如果内容作者与登录用户具有相同的实体引用值，则在视图中显示内容

Question

标题几乎解释了它。我有一个名为“公司”的用户实体引用字段（节点）。我想在一个视图中显示所有内容，其中该内容的作者与登录用户具有相同的“公司”值。我对实现此目标的关系有点困惑。

好的，我能够根据 Kien 的回答来完成这项工作。如果使用实体参考 7-x.1.5 而不是 'field_company_nid'，则正确的列名称是 'field_company_target_id'


<pre><code>/**
 * Implements hook_views_query_alter().
 */
function your_module_views_query_alter(&$view, &$query) {
  if ($view->name == 'article' && $view->current_display == 'page') { // replace 'article' and 'page' by your view's machine name and display's machine name
    $author_join = new views_join();
    $author_join->table = 'field_data_field_company'; // 'field_company' is machine name of the user's field referencing to Company
    $author_join->field = 'entity_id';
    $author_join->left_table = 'node';
    $author_join->left_field = 'uid';
    $author_join->type = 'left';
    $view->query->add_relationship('author_company', $author_join, 'node');

    $current_join = new views_join();
    $current_join->table = 'field_data_field_company'; // 'field_company' is machine name of the user's field referencing to Company
    $current_join->field = 'field_company_target_id';
    $current_join->left_table = 'author_company';
    $current_join->left_field = 'field_company_target_id';
    $current_join->type = 'left';
    $view->query->add_relationship('current_company', $current_join, 'author_company');

    global $user;
    $view->query->where[1]['conditions'][] = array(
      'field' => 'current_company.entity_id',
      'value' => $user->uid,
      'operator' => '='
    );
  }
}

Answer 1

你想要的似乎有点难以通过简单地设置视图来完成。您可能需要一些自定义代码来更改视图查询。下面是一种使用 hook_views_query_alter():

的方法

创建自定义模块（假设是your_module）
在文件 your_module.module 中，执行 hook_views_query_alter():

/**
 * Implements hook_views_query_alter().
 */
function your_module_views_query_alter(&$view, &$query) {
  if ($view->name == 'article' && $view->current_display == 'page') { // replace 'article' and 'page' by your view's machine name and display's machine name
    $author_join = new views_join();
    $author_join->table = 'field_data_field_company'; // 'field_company' is machine name of the user's field referencing to Company
    $author_join->field = 'entity_id';
    $author_join->left_table = 'node';
    $author_join->left_field = 'uid';
    $author_join->type = 'left';
    $view->query->add_relationship('author_company', $author_join, 'node');

    $current_join = new views_join();
    $current_join->table = 'field_data_field_company'; // 'field_company' is machine name of the user's field referencing to Company
    $current_join->field = 'field_company_nid';
    $current_join->left_table = 'author_company';
    $current_join->left_field = 'field_company_nid';
    $current_join->type = 'left';
    $view->query->add_relationship('current_company', $current_join, 'author_company');

    global $user;
    $view->query->where[1]['conditions'][] = array(
      'field' => 'current_company.entity_id',
      'value' => $user->uid,
      'operator' => '='
    );
  }
}

启用您刚刚创建的模块。

如果内容作者与登录用户具有相同的实体引用值，则在视图中显示内容

Show Content in a View if content Author has the same entity reference value as the logged in user

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drupal-7

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