By sej123 on

Hi

I have content type 1 with vocabulary 1 and content type 2 with vocabulary 2.
Vocabulary 1 and 2 have same term names.
Now, I want to create a view that is displayed on content type 2 that picks up the term name from current node (of content type 2) and searches for the same term name in vocabulary 1(of content type 1) and then outputs the desired fields, say imagefield.
Something like,

Select imagefield from content type 1 WHERE vocabulary 2 -> term (of current node) = vocabulary 1 -> term .
[Sorry for the wrong syntax.]

Have I convinced the point?

Please help.

Thanks in advance.

⋅ Categories: Drupal 6.x

Comments

sej123’s picture

sej123 commented

I have found something -
I added argument handling code for argument type- Taxonomy term as - 

$node = node_load(arg(1));
if ($node){
foreach($node->taxonomy as $term){$terms[]=$term->tid;}
return implode('+',$terms);
}else{return;}

I am using url-aliases for my content types. Is it because of this that the arguments are not working?

I am really lost!I
I want to create a view that takes the taxonomy term of current node and searches for other nodes that share same taxonomy term. The other nodes are of different content type with different vocabulary but same term name.

I would be highly obliged for any help.

El Viejo’s picture

El Viejo commented

Did you get this resolved? I'm working on the very same issue.

sej123’s picture

sej123 commented

Can anyone please guide me??

yrre7’s picture

yrre7 commented

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Sinan Erdem’s picture

Sinan Erdem commented

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