Submit AngularJS form with CodeIgniter

Question

Let me preface this by saying I'm new to AngularJS & CodeIgniter. Trying to create a single page app. In the center it displays news stories. On the side is a form to submit new stories. I'm to the point where the stories display fine from the DB. But when I try to create new stories from the form on the side, it enters 0 into the DB. I assume I have an issue with my function in the model. And it's probably as basic as not having the information encoded correctly to pass between MVC.

index.php

<div class="span2" ng-controller="NewStoryCtrl">
  <h4>New Story</h4>
  <form name="newStory">
      <label for="title">Title: </label>
      <input type="text" name ="title" ng-model="news.title" required>
      <label for="text">Text: </label>
      <textarea type="text" name="text" ng-model="news.text"required></textarea>
      <button ng-click="createNews()">Submit</button>
</form>

controller.js

function NewStoryCtrl($scope, $http) {
$scope.news = {title:$scope.title, text:$scope.text};
$scope.createNews = function(){
    $http.post('/ci/index.php/news/create/',$scope.news);
};}

news.php (controller)

public function create() {
    $this->news_model->set_news();
}

news_model.php (model)

public function set_news() {
$this->load->helper('url');

$slug = url_title($this->input->post('title'), 'dash', TRUE);

$data = array(
    'title' => $this->input->post('title'),
    'slug' => $slug,
    'time' => time(),
    'text' => $this->input->post('text')
);

return $this->db->insert('news', $data);
}

The stuff in the model is leftover from my initial CI news tutorial. That's why I assume the error is here. What's the best way to pass the information from the controllers.js to the model?

EnigmaRM · Accepted Answer · 2013-03-13 22:13:14Z

6

As expected, my issue was with not getting the right type of data. Controller was expecting a variable, but in Angular controller I was passing it as JSON. I hadn't ever gone through decoding.

news.php (controller)

public function create() {
   $data = json_decode(file_get_contents('php://input'), TRUE);
   $this->news_model->set_news($data);
}

And then in the model, I just needed to pass it as a parameter to set_news(). I ended up changing some variable names, just for personal clarification.

*news_model.php*

public function set_news($data) {
    $this->load->helper('url');
    $slug = url_title($data['title'], 'dash', TRUE);

    $retval = array(
        'title' => $data['title'],
        'slug' => $slug,
        'time' => time(),
        'text' => $data['text']
    );

    return $this->db->insert('news', $retval);
}

answered Mar 13, 2013 at 22:13

EnigmaRM

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1 Comment

Razvan Zamfir Over a year ago

Very useful information. Could you post all the code related to this form in one answer?

artless-noise-bye-due2AI · Accepted Answer · 2013-07-05 12:42:29Z

1

I haven't used Angular.js, but your data format is probably correct. For example, Backbone sends a single $_POST variable called model with json encoded data, as I have used it.

It is imperative that you use Firebug, Webdev or other tools to see what is going on when you try to do AJAX work like this; otherwise you will go crazy . Look at the variable being sent to your backend - it is probably described, an encoded single var you will need to collect, json_decode, CLEAN & VALIDATE and then use.

edited Jul 5, 2013 at 12:42

artless-noise-bye-due2AI

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answered Mar 12, 2013 at 21:37

jmadsen

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2 Comments

EnigmaRM Over a year ago

Thanks. And I have been using Firebug. a JSON file is being passed. But I'm not familiar with collecting or decoding json. In the controller, how do I call it? just "news"? I've encoded things before. That makes more sense to me as I already have a PHP variable.

EnigmaRM Over a year ago

Found a solution: $data = json_decode(file_get_contents('php://input'), TRUE);

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