【问题标题】:Drupal 7, validate node as per interfaceDrupal 7,根据接口验证节点
【发布时间】:2020-03-19 16:26:03
【问题描述】:

我创建了一个 web 服务,它工作正常,我正在执行节点创建操作,它工作正常。 我需要以与在接口插入表单中验证相同的方式验证要保存的节点。

我试过了

drupal_form_submit($nodeType . '_node_form', $form_state, (object) $node);

它的节点引用字段总是给我错误

您能否建议其他方法来执行与以编程方式创建的节点上的接口相同的验证?

节点引用字段的错误是:

" field_ente : 这个条目不能 参考。 "

节点 (6310) 正确存在,如果我尝试执行 node_save,它会正确保存

完整功能如下

function my_ws_resource_create($field_nome = '', $field_cognome = '', $field_codice_fiscale = '', $field_data_di_nascita = '', $field_ente= '')
{

    module_load_include('inc', 'node', 'node.pages');
    global $user;
    $nodeType = 'contatti';

    $node = new stdClass();
    $node->type = $nodeType;
    $node->uid = $user->uid;
    $node->status = 1;
    $node->revision = 1;
    $node->promote = 0;
    $node->comment = 0;

    node_object_prepare($node);

    $node->field_cognome['und'][0]['value'] = $field_cognome;
    $node->field_nome['und'][0]['value'] = $field_nome;
    $node->field_codice_fiscale['und'][0]['cck_codicefiscale'] = $field_codice_fiscale;
    $node->field_data_di_nascita['und'][0]['value'] = $field_data_di_nascita;
    $node->field_categoria_contatto['und'][0]['tid'] = '66';



    // $node->field_ente = array('und' => array(array('nid'=> $field_ente )));
    // this field causes the error
    $node->field_ente = array('und' => array(array('nid'=> '6310')));



    $node->field_simplenews_term['it'][0]['tid'] = '13660';

    $form_state = array();      
    $form_state['values']['type'] = $nodeType;    
    $form_state['values']['name'] = $user->name;    
    $form_state['values']['status'] = 1;
    $form_state['values']['promote'] = 1;
    $form_state['values']['sticky'] = 0;

    $form_state['values']['op'] = t('Save');
    drupal_form_submit($nodeType . '_node_form', $form_state, (object) $node);

    if ($errors = form_get_errors()) {
        return services_error(implode(" ", $errors), 406, array('form_errors' => $errors));
    }
    return 'Creation successful';
}

【问题讨论】:

    标签: php validation drupal-7 nodereference


    【解决方案1】:

    我在以下方面取得了成功(删除 $form_state 并替换 drupal_form_submit):

    if ($node = node_submit($node)) {
      node_save($node);
      // Success!
    }
    else {
      // Fail :(
    }
    

    【讨论】:

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