How to display the selected value from the drop-down box?

Question

The following code is displaying the states, but when I click the submission button, I want to display the selected state and city. I am not able to display it. How can I achieve this?

$form['state1'] = array(
  '#type' => 'select',
  '#title' => t('State'),
  '#options' => $options_state,
  '#required' => TRUE,
  '#ajax' => array(
    'callback' => '_ajaxfunction',
    'wrapper' => 'divaroundseconddropdown'
  ),
);


$form['city_element_wrapper'] = array(
    '#prefix' => '<div id="divaroundseconddropdown">',
    '#suffix' => '</div>',
);


function _ajaxfunction($form, $form_state) { 
  $key = !empty($form_state['values']['state1']) ? $form_state['values']['state1'] : 1;
  $query_city=db_select('city', 'c');
  query_city->fields('c',array('District_Code','District_Name','State_Code'));
  $query_city->condition('State_Code', $key, '=');

  $query_city->orderBy('District_Name','ASC');
  $city_results = $query_city->execute();
  $options_city = array();

  foreach ($city_results as $city_record) {
    // This is correct.
    $options_city[$city_record->District_Code]=t($city_record->District_Name);
  }

  // Render city setting.
  $form['city_element_wrapper']['city'] = array(
    '#type' => 'select',
    '#title' => t('District'),
    '#options' => $options_city,
    '#required' => TRUE,);

  return $form['city_element_wrapper']; 
}

The following code is not working properly:

function form_test_form_submit($form, $form_state) {
  echo $prog_state = $form_state['values']['state1'];
  echo $data1 = $form_state['values']['city_element_wrapper'];
}

Answer 1

form_test_form_submit is called in php that's not displayed to user. Form is processed, it's values saved, and user is taken to desired page. Now you have 2 options:

• In submit handler set$form_state['redirect'] to a page you like. Save values to a database. On target page read them and display.
• In validation handler add markup elements with selected values, if you want to emphasize selection but leave user on a form.

Answer 2

Form:

$form['field_abt_aff_mainact'] = array(
  '#type' => 'select',
  '#multiple' => TRUE,
  '#size' => 5,
  '#title' => t('Main Activity:'),
  '#options' => array(
    '0' => 'select',
    '1' => 'India',
    '2' => 'U.S',
    '3' => 'UK'
  ),
  '#default_value' =>$rels,
);

Insertion to database:

$field_abt_aff_mainact = array();

foreach ($form_state['values']['field_abt_aff_mainact'] as $value) {
  $field_abt_aff_mainact[] =$value;
}
        
$insertion = db_insert('incident_occupation_disease')
  ->fields(array('occupation_work_main_act' => implode(',', $field_abt_aff_mainact)))
  ->condition('main_incident_id', $inst_id, '=')
  ->execute();

Display Default values:

To show default value of multiple select:

// Fetch values from databse 
$fetching = db_select('incident_occupation_disease', 'ifb')
  ->fields('ifb')
  ->condition('main_incident_id', $inst_id, '=')
  ->execute()
  ->fetchAll();
    
foreach ($defautvalues as $default) {   
  $occupation_work_main_act[] = $default->occupation_work_main_act;
}
          
$occupation_work_main_act = array();
$resultsed = explode(",", $occupation_work_main_act[0]);
$rels = array();
foreach ($resultsed as $rel) {
  $rels[]=$rel;
}

Here, $rels array will be use in Form API as default/selected value.