Question

我使用dupal API在drupal中为表单创建了一个自定义模块，并在我的数据库中创建了一个表我想要的是保存下拉框的值而不是它的索引到数据库...我的.module文件是：

<?php
function form_test_menu() {
$items['formtest'] = array(
    'title' => 'Form Test',
    'page callback' => 'drupal_get_form',
    'page arguments' => array('form_test_form'),
    'access callback' => TRUE,
);
return $items;

}

function form_test_form($form,&$form_submit) {
$form['name'] = array(
    '#title' => t('Name'),
    '#type' => 'textfield',
    '#required' => TRUE,
);
$form['college'] = array(
    '#title' => t('College/University'),
    '#type' => 'textfield',
    '#required' => TRUE,
);
$form['education'] = array(
    '#title' => t('Education'),
    '#type' => 'select',
    '#description' => 'Select your higher education .',
    '#options' => array(t('--- SELECT ---'), t('B.tech'), t('MCA'), t('MBA'),t('Graduate')),
    '#required' => TRUE,
);
$form['percentage'] = array(
    '#title' => t('Percentage/CGPA'),
    '#type' => 'textfield',
    '#required' => TRUE,
);
$form['application'] = array(
    '#title' => t('Job Applied'),
    '#type' => 'select',
    '#options' => array(t('---select---'),t('Web Developer'),t('Web Designer'),t('SEO'),t('Marketing')),
    '#required' => TRUE,
);


$form['submit'] = array(
    '#value' => 'Submit',
    '#type' => 'submit',
);
return $form;

}

function form_test_form_submit($form, $form_state) {
global $user;
// Here u can insert Your custom form values into your custom table.
db_insert('drupal')
    ->fields(array(
        'name' => $form_state['values']['name'],
        'college' => $form_state['values']['college'],
        'education' => $form_state['values']['education'],
        'percentage' => $form_state['values']['percentage'],
        'application' => $form_state['values']['application'],

    ))->execute();
drupal_set_message("successfully saved data");

} ？＆GT;

所有都在一个文件（.module）....任何帮助将不胜感激。谢谢。

Answer 1

解决方案是将options数组从一个函数传递到另一个函数。要实现这一点，请使用＆＃34; form_test_form＆＃34;的第二个参数。和＆＃34; form_test_form_submit＆＃34;功能

function form_test_form($form,&$form_submit) {
  $form_submit['myoptions'] = array(' select', 'option1' .....
}

然后

function form_test_form_submit($form,&$form_state) {    
 $myoptions = $form_state['myoptions']; etc ...
}

Drupal中的表单提交

1 个答案: