $this->db->select('count(*) as count'); 
    $this->db->from('mas_user'); 
    $this->db->where('login_id',$username); 
    return $this->db->get()->result(); 
} 

MySQL表

user_creation_id login_id password 
    1     admin  admin123 

ajax : 1 
name :admin 

响应

{"response":{"userid":[]}} 
status 200 ok 

回答

0

看来问题就在你的login.php你问ajax等于真,whe ñAJAX实际上是等于1

使用以下两个选项

一个在你的login.php尝试

if ($this->input->post('ajax') == 1) 

或在您的login_view.php

var name = { 
ajax: true, 
name: $('#login-username').val() 
} 
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