CODEWITHSUNDEEP
mysqlsql
Ted
Ted

Reputation: 4166

MySQL: count with group by and max limit

I need to count taggers efforts, by count how many tags added by each tagger, with maximum tag for each article = 5 tags.

ex:

  1. If tagger added 4 tags to one Article count 4.
  2. If tagger added 5 tags to one Article count 5.
  3. If tagger added 6 tags to one Article count 5.
  4. If tagger added 10 tags to one Article count 5.

table structure article_tags

id    user_id    article_id    tag_id
1        1            1            1
2        1            1            2
3        1            2            1
4        2            3            1
5        2            3            2

Is there advanced use of Group by with limit, or something like this? thx,

update:

I need to return how many tags added to articles, with max 5 for a single article, (so If someone added 30 tags for 30 articles = 30

Upvotes: 2

Views: 134

Answers (4)

Gordon Linoff
Gordon Linoff

Reputation: 1269803

Is this what you want?

select user_id, least(count(*), 5)
from article_tags
group by user_id;

EDIT:

I think you want two levels of aggregation:

select article_id, sum(cnt) as restricted_tag_count
from (select user_id, article_id, least(count(*), 5) as cnt
      from article_tags
      group by user_id, article_id
     ) ua
group by article_id;

Upvotes: 7

Ted
Ted

Reputation: 4166

Solution based on 'Gordon Linoff'

SQL query:

select t.user_id,sum(t.total_tags) total
from
(
    select user_id, least(count(*), 5) total_tags
    from article_tags
    group by user_id,article_id
) t
group by t.user_id

thx,

Upvotes: 0

Ankit Bajpai
Ankit Bajpai

Reputation: 13509

You can try something like this:-

SELECT user_id, CASE WHEN COUNT(tag_id) <= 5 THEN COUNT(tag_id)
                ELSE 5 END AS TOTAL_TAGS
FROM YOUR_TABLE
GROUP BY user_id

Hope this helps.

Upvotes: 0

Mahesh Madushanka
Mahesh Madushanka

Reputation: 2998

  SELECT USER_ID,ARTICAL_ID,if(count(*)>5,5,count(*)) as tags
 from article_tags group by USER_ID,ARTICAL_ID;

Upvotes: 0

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