Better code to show uncommented resources (Quip)

217 Posts

kudykam Reply #1, 13 years, 4 months ago

Hi, I am trying to list resources (like getresources) of uncommented resources created by certain user.
After login user can create new resource and also should post comment. For that reason I need to have page showing list of resources that are not commented by the author of resource. What I manage to create is this: (which in not very optimal as you can see)

[[!getPage?   &elementClass=`modSnippet`  &element=`getResources`  &parents=`-1`  &resources=`[[!GetUserProducts? &depth=`3` &count=`0` &parent=`2`  &commentList=`[[!QuipCount?  &type=`user` &user=`active` &listId=`1`]]`]]`  &limit=`20`  &pageVarKey=`page` &tpl=`kategorie-tpl` ]]

QuipCount is original snippet for Quip modified to list resource ID instead of number of resource, posted by logged user.
GetUserProducts returns list of IDs from parent 2 created by logged user and exclude those in commentList.
Getpage is only for creating template for resources....

GetUserProducts snippet:

<?php
/*
* $parent - where to look
* $depth - how deep
* $commentList - list of comment by logged user
*/


$ids = $modx->getChildIds($parent,$depth,array('context' => 'web'));
$c = $modx->newQuery('modResource');
$c->leftJoin('modUser','Author','modResource.createdby = Author.id');
$user = $modx->user->get('id');
$commentList = explode(",",$commentList); 
$c->where(array(
    'id:IN' => $ids,
    'id:NOT IN' => $commentList,
    'isfolder' => '0',
    'published' => 1,
    'deleted' => 0,
    'Author.id' => $user,

));

if ($count) {
  return $modx->getCount('modResource',$c); //return only count
  }
$docs = $modx->getCollection('modResource',$c);
foreach ($docs as $doc) {
    $output .= $doc->get('id') . ',';   
}
return substr($output, 0, -1) ; //remove last comma

Problem is that I dont know how to combine all snippets together and make it optimized....

This question has been answered by kudykam. See the first response.

5,888 Posts

Bruno17 Reply #2, 13 years, 4 months ago

what about a counter-TV, which updates on each comment-add/remove.
Than its easy to filter/sort by comment-counts

-------------------------------

you can buy me a beer, if you like MIGX

http://webcmsolutions.de/migx.html

Thanks!

217 Posts

kudykam Reply #3, 13 years, 4 months ago

Do you mean each time the author add comment to his resource snippet would save id of this resource to his profile?
So except using quipcount snippet I would just get ids form extended fields....
Do you think this would speed things up?

discuss.answer

217 Posts

kudykam Reply #4, 13 years, 4 months ago

Problem solved this way:
I have extended field for each user with id of each commented page...
So I modified GetUserProducts snippet to just deduct these resources from final output...

<?php
/*
* $parent - where to look
* $depth - how deep
* $count - 1- return number, 0-return list of id
*/


$user = $modx->user->get('id');
if (empty($user)) return "Not logged in!";
$ids = $modx->getChildIds($parent,$depth,array('context' => 'web'));
$c = $modx->newQuery('modResource');
$c->leftJoin('modUser','Author','modResource.createdby = Author.id');

$profile = $modx->user->getOne('Profile');
$fields = $profile->get('extended');
$commentList = explode(",",$fields['Quip']); // get id of all commented resources
$c->where(array(
    'id:IN' => $ids,  
    'id:NOT IN' => $commentList, 
    'isfolder' => '0',
	'published' => 1,
    'deleted' => 0,
    'Author.id' => $user,

));

if ($count) {
  return $modx->getCount('modResource',$c); 

  }
$docs = $modx->getCollection('modResource',$c);


foreach ($docs as $doc) {
    $output .= $doc->get('id') . ',';  

}
return substr($output, 0, -1) ;

Answered Better code to show uncommented resources (Quip)