Question

我有一个带有id的实体表，以及一个类别（几个不同的值允许使用NULL）来自3个不同年份（类别可以从1年到另一年不同），采用“宽”表格格式：

| ID  | CATEG_Y1 | CATEG_Y2 | CATEG_Y3 |
+-----+----------+----------+----------+
| 1   | NULL     | B        | C        |
| 2   | A        | A        | C        |
| 3   | B        | A        | NULL     |
| 4   | A        | C        | B        |
| ... | ...      | ...      | ...      |

我想简单地按类别计算实体数量，按类别分组，单独计算年份：

+-------+----+----+----+
| CATEG | Y1 | Y2 | Y3 |
+-------+----+----+----+
| A     | 6  | 4  |  5 | <- 6 entities w/ categ_y1, 4 w/ categ_y2, 5 w/ categ_y3  
| B     | 3  | 1  | 10 |
| C     | 8  | 4  |  5 |
| NULL  | 3  | 3  |  3 |
+-------+----+----+----+

我想我可以通过将值分组到另一列并UNION ALL结果来实现，但我想知道是否有更快速的＆amp;方便的方式，如果我有更多的列/年可以管理（例如20-30个不同的值），它是否可以推广

Answer 1

有点笨拙，但可能有人有更好的主意。查询首先收集所有不同的类别（from部分中的union-query），然后使用select部分中的专用子查询计算出现的次数。如果已经有一个表已经定义了可用的类别，那么可以省略union-part（我假设categ_y1是这种主类别表的外键）。希望没有很多拼写错误：

select categories.cat,
       (select count(categ_y1) from table ty1 where select categories.cat = categ_y1) as y1,
       (select count(categ_y2) from table ty2 where select categories.cat = categ_y2) as y2,
       (select count(categ_y3) from table ty3 where select categories.cat = categ_y3) as y3
from ( select categ_y1 as cat from table t1
 union select categ_y2 as cat from table t2
 union select categ_y3 as cat from table t3) categories

Answer 2

使用jsonb函数将数据（来自问题）转换为以下格式：

select categ, jsonb_object_agg(key, count) as jdata
from (
    select value as categ, key, count(*)
    from my_table t,
    jsonb_each_text(to_jsonb(t)- 'id')
    group by 1, 2
    ) s
group by 1
order by 1;

 categ |                     jdata                     
-------+-----------------------------------------------
 A     | {"categ_y1": 2, "categ_y2": 2}
 B     | {"categ_y1": 1, "categ_y2": 1, "categ_y3": 1}
 C     | {"categ_y2": 1, "categ_y3": 2}
       | {"categ_y1": 1, "categ_y3": 1}
(4 rows)

对于已知（静态）年数，您可以轻松解压缩jsonb列：

select categ, jdata->'categ_y1' as y1, jdata->'categ_y2' as y2, jdata->'categ_y3' as y3 
from (
    select categ, jsonb_object_agg(key, count) as jdata
    from (
        select value as categ, key, count(*)
        from my_table t,
        jsonb_each_text(to_jsonb(t)- 'id')
        group by 1, 2
        ) s
    group by 1
    ) s
order by 1;

 categ | y1 | y2 | y3 
-------+----+----+----
 A     | 2  | 2  | 
 B     | 1  | 1  | 1
 C     |    | 1  | 2
       | 1  |    | 1
(4 rows)

要获得完全动态的解决方案，您可以使用Flatten aggregated key/value pairs from a JSONB field中描述的函数create_jsonb_flat_view()。

Answer 3

我会这样做，因为使用union all后跟聚合：

select categ, sum(categ_y1) as y1, sum(categ_y2) as y2,
       sum(categ_y3) as y3
from ((select categ_y1, 1 as categ_y1, 0 as categ_y2, 0 as categ_y3
       from t
      ) union all
      (select categ_y2, 0 as categ_y1, 1 as categ_y2, 0 as categ_y3
       from t
      ) union all
      (select categ_y3, 0 as categ_y1, 0 as categ_y2, 1 as categ_y3
       from t
      )
     )
group by categ ;

按多列

3 个答案: