I am using Couchbase 3.0 and I have a use case where I want to perform sorting, but as i cam through couchbase does not provide Sorting on Values but it provides Sorting on Keys, But as I use descending(true) it is returning me empty list. And on the other hand If I just simply use it without descending then it is giving me all the docs related.
My Map function is :
function (doc, meta) {
emit([meta.id,doc.latest.sortData],null);
}
}
Now my use case is that I want to perform a match query on meta.id and then for all the matched cases sort the data and then find out the top values.
The code that i am using to so is :
ViewQuery.from(DesignDocName, viewName).startKey(JsonArray.from(write(List("something","")))).descending(true).limit(5).stale(Stale.FALSE))
If I remove the descending parameter then I get the related rows but they are not sorted. So could you please provide me a way in which this can be done.
Any Help is appreciated
Thanks!
I was reading Couchbase's docs and stumbled upon this line:
Because selection is made after sorting the view results, if you
configure the results to be sorted in descending order and you are
selecting information using a key range, then you must also reverse
the startkey and endkey parameters.
see here: http://docs.couchbase.com/admin/admin/Views/views-querying.html#ordering
Related
I am trying to sort the result of my mongoDB query and add a ranking based on that sorting. Currently I only call .find().sort({total: 1}) and this gives me the correct ordering of the documents. But is it possible to "add a field" based on that sorting (basically a ranking field, starting from 1 and counting up)? I tried googling but didnt found anything that suits for this purpose.
Thanks in advance.
I need Algolia to always return me 5 results from a full text search even if the query text itself bears little or no relevance to the actual returned results. Before someone suggests it, I have already tried to set the removeWordsIfNoResults option to all of it's possible modes and this still doesn't guarantee that I get my 5 results.
The purpose of this is to create a 'relevant entities' sidebar where the name of the current entity is used to search for other entities.
Any suggestions?
Using the removeWordsIfNoResults=allOptional query parameter is indeed a good way to go -> because all query words are required to match an object by default, fallbacking to "optional" is a good way to still retrieve results if one you the query words (or the combination of words) doesn't match anything.
index.search(query, { removeWordsIfNoResults: 'allOptional' });
Another solution is to always consider all query words as optional (not only as a fallback); to make sure the query foo bar baz is interpreted as OPT(foo) AND OPT(bar) AND OPT(baz) <=> foo OR bar OR baz. The difference is that this query will retrieve more results than the previous one because 1 single matching word will be enough to retrieve the object.
index.search(query, { optionalWords: query });
That being said, there is no way to force the engine to retrieve "at least" 5 results. What I would recommend is to have a small frontend logic:
- do the query with removeWordsIfNoResults or optionalWords
- if the engines returns less than 5 results, do another query
I know that the ATTR function is used for aggregation, but can someone explain it in simple terms?
In the most simplest of terms, ATTR returns a value if it is unique, otherwise it returns "*". I think you'll find this link helpful with examples.
https://www.interworks.com/blog/tcostello/2014/05/15/attr-tableaus-attribute-function-explained
You cannot mix aggregate and non aggregate comparisons in tableau, you have to use ATTR, for e.g. if ATTR(segment) ='Corporate' then sum(sales)
ATTR is like using an already aggregated field for comparison with another aggregated field. Measures are taken as aggregated fields while dimension's aren't. If you have created a field which is already aggregated and still you want to use this field as measure it will be shown as ATTR as it cannot be further aggregated but is behaving like it has.
I need to retrieve a set of answers according to 2 attributes.
This is what i want to do:
# where liker_ids is an array and user_id is a bson in the answer document
feed_answers=Answer.any_in(:liker_ids=>to_use,:user_id.in=>to_use).desc()map{|a|a}
What i ended up doing:
# to use
to_use=[some array of ids]
friend_answers=Answer.any_in(:liker_ids=>to_use).map{|a|a}
liked_answers=Answer.where(:user_id.in=>to_use).map{|a|a}
feed_answers=(friend_answers+ liked_answers).sort{|x,y| y.created_at<=>x.created_at}
The problem is that I do not not know how to combine the 2 queries into one. I have been trying out various combinations, but nothing seems to work. and my hacked together method is highly inefficient of course.
You should do(Missing parameter to desc):
Answer.any_in(:liker_ids=>to_use, :user_id.in=>to_use).desc(:created_at)
But the any_in here is not correctly used, it behaves similar to where in this situation. You probably want or:
Answer.or(:liker_ids=>to_use).or(:user_id.in=>to_use).desc(:created_at)
# or
Answer.any_of({:liker_ids=>to_use}, {:user_id.in=>to_use}).desc(:created_at)
# or
Answer.or({:liker_ids=>to_use}, {:user_id.in=>to_use}).desc(:created_at)
You don't need that map at the end of criteria chain, mongoid criteria are lazy loaded, when they encounter a method which criteria do not respond to. They can also leverage mongodb cursors, so it is advised not to use map if it is not necessary. You should use Criteia#only or Criteria#without if you want to retrieve subset of fields from mongodb.
According to the documentation, Zend Lucene is supposed to sort lexicographically. I am finding this is not the case. If I have a query 'avg:[050 TO 300]', yes it will return all values in that range, but it will sort them according to the document id, not the value.
I have found that the find() function can accept additional parameters, allowing me to sort by a specific column (eg $hits = $index->find($query, 'avg', SORT_NUMERIC, SORT_ASC);). However, I am creating $query dynamically and do not want to sort every search by 'avg'.
How do I force Lucene to sort the results automatically, lexicographically, when I do a range search? And if that's not possible, how do I dynamically add a sort field to the find function?
Why don't you sort $hits by yourself after getting the result from $index->find(...)? Ok this looks like a workaround and will be time-consuming for very large resultsets, but I guess that this is the easiest way in most cases.