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I'm using a content type for Clients and another content type for Tasks. Each client has many tasks and each task belongs to one client. Both types are connected by a reference field defined for the Task content type that points to a Client node.

I know how to display a Client's information together with a list of all Tasks for that client. I'm using a view in a block that is parameterized with the Client node ID. That was easy.

However, in order to create a new Task for a Client, I need to display the Client's information together with the create new task form.

Is there any way I can do that in Drupal 7?

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If you need to put that info in a block near the form, you can create your custom hook like

function mymodule_menu(){
  $items = array();
  $items['mynode/add/%/%'] = array(
      'title' => t('Create content'),
  'description' => 'Calls to a drupal_get_form',
  'access callback' => 'user_access',
  'page callback' => 'my_function',
  'page arguments' => array(2,3),
  'access arguments' => array('access content'),
  'type' => MENU_CALLBACK,
 );


 return $items;
}

So now you have a page called /mynode/$type/$id_user

In that page you can render the form of that content type using drupal_get_form() and returning it rendered. Now in the blocks config you can add your block view to the pages called /mynode/mytype/* , getting the user id argument from the url will let you render that view with the user data.

  • Thank you very much! I'll give this a try and I'll come back here to report what happened. – David Nov 7 '12 at 13:22
  • If you have any problem, tell me. – rafinskipg Jan 4 '13 at 13:59

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