Data check in database if already exist in codeigniter through ajax

Question

I want to check data in database through ajax in codeigniter how can get check name already exist through ajax

controller

public function insert(){
    $user = array('Name' => $this->input->post('name'), 
                  'Cnic' => $this->input->post('cnic'),
                  'Phone' => $this->input->post('phone'),
                  'Address' => $this->input->post('address'),
        );
    $this->load->model('suppliermodel');
    if($this->suppliermodel->checkData($user['Name'])){
        if ($this->session->userdata('user_id'))
          $detail = 'Already exist';
      $this->load->view('admin/includes/header');
      $this->load->view('admin/includes/sidemenu');
      $this->load->view('admin/add_supplier',['exist'=>$detail]);
      $this->load->view('admin/includes/footer');
      $this->load->view('admin/includes/add_supplier_footer');

    }
    else{
     $this->suppliermodel->add($user);  
    }

}

model

public function checkData()
{
  $name = $this->input->post('name');
  $this->db->select('*');
  $this->db->where('Name', $name);
  $this->db->from('suppliers');
  $query = $this->db->get();
  if($query->num_rows() >0){
    return $query->result();
  }
  else{
    return $query->result();
    return false;
  }
}

what is ajax code and controller function

this if($this->suppliermodel->checkData($user['Name'])){ should be if(count($this->suppliermodel->checkData($user['Name'])) > 0){ — devpro
– devpro, Commented Sep 5, 2016 at 12:27
you are passing arg in the function checkData in controller but not in modal — Ish
– Ish, Commented Sep 5, 2016 at 12:30

Community · Accepted Answer · 2020-06-20 09:12:55Z

1

How about this?

Controller

public function insert(){

    // set a rule that make the Name field is unique by 'set_rules'
    $this->form_validation->set_rules('Name', 'Name Field', 'required|is_unique[suppliers.name]');
    //$this->form_validation->set_rules('[Other Field]', '[Field Name to Display]', '[Restriction]');

    // if the field cannot pass the rule
    if ($this->form_validation->run() === FALSE) {

        $errors = array();

        foreach ($this->input->post() as $key => $value) {
            // Add the error message for this field
            $errors[$key] = strip_tags(form_error($key));
        }
        // Clear the empty fields (correct)
        $response['errors'] = array_filter($errors); 
        $response['status'] = false;

    }
    else {
        // otherwise, call the model
        $result = $this->suppliermodel->add($user);  

        if ( $result ) {
            $response['status'] = true;
        }
        
    }

    echo json_encode($response);
}

JavaScript

$.ajax({
    url: '//localhost/insert',
    data: {
        Name: $('input[name=Name]').val(),
        Cnic: $('input[name=Cnic]').val(),
        Phone: $('input[name=Phone]').val(),
        Address: $('input[name=Address]').val()
    },
    dataType: 'JSON',
    type: 'POST',
    error: function(xhr) {
        alert('request failed!');
    },
    success: function(response) {

        var response = $.parseJSON(JSON.stringify(response));

        if (response.status != true) {

            $.each(response.errors, function(field, i) {
                alert( field+ ' errors with ' + i)
            });
        }
        else {

            alert('success!');
        }
    }
});

Use is_unique[table.fieldToCompare] to make the field is always unique.

Wish it helps.

set_rules restrictions, see the Codeigniter User Guide Form validation

If fails, the controller would return a set of JSON, with field and error message. Then you can handle it in $.ajax's success.

edited Jun 20, 2020 at 9:12

CommunityBot

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answered Sep 5, 2016 at 12:46

Benyi

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4 Comments

salman Over a year ago

i want it through ajax

Benyi Over a year ago

@salman I've updated the answer contains the code that you want. If fails, the controller returns a set of JSON containing status and errors message. Then you can handle it in JavaScript code. I use this as my solution.

Benyi Over a year ago

The main idea of the answer is that you don't have to make a function again your own by use set_rules method provided by Codeigniter. If form validation fails, return an encoded error message as JSON so that you can handle yourself, anyway you want.

devpro Over a year ago

@salman: just review this: stackoverflow.com/questions/23172903/ajax-json-unexpected-token ... chk this, or enjoy kro dost

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