CODEWITHSUNDEEP
javascriptphpajaxcodeigniter
jldavis76
jldavis76

Reputation: 822

AJAX call in CodeIgniter not working as expected

I have an ajax call from my javascript file that is attempting to post data to a specific function in my controller. When I view the data in the success function of my call, I get the html of a totally unrelated view as the returned data, and I am unsure why.

My AJAX call:

function processResults(task_id){
  var finalResults = localStorage.getItem('results');
  console.log(finalResults);

  $.ajax({
    type: 'POST',
    url: 'task/getResults',
    data: {'answers': finalResults},
    success: function(data){
      console.log(data);
    }
 });
}

My Controller function:

public function getResults(){

  $this->load->model('testResults');

  $finalResults = $this->input->post('answers');

  $finalResults = json_decode($finalResults, true);

  if ($taskTestId != '') {
    $this->testAnswers->insertTaskData($finalResults);
  }
}

Is the url in the ajax call incorrect? I'm not really sure why it is outputting the html of a completely unrelated view. Any help would be appreciated!

Upvotes: 0

Views: 659

Answers (2)

Jordi
Jordi

Reputation: 13

Tpojka answer will work.

But there is a more elegant way to return a JSON objet with CI 

$this->output
        ->set_content_type('application/json')
        ->set_output(json_encode(array('foo' => 'bar')));

http://www.codeigniter.com/userguide3/libraries/output.html#CI_Output::set_content_type

Upvotes: 0

Tpojka
Tpojka

Reputation: 7111

Last line controller/method passed in AJAX should be echo of something. At the end you would have something like:

if ($expression) {
    echo json_encode($something);
} else {
    echo json_encode($something_else);
}

Upvotes: 1

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