Postgres table 选择多列并将结果（列）动态转换为行 - 将列转置为行

Question

我有以下类型的table（伤害）结构

  Rowid    damageTypeACount      damageTypeBCount damageTypeCCount
  1        23                     44              33

而且我还需要读取这些 table 行，结果为（标题、id 是手动设置的），将列转换为行但具有附加属性

  id           damagecountsum          label
    1           23                  Damage Type A
    2           44                  Damage Type B
    3           33                  Damage Type C

我做了以下查询并且它有效但想知道是否有更好的方法

     SELECT 1 as id,
             damageTypeACount as damagecount,
            "Damage Type A Count" as label
     FROM Damage where rowId=1
      UNION ALL
     SELECT 2 as id,
            damageTypeBCount as damagecount,
            "Damage Type B Count" as label
     FROM Damage where rowId=1
     UNION ALL
     SELECT 3 as id,
           damageTypeCCount as damagecount,
            "Damage Type C Count" as label
     FROM Damage where rowId=1

上面的查询按预期工作，但我想知道是否可以在单个 select 语句中将列转换为行

Answer 1

您可以使用横向联接取消透视：

select x.*
from damage d
cross join lateral (values
    (d.rowId, d.damageTypeACount, 'Damage Type A'),
    (d.rowId, d.damageTypeBCount, 'Damage Type B'),
    (d.rowId, d.damageTypeCCount, 'Damage Type C')
) as x(id, damagecount, label)

这会重新影响每个生成的行的原始 id。您还可以使用 row_number():

生成新的 ID

select row_number() over(order by id, label) id, x.*
from damage d
cross join lateral (values
    (d.rowId, d.damageTypeACount, 'Damage Type A'),
    (d.rowId, d.damageTypeBCount, 'Damage Type B'),
    (d.rowId, d.damageTypeCCount, 'Damage Type C')
) as x(rowId, damagecount, label)

如果需要，您可以使用 where 子句过滤结果集：

where d.rowId = 1

Postgres table 选择多列并将结果（列）动态转换为行 - 将列转置为行

Postgres table selecting multiple columns and dynamically converting the result (column) into row - transposing column to rows

sql

postgresql

union

unpivot