没有退出功能，PHP无法返回值

Question

I need help. I am getting problem in returning value from Codeigniter. Whenever, I use exit; after echo it work fine but whenever i try return true it's dosen't work.

Same as i have comment code in PHP code. if i use exit after echo it works but if i don't do that it returns nothing

Ajax Request

$('#social-form').on('submit', function(e){

    e.preventDefault();
    var str = $( "#social-form" ).serialize();
    if (str === '') {
        swal("Please Fill All Fields");
    } else {
        $.ajax({
            type: "POST",
            url: baseUrl + "/admin/social/",
            data: str
        })
        .done(function (data) {
                console.log(data);
                swal("Information", data, "info");
            })
        .error(function () {
            swal("Oops", "We couldn't connect to the server!", "error");
        });
    }
});

Codeigniter-3

public function social(){
    $name = $this->input->post('name');
    $profile = $this->input->post('profile');
    $this->form_validation->set_rules('name', 'name', 'required|trim');
    $this->form_validation->set_rules('profile', 'profile', 'required|trim');
    if ($this->input->post() && $this->form_validation->run() != FALSE) {
        $this->load->model('Social_model','social');
        $this->social->update($name,$profile);
        echo 1;
        //exit;
        //return true;
    }
    else
    {
        echo 0;
        //exit;
        //return false;
    }
}

Answer 1

CodeIgniter has a layout, so after outputting a response there could be views that are outputted after your response, such as a footer or a debug bar.

Try using your console to see the status code of the response. Also note that it isn't bad practice in CodeIgniter to exit after AJAX calls, so perhaps you should just write a AJAX response helper which does all that for you (like setting the header and adding the exit ).

Answer 2

You probably need to be more specific about what you echo. This is one of several possible solutions.

controller

 public function social(){
    $name = $this->input->post('name');
    $profile = $this->input->post('profile');
    $this->form_validation->set_rules('name', 'name', 'required|trim');
    $this->form_validation->set_rules('profile', 'profile', 'required|trim');
    if ($name && $this->form_validation->run() != FALSE) {
        $this->load->model('Social_model','social');
        $this->social->update($name,$profile);
        $out = json_encode(array('result' => 'success'));
    }
    else
    {
        $out = json_encode(array('result' => 'failed'));
    }
        echo $out;
}

javascript

$('#social-form').on('submit', function (e) {
    e.preventDefault();
    var str = $("#social-form").serialize();
    if (str === '') {
        swal("Please Fill All Fields");
    } else {
        $.ajax({
            type: "POST",
            url: baseUrl + "/admin/social/",
            data: str,
            dataType: 'json'
        })
          .done(function (data) {
              console.log(data);
              if (data.result === 'success') {
                  swal("Information", "Success", "info");
              } else {
                  swal("Information", "Failed", "info");
              }
          })
          .error(function () {
              swal("Oops", "We couldn't connect to the server!", "error");
          });
    }
});