Get n grouped categories and sum others into one

问题

I have a table with the following structure:

Contents (
  id
  name
  desc
  tdate
  categoryid
  ...
)

I need to do some statistics with the data in this table. For example I want to get number of rows with the same category by grouping and id of that category. Also I want to limit them for n rows in descending order and if there are more categories available I want to mark them as "Others". So far I have come out with 2 queries to database:

Select n rows in descending order:

SELECT COALESCE(ca.NAME, 'Unknown') AS label
    ,ca.id AS catid
    ,COUNT(c.id) AS data
FROM contents c
LEFT OUTER JOIN category ca ON ca.id = c.categoryid
GROUP BY label
    ,catid
ORDER BY data DESC LIMIT 7

Select other rows as one:

SELECT 'Others' AS label
    ,COUNT(c.id) AS data
FROM contents c
LEFT OUTER JOIN category ca ON ca.id = c.categoryid
WHERE c.categoryid NOT IN ($INCONDITION)

But when I have no category groups left in db table I still get an "Others" record. Is it possible to make it in one query and make the "Others" record optional?

回答1:

The specific difficulty here: Queries with one or more aggregate functions in the SELECT list and no GROUP BY clause produce exactly one row, even if no row is found in the underlying table.

There is nothing you can do in the WHERE clause to suppress that row. You have to exclude such a row after the fact, i.e. in the HAVING clause, or in an outer query.

Per documentation:

If a query contains aggregate function calls, but no GROUP BY clause, grouping still occurs: the result is a single group row (or perhaps no rows at all, if the single row is then eliminated by HAVING). The same is true if it contains a HAVING clause, even without any aggregate function calls or GROUP BY clause.

It should be noted that adding a GROUP BY clause with only a constant expression (which is otherwise completely pointless!) works, too. See example below. But I'd rather not use that trick, even if it's short, cheap and simple, because it's hardly obvious what it does.

The following query only needs a single table scan and returns the top 7 categories ordered by count. If (and only if) there are more categories, the rest is summarized into 'Others':

WITH cte AS (
   SELECT categoryid, count(*) AS data
        , row_number() OVER (ORDER BY count(*) DESC, categoryid) AS rn
   FROM   contents
   GROUP  BY 1
   )
(  -- parentheses required again
SELECT categoryid, COALESCE(ca.name, 'Unknown') AS label, data
FROM   cte
LEFT   JOIN category ca ON ca.id = cte.categoryid
WHERE  rn <= 7
ORDER  BY rn
)
UNION ALL
SELECT NULL, 'Others', sum(data)
FROM   cte
WHERE  rn > 7         -- only take the rest
HAVING count(*) > 0;  -- only if there actually is a rest
-- or: HAVING  sum(data) > 0

You need to break ties if multiple categories can have the same count across the 7th / 8th rank. In my example, categories with the smaller categoryid win such a race.
Parentheses are required to include a LIMIT or ORDER BY clause to an individual leg of a UNION query.
You only need to join to table category for the top 7 categories. And it's generally cheaper to aggregate first and join later in this scenario. So don't join in the the base query in the CTE (common table expression) named cte, only join in the first SELECT of the UNION query, that's cheaper.
Not sure why you need the COALESCE. If you have a foreign key in place from contents.categoryid to category.id and both contents.categoryid and category.name are defined NOT NULL (like they probably should be), then you don't need it.

The odd `GROUP BY true`

This would work, too:

...

UNION ALL
SELECT NULL , 'Others', sum(data)
FROM   cte
WHERE  rn > 7
GROUP BY true;

And I even get slightly faster query plans. But it's a rather odd hack ...

SQL Fiddle demonstrating all.

Related answer with more explanation for the UNION ALL / LIMIT technique:

Sum results of a few queries and then find top 5 in SQL

回答2:

The quick fix, to make the 'Others' row conditional would be to add a simple HAVING clause to that query.

HAVING COUNT(c.id) > 0

(If there are no other rows in the contents table, then COUNT(c.id) is going to be zero.)

That only answers half the question, how to make the return of that row conditional.

The second half of the question is a little more involved.

To get the whole resultset in one query, you could do something like this

(this is not tested yet; desk checked only.. I'm not sure if postgresql accepts a LIMIT clause in an inline view... if it doesn't we'd need to implement a different mechanism to limit the number of rows returned.

  SELECT IFNULL(t.name,'Others') AS name
       , t.catid                 AS catid
       , COUNT(o.id)             AS data 
    FROM contents o
    LEFT 
    JOIN category oa
      ON oa.id = o.category_id
    LEFT
    JOIN ( SELECT COALESCE(ca.name,'Unknown') AS name
                , ca.id                       AS catid
                , COUNT(c.id)                 AS data
             FROM contents c
             LEFT
             JOIN category ca
               ON ca.id = c.categoryid
            GROUP 
               BY COALESCE(ca.name,'Unknown')
                , ca.id
            ORDER
               BY COUNT(c.id) DESC
                , ca.id DESC
            LIMIT 7
         ) t
      ON ( t.catid = oa.id OR (t.catid IS NULL AND oa.id IS NULL)) 
   GROUP
      BY ( t.catid = oa.id OR (t.catid IS NULL AND oa.id IS NULL)) 
       , t.catid
   ORDER
      BY COUNT(o.id) DESC
       , ( t.catid = oa.id OR (t.catid IS NULL AND oa.id IS NULL)) DESC
       , t.catid DESC
   LIMIT 7

The inline view t basically gets the same result as the first query, a list of (up to) 7 id values from category table, or 6 id values from category table and a NULL.

The outer query basically does the same thing, joining content with category, but also doing a check if there's a matching row from t. Because t might be returning a NULL, we have a slightly more complicated comparison, where we want a NULL value to match a NULL value. (MySQL conveniently gives us shorthand operator for this, the null-safe comparison operator <=>, but I don't think that's available in postgresql, so we have to express differently.

     a = b OR (a IS NULL AND b IS NULL)

The next bit is getting a GROUP BY to work, we want to group by the 7 values returned by the inline view t, or, if there's not matching value from t, group the "other" rows together. We can get that to happen by using a boolean expression in the GROUP BY clause.

We're basically saying "group by 'if there was a matching row from t'" (true or false) and then group by the row from 't'. Get a count, and then order by the count descending.

This isn't tested, only desk checked.

回答3:

You can approach this with nested aggregation. The inner aggregation calculates the counts along with a sequential number. You want to take everything whose number is 7 or less and then combine everything else into the others category:

SELECT (case when seqnum <= 7 then label else 'others' end) as label,
       (case when seqnum <= 7 then catid end) as catid, sum(cnt)
FROM (SELECT ca.name AS label, ca.id AS catid, COUNT(c.id) AS cnt,
             row_number() over (partition by ca.name, catid order by count(c.id) desc) as seqnum
      FROM contents c LEFT OUTER JOIN
           category ca
           ON ca.id = c.categoryid
      GROUP BY label, catid
     ) t
GROUP BY (case when seqnum <= 7 then label else 'others' end),
         (case when seqnum <= 7 then catid end) 
ORDER BY cnt DESC ;

来源：https://stackoverflow.com/questions/29560381/get-n-grouped-categories-and-sum-others-into-one

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sql