PostgreSQL

Question

我有两个表：Subject 和 Journal 如下：

Subject
 id | name
----------
  1 | fruit
  2 | drink
  3 | vege
  4 | fish

和

Journal
 id | subj | reference | value
------------------------------
  1 |    1 |       foo |    30
  2 |    2 |       bar |    20
  3 |    1 |       bar |    35
  4 |    1 |       bar |    10
  5 |    2 |       baz |    25
  6 |    4 |       foo |    30
  7 |    4 |       bar |    40
  8 |    1 |       baz |    20
  9 |    2 |       bar |     5

我想用 subj 和 reference 对 Journal.value 组求和。

我知道 group by 子句是为了这个目的，但我希望输出如下：

reference | subj_1 | subj_2 | subj_3 | subj_4
          |  fruit |  drink |   vege |   fish (even better)
---------------------------------------------
      foo |     30 |        |        |     30
      bar |     45 |     25 |        |     40
      baz |     20 |     25 |        |

这可能吗？

Answer 1

这会产生您想要的结果：

SELECT *
FROM   crosstab(
   'SELECT reference, subj, sum(value)
    FROM   journal
    GROUP  BY 1, 2
    ORDER  BY 1, 2'

  , $$VALUES (1), (2), (3), (4)$$
   ) AS ct (reference text, fruit int, drink int, vege int, fish int);

db<>fiddle here

除了排序顺序，这看起来很随意？

详细解释和说明：

PostgreSQL Crosstab Query

Answer 2

您可以根据当前数据生成Sql报表。

然后使用生成的 Sql 语句

示例数据：

create table Subject (
 id serial primary key, 
 name varchar(30) not null
 );

insert into Subject (id, name) values
 (1 ,'fruit')
,(2 ,'drink')
,(3 ,'vege')
,(4 ,'fish');
 
create table Journal (
 id int, 
 subj int, 
 reference varchar(30), 
 value int
);

insert into Journal   
(id, subj, reference, value) values
 (1, 1, 'foo', 30)
,(2, 2, 'bar', 20)
,(3, 1, 'bar', 35)
,(4, 1, 'bar' ,10)
,(5, 2, 'baz', 25)
,(6, 4, 'foo', 30)
,(7, 4, 'bar', 40)
,(8, 1, 'baz', 20)
,(9, 2, 'bar', 5);

生成语句：

SELECT $f$SELECT * FROM crosstab(
     $$SELECT DISTINCT ON (1, 2)
       j.reference, 'subj_'||j.subj||'_'||s.name AS data_type, SUM(j.value) AS val
       FROM Journal j
       JOIN Subject s ON s.id = j.subj
       GROUP BY j.reference, j.subj, s.name
       ORDER BY j.reference$$

    ,$$VALUES ($f$     || string_agg(quote_literal(data_type), '), (') || $f$)$$)
AS x (reference text, $f$ || string_agg(quote_ident(data_type), ' int, ') || ' int)'
AS Stmt
FROM  (SELECT concat('subj_', id, '_', name) AS data_type FROM Subject) x

| stmt                                                                                                                                                                                                                                                                                                                                                                                                                                                                                             |
| :----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |
| SELECT * FROM crosstab(<br>     $$SELECT DISTINCT ON (1, 2)<br>       j.reference, 'subj_'||j.subj||'_'||s.name AS data_type, SUM(j.value) AS val<br>       FROM Journal j<br>       JOIN Subject s ON s.id = j.subj<br>       GROUP BY j.reference, j.subj, s.name<br>       ORDER BY j.reference$$<br><br>    ,$$VALUES ('subj_1_fruit'), ('subj_2_drink'), ('subj_3_vege'), ('subj_4_fish')$$)<br>AS x (reference text, subj_1_fruit int, subj_2_drink int, subj_3_vege int, subj_4_fish int) |

运行这

SELECT * FROM crosstab(
     $$SELECT DISTINCT ON (1, 2)
       j.reference, 'subj_'||j.subj||'_'||s.name AS data_type, SUM(j.value) AS val
       FROM Journal j
       JOIN Subject s ON s.id = j.subj
       GROUP BY j.reference, j.subj, s.name
       ORDER BY j.reference$$

    ,$$VALUES ('subj_1_fruit'), ('subj_2_drink'), ('subj_3_vege'), ('subj_4_fish')$$)
AS x (reference text, subj_1_fruit int, subj_2_drink int, subj_3_vege int, subj_4_fish int)

reference | subj_1_fruit | subj_2_drink | subj_3_vege | subj_4_fish
:-------- | -----------: | -----------: | ----------: | ----------:
bar       |           45 |           25 |        null |          40
baz       |           20 |           25 |        null |        null
foo       |           30 |         null |        null |          30

db<>fiddle here

PostgreSQL - 按两列分组并使用一列作为结果列

PostgreSQL - Group By Two Columns And Use One As Column For Result

sql

pivot