Drupal is up and running but how do I ...?

"recent posts" bug cornered .... server issue?

Drupal 4.6.6 - fresh clean load (only have 5 test pages)

I been chasing this bug for awhile now -- I'm sure what is causing it -

------ but I know what is not causing it and I'm hoping someone can guess what I need to do to turn it on.

Fresh clean load
Disabled Comments/Enabled Comments/Disabled Comments
Didn't load PathAuto.module
Have not deleted any pages

Turned off User Paths
Turned off Clean URLS
(wasn't getting a ModRewrite error when they were enabled and they would display properly - but disabled just the same for now)

make users get their blogs

i have drupal already running and perfectly installed.

now, i want a feature that gets all my users to get their own blogs. where do i configure that?

thanks.

Alternative to drupal_goto()

Is there another means of redirecting urls other than using the drupal_goto() function

I'm redirecting a 401 error to a static page containing the function but now it seems to be heading straight to the page with outout doing basic authentication on the file

removing the 401 brings back the login in screen

External Link to Open Inside Drupal

Hello - I have searched everywhere but have not found how to open an external link (web page) inside your drupal pages (in the main body) ... any hints? Thank You in advance.

Make aggregator category view display only 1 category when running in a page

Think I posted in wrong forum before -

Hi been trying to make the aggregator category view run on a page but only displaying results for 1 of the categories have got the code -

$result = db_query('SELECT c.cid, c.title, c.description FROM {aggregator_category} c LEFT JOIN {aggregator_category_item} ci ON c.cid = ci.cid LEFT JOIN {aggregator_item} i ON ci.iid = i.iid GROUP BY c.cid, c.title, c.description');
  $output = "<div id=\"aggregator\">\n";

  while ($category = db_fetch_object($result)) {
    $output .= '<h2>'. check_plain($category->title) ."</h2>\n";
    if (variable_get('aggregator_summary_items', 3)) {
      $list = array();
      $items = db_query_range('SELECT i.title, i.timestamp, i.link, f.title as feed_title, f.link as feed_link FROM {aggregator_category_item} ci LEFT JOIN {aggregator_item} i ON i.iid = ci.iid LEFT JOIN {aggregator_feed} f ON i.fid = f.fid WHERE ci.cid = %d ORDER BY i.timestamp DESC', $category->cid, 0, variable_get('aggregator_summary_items', 3));
      while ($item = db_fetch_object($items)) {
        $list[] = theme('aggregator_summary_item', $item);
      }
      $output .= theme('item_list', $list);
    }
    $output .= '<div class="links">'. theme('links', array(l(t('more'), 'aggregator/categories/'. $category->cid))) ."</div>\n";
  }
  $output .= '</div>';

  return $output;

which displays the default all categories view, how is it possible to make this display for only 1 of the categories ie where cid = 1 ?

Forms question

Hello all,
This may be a really easy question but I can't seem to find an answer to it. If I want to have a custom form in my site, I know I will have to use the Forms API, but how do I get the form to display?

Will I need to create a custom module?

Is there a tutorial anywhere that kind of walks you through creating a new site, then customizing it?

Thanks.

Chris

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