CODEWITHSUNDEEP
javascriptphpjqueryajaxcodeigniter
kev_m
kev_m

Reputation: 325

Codeigniter ajax form validation error

I have researched this code that will make my validation with ajax.

My controller is:

public function create(){


    $data = array('success' => false, 'messages' => array());

    $this->form_validation->set_rules('PROVINCE','Province Name','trim|required|max_length[30]|callback_if_exist');
    $this->form_validation->set_error_delimiters('<p class="text-danger"','</p>');

    if($this->form_validation->run($this)){

        $data['success'] = true;
    }else{
        foreach ($_POST as $key => $value) {
            # code...
            $data['messages']['key'] = form_error($key);
        }
    }

    echo json_encode($data);
}

And my javascript is:

<script>
    $('#form-user').submit(function(e){
        e.preventDefault();

        var me = $(this);

        // perform ajax
        $.ajax({
            url: me.attr('action'),
            type: 'post',
            data: me.serialize(),
            dataType: 'json',
            success: function(response){
                if (response.success == true){
                    alert('success');
                }else{
                    $.each(response.messages, function(key, value) {
                        var element = $('#' + key);

                        element.closest('div.form-group')
                        .removeClass('has-error')
                        .addClass(value.length > 0 ? 'has-error' : 'has-success')
                        .find('.text-danger')
                        .remove();

                        element.after(value)

                    });
                }
            }

        });


    });


</script>

At first my if else statement was:

            if (response.success == true){
                alert('success');
            }else{
               alert('failed');

                });

But when i put the codes:

  $.each(response.messages, function(key, value) {
                        var element = $('#' + key);

                    element.closest('div.form-group')
                    .removeClass('has-error')
                    .addClass(value.length > 0 ? 'has-error' : 'has-success')
                    .find('.text-danger')
                    .remove();

                    element.after(value)

The button doesn't work anymore if the validation fails.

Upvotes: 0

Views: 511

Answers (1)

charlietfl
charlietfl

Reputation: 171669

You are hard coding the string 'key' in the php loop where you actually want the variable $key

Change

$data['messages']['key']

To

$data['messages'][$key]

Upvotes: 1

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