Question

运行PostgreSQL 7.x（是的，我正在升级）

示例：

SELECT
    CASE 
        WHEN "substring"(t."Field"::text, 0, 3) = '01'::text THEN 'group one'::text
        WHEN "substring"(t."Field"::text, 0, 4) = '123'::text THEN 'group one'::text
        WHEN "substring"(t."Field"::text, 0, 5) = '4567'::text THEN 'group two'::text
        WHEN "substring"(t."Field"::text, 0, 6) = '99999'::text THEN 'group three'::text
        WHEN "substring"(t."Field"::text, 0, 3) = '07'::text THEN 'group three'::text
        ELSE NULL::text
    END AS new_field,
    CASE 
        WHEN "substring"(t."Field"::text, 0, 3) = '01'::text THEN 'subgroup a'::text
        WHEN "substring"(t."Field"::text, 0, 4) = '123'::text THEN 'subgroup a'::text
        WHEN "substring"(t."Field"::text, 0, 5) = '4567'::text THEN 'subgroup a'::text
        WHEN "substring"(t."Field"::text, 0, 6) = '99999'::text THEN 'subgroup a'::text
        WHEN "substring"(t."Field"::text, 0, 3) = '07'::text THEN 'subgroup b'::text
        ELSE NULL::text
    END AS another_new_field,...

有没有办法让一个case语句给出两个字段作为结果，因为处理的数据是相同的，它只是不同的标签。

Answer 1

我不熟悉Postgre，但也许您可以尝试使用公用表表达式来构建索引查找条件的表以及组和子组的输出结果，这可能看起来像这样（当然条件可能会来来自实际的表......）：

with Lookup as (
    select 0 as start, 3 as end,
        '01' as match, 'group one' as sgroup, 'subgroup a' as subgroup union
    select 0 as start, 4 as end,
        '123' as match, 'group one' as sgroup, 'subgroup a' as subgroup union
    select 0 as start, 5 as end,
        '4567' as match, 'group two' as sgroup, 'subgroup a' as subgroup union
    select 0 as start, 6 as end,
        '99999' as match, 'group three' as sgroup, 'subgroup a' as subgroup union
    select 0 as start, 3 as end,
        '07' as match, 'group three' as sgroup, 'subgroup b' as subgroup union
)
select
    L.sgroup as new_field,
    L.subgroup as another_new_field
from table T
left join Lookup L on "substring"(T."Field"::text, L.start, L.End) = L.match

然后您只需要定义一次条件，并匹配它们一次以输出两个字段。正如@Tom在评论中所表达的那样，这种情况在SQL中没有得到很好的处理，因此你会得到一个像上面这样的“hacky”解决方案。

最诚挚的问候，

CASE语句结果为多个字段

1 个答案: