codeigniter

问题 I'm trying to log in using verify_password. I had this working after I updated to php5.5 now it's not working. I don't know what I have done to break it. Here's the Model <?php class Login_model extends CI_Model { public function __construct() { // Call the CI_Model Constructor parent::__construct(); $this -> load -> database(); } public function login($email, $password) { // SELECT id, email, password FROM user_registration WHERE email = $email & password =$password LIMIT 1 $this -> db ->

问题 i am novice in codeigniter. I am trying to insert multiple news that contains some check-box where its values are looped from a database. But i don't know what should code i write for this in controller & Model to insert values. can any one suggest or help me to write code? view(content.php) <?php foreach($result as $aresult) { ?> <input type="checkbox" name="category_name[<?php echo $aresult->category_name; ?>]" value="<?php echo $aresult->category_name;?>" /> <?php echo $aresult->category

问题 I'm using a cron that gets accesses a URL to run a scheduled process. I keep getting successful emails... is there a way for me to only get emails if the wget request fails? wget http://www.domain.com/cron/dailyEmail 2>&1; 回答1: wget --quiet http://www.domain.com/cron/dailyEmail || echo "wget failed" (Note that an empty response is not a failure.) 回答2: If you're running wget directly in the cron setup, then no. You can't conditionally redirect output. You could, however, put the wget command

问题 This question already has answers here : php : how to pass database value into a javascript/jquery function (3 answers) Closed 6 years ago . i have to get output in this format var sampleTags = ['c++', 'scala']; My javascript function is: <script> $(document).ready(function(){ $(function(){ var sampleTags; $.ajax({ url:"<?php echo base_url('ajax_get_tags/gettags'); ?>" }).done(function(data) { if (data) { sampleTags = data; } }); ...................... ....................... $(function(){

问题 I am developing a project on codeigniter , I am facing difficulty while creating pagination of records. Actually, when I user $this->db->get() then its okay, pagination work fine. But when do pagination using $this->db->query() then pagination is nor working, following is the code in my controller . $this->load->library('pagination'); $pagination_config['base_url'] = base_url().'welcome/index'; $pagination_config['per_page'] = 3; $pagination_config['num_links'] = 3; $pagination_config['total

问题 With the new version of CodeIgniter; you can only set rules in a static form_validation.php file. I need to analyze the posted info (i.e. only if they select a checkbox). Only then do I want certain fields to be validated. What's the best way to do this, or must I use the old form validation class that is deprecated now? 回答1: You cannot only set rules in the config/form_validation.php file. You can also set them with: $this->form_validation->set_rules(); More info on: http://codeigniter.com

问题 Since I have more than 20k~ items to show in the DataTable I want to populate it according to some parameters to avoid big lag. // Clears the DataTable to avoid duplication of items $('#contacts-table > tbody').empty(); var category_id = 5; // Just for tests $.post('ajax_getAll', {category_id: category_id}, function(response){ // I retrieve the 'response' as json_encoded var json = JSON.parse(response); }); If I were not using DataTables , in the traditional way I would just do (to populate

问题 I have a website which has 2 parts - lets say A and B. Now I want to move part A to a new domain. Currently, the URLs are of the form: www.xyz.com/A/controller/function - for part A www.xyz.com/A/B/controller/function - for part B my new links will be: www.abc.com/controller/function - for part A www.xyz.com/B/controller/function - for part B Can you suggest me a good way to handle these redirects? I am using Codeigniter Framework. 回答1: Are you using the same CI install for both parts? If so

问题 Looking at http://instagram.com/developer/authentication/ - The recommended way of communication is serverside. I understand how I do step1 and step2. But in step3 they want me POST data directly serverside? (I understand how I could do a post request from jQuery) But is it even possible to do directly from PHP/CodeIgniter? This is basically step3: In the previous step, you’ll have received a code which you’ll have to exchange in order to receive an access_token for the user. In order to make

问题 I Have used this ignited_datatables https://github.com/IgnitedDatatables/Ignited-Datatables in my CodeIgniter project . It successfully returns data from the database . But the main problem is that when I add a new record the newly inserted data is not shown in the table automatically and when I refresh the page then the newly inserted data is then shown in the table . so I want to show that data automatically through ajax and I have not to refresh the page. My View <div class="box-body table