PHP Snippet: If node is specific node-type and current logged in user is node author

Question

I would like to combine these two php codes to display a block only on nodes of a special node-type and when the current logged in user is the author of the node. Any help very much appreciated!

$types = array('middle_content' => 1);
if ($node = menu_get_object()) {
  if (isset($types[$node->type])) {
    return TRUE;
  }
}

as found on Block on specific pages and node types

global $user;
if (arg(0) == 'node' && is_numeric(arg(1))) {
  // Get the nid
  $nid = arg(1);

  // Load the node if you need to
  $node = node_load($nid);
}

if($node->uid == $user->uid){
return TRUE;
} else {
return FALSE;
}

as found on How can I check if the author of the current node is the current logged in user?

Answer 1

Assuming Drupal 7, if you're using PHP Filter, go to the block's settings -> Visibility settings -> Pages -> Show block on specific pages. Select "Pages on which this PHP code returns TRUE (experts only)." Use the following code replacing my_content_type with the machine name of your content type.

<?php 
  $node = menu_get_object();
  global $user;
  if (isset($node) && $node->type == 'my_content_type' && $node->uid == $user->uid) {
    return TRUE;
  }
?>

Alternatively, you can select one or more content types under Visibility settings -> Content types. Then you'd only need to check for the node's author, thusly:

<?php 
  $node = menu_get_object();
  global $user;
  if (isset($node) && $node->uid == $user->uid) {
    return TRUE;
  }
?>

Note the security issues presented by the PHP filter module.