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I have a view displaying a list a menu of all the content for a node. It contains a list of all the fields and views on the page, with the output rewritten as a link eg link to field A . I can do this fine for each field but I also have a view 'teachers' exposed as a block. If the view has content I want to have a link to it in the menu. I'm trying to use Views PHP to code if the view has content, echo '' What function do I call to see if the view exists and has content? I've tried if(views_get_view('teachers) == TRUE){ echo "";} but this always returns TRUE even if there is no content for that view for a specific node.

I want to do something similar to

2

You can look at views_get_view_result() and do something like this:

$result=views_get_view_result('teachers');
if (isset($result[0])) {
  // we have results
} else {
  // we don't
}
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It can be done simpler without querying for a result

$view = views_get_view('my_view_name');
if($view) {
  //continue logic..
}

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