Postgresql

Question

我有以下 table:

我想旋转它并得到以下结果：

  Table_name        Column_name      Values         ddmf_etl_id
"PROFITCENTERID"    "CO_AREA"       "DE02"           16807487
"PROFITCENTERID"    "PROFIT_CTR"    "0000001119"     16807487
"PROFITCENTERID"    "CO_AREA"       "DE02"           16807488
"PROFITCENTERID"    "PROFIT_CTR"    "0000001120"     16807488

不幸的是，我未能使用以下 select 执行此操作，因为 CROSSTAB 的参数数量有限 (3)：

SELECT  'PROFITCENTERID' AS table_name,t.*
FROM   

crosstab(

        'SELECT unnest(''{CO_AREA,PROFIT_CTR}''::text[]) AS col
             , row_number() OVER ()
             , unnest(ARRAY[CO_AREA::text,PROFIT_CTR::text]) AS val
        FROM   "1".PROFITCENTER_PROFITCENTERID'

   ) t (column_name text,  values text);

select 语句的结果（错误）：

在 Postgresql 中是否有其他替代交叉表的方法？或者也许可以用 Crosstab 做到这一点？

感谢任何帮助！谢谢！

Answer 1

使用unnest()将每行加倍并select适当的值：

select
    'profitcenterid' as table_name,
    unnest(array['co_area', 'profit_ctr']) as column_name,
    unnest(array[co_area, profit_ctr]) as value,
    ddmf_etl_id as ddmf_etl_id
from profitcenter_profitcenterid

   table_name   | column_name |   value    | ddmf_etl_id 
----------------+-------------+------------+-------------
 profitcenterid | co_area     | de02       |    16807487
 profitcenterid | profit_ctr  | 0000001119 |    16807487
 profitcenterid | co_area     | de02       |    16807488
 profitcenterid | profit_ctr  | 0000001120 |    16807488
(4 rows)

Live demo in rextester.

但是，这对获得预期的 json 结果帮助不大。您想获取任意键名，因此无法根据子查询结果自动生成对象。它们可以用原语 jsonb_build_object():

创建

select to_jsonb(s)
from (
    select
        'profitcenterid' as table_name,
        17 as type,
        '' as value,
        jsonb_build_array(
            jsonb_build_object('name', 'co_area', 'value', co_area, 'type', 0, 'children', '[]'::jsonb),
            jsonb_build_object('name', 'profit_ctr', 'value', profit_ctr, 'type', 0, 'children', '[]'::jsonb)
        ) as children
    from profitcenter_profitcenterid
    ) s

Example in rextester.

附带说明一下，预期的结构似乎不必要地复杂，简单是一种美德。

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