how to display view block in node page

Question

I have a block in view that is displaying the user details, i want to display that block in a nodes created by the respective users.

i am successfully using this block in user page now i want it on node page.

please help me.

For another block which is also coming from view is worked fine with this code Show block on specific pages:

<?php

//Read URL
$path=$_GET['q'];

//If URL is node page

if ( strpos($path,'node')===0){
//Parse URL to get nid
$links=explode("/",$_GET['q']);
$nid=$links[1];

//Load node

$node=node_load($nid);

//Display block only if node is of certain content type
if($node->type=='user_art'){
return TRUE;
}
}

return FALSE;
?>

Thanks

Answer 1

It is better to check the arguments from arg() function, this way:

<?php
// PHP VISIBILITY CODE:
// If URL is node page

if (arg(0) == 'node' && is_numeric(arg(1))){
  $nid = arg(1);

  // Load node:
  $node=node_load($nid);

  // Display block only if node is of certain content type
  if ($node->type=='user_art') {
    return TRUE;
  }
}

return FALSE;
?>

Maybe the problem is with the views argument to get $node->uid. Try this:

More info: https://groups.drupal.org/node/129979#comment-751473

1 - Create a new Block View to view Users (not Content)

2 - Filter Criteria should say User: Active (Yes)

3 - There is a default sort Criteria item. Remove it since it doesn't matter.

4 - Add a Relationship: Choose "User: Content authored"

5 - Add a Contextual Filter: Chose "Content: Nid"

6 - In the configuration options for the newly added Content: Nid, choose these options:
WHEN THE FILTER VALUE IS NOT AVAILABLE:
Provide default value
In the Select box, choose Content ID From URL.

WHEN THE FILTER VALUE IS AVAILABLE OR A DEFAULT IS PROVIDED
Specify validation criteria
Validator select box: Content
Content Types: choose the content types you want the view to work with.
Filter Value Format: Node ID
Action to take if the filter value does not validate: Hide View

7 - Save the View.

8 - Go to Structure -> Blocks and configure the block to show on the types of content you want it to show on and position it wherever you want it to appear.

9 - Done.

Answer 2

I think what you are trying to accomplish is something like this:

<?php

  global $user;

  $node=menu_get_object();

  if (isset($node->uid) && $node->uid==$user->uid) {
    return TRUE;
  } 

  return FALSE;

?>

The above uses the menu_get_object() function to look at what, if any, node page you are currently on, and if the user who created that node is the same as the current user, it returns TRUE. Otherwise, if you aren't on such a node page, it returns FALSE.

I might add this may be a better way to accomplish your logic for displaying a block on only a certain type of node, eg something like this:

<?php

  $node=menu_get_object();

  if (isset($node->type) && $node->type=='WHATEVER') {
    return TRUE;
  } 

  return FALSE;

?>