从路径枚举列创建一个透视/扁平层次结构表答案

【问题标题】：Create a pivoted/flattened hierarchies table from a path enumeration column从路径枚举列创建一个透视/扁平层次结构表
【发布时间】：2022-11-16 23:46:21
【问题描述】：

我有一个带有路径枚举列（沿袭列）的层次结构表

我的表：

path	desc	lvl
D	Chicago	1
DADB	Los Angeles	2
DADB761	Miami	3
DADB761G93	Detroit	4
DADB761G93276	San Francisco	5
DADB761G93277	Seattle	5

期望的结果：

desc1	desc2	desc3	desc4	desc5	code1	code2	code3	code4	code5
Chicago	Los Angeles	Miami	Detroit	San Francisco	D	ADB	761	G93	276
Chicago	Los Angeles	Miami	Detroit	Seattle	D	ADB	761	G93	277

假设递归 CTE 是最佳路径，我如何生成所需的结果表？

重现表的 SQL 语句：

create table dbo.#mytable (
    [path] VARCHAR(13),
    [desc] VARCHAR(70),
    [lvl] SMALLINT
)

insert into #mytable([path],[desc],[lvl]) 
values('D','Chicago',1),('DADB','Los Angeles',2),('DADB761','Miami',3),('DADB761G93','Detroit',4),
('DADB761G93276','San Francisco',5),('DADB761G93277','Seattle',5)

【问题讨论】：

标签： sql sql-server common-table-expression hierarchy recursive-query

【解决方案1】：

通常，您会使用递归 CTE 中多行的数据构建 [path] 列。在这种情况下，您正在“展平”多行表格。如果您有超过 5 个级别，这不会自动扩展。您必须为每个级别添加一列。话虽如此，您可以像这样做您要问的事情：

SQL Fiddle

MS SQL Server 2017 架构设置:

CREATE TABLE mytable (
    [path] VARCHAR(13) null,
    [desc] VARCHAR(70) null,
    [lvl] SMALLINT null
)

INSERT INTO mytable ([path],[desc],[lvl]) 
VALUES
  ('D','Chicago',1)
  ,('DADB','Los Angeles',2)
  ,('DADB761','Miami',3)
  ,('DADB761G93','Detroit',4)
  ,('DADB761G93276','San Francisco',5)
  ,('DADB761G93277','Seattle',5)

查询 1:

WITH destruct AS (

  SELECT DISTINCT 
    CAST([path] as nvarchar(512)) as code1
    , CAST('' as nvarchar(512)) as code2
    , CAST('' as nvarchar(512)) as code3
    , CAST('' as nvarchar(512)) as code4
    , CAST('' as nvarchar(512)) as code5
    , CAST([desc] as nvarchar(512)) as desc1
    , CAST('' as nvarchar(512)) as desc2
    , CAST('' as nvarchar(512)) as desc3
    , CAST('' as nvarchar(512)) as desc4
    , CAST('' as nvarchar(512)) as desc5
    , mt.[path]
    , mt.lvl
  FROM mytable as mt
  WHERE lvl = 1

  UNION ALL

  SELECT
    CAST(d.code1 as nvarchar(512)) as code1
    , CAST(CASE WHEN d.lvl + 1 = 2 THEN RIGHT(mt.[path], LEN(mt.[path])-LEN(d.[path])) ELSE d.code2 END as nvarchar(512)) as code2
    , CAST(CASE WHEN d.lvl + 1 = 3 THEN RIGHT(mt.[path], LEN(mt.[path])-LEN(d.[path])) ELSE d.code3 END as nvarchar(512)) as code3
    , CAST(CASE WHEN d.lvl + 1 = 4 THEN RIGHT(mt.[path], LEN(mt.[path])-LEN(d.[path])) ELSE d.code4 END as nvarchar(512)) as code4
    , CAST(CASE WHEN d.lvl + 1 = 5 THEN RIGHT(mt.[path], LEN(mt.[path])-LEN(d.[path])) ELSE d.code5 END as nvarchar(512)) as code5
    , CAST(d.desc1 as nvarchar(512)) as desc1
    , CAST(CASE WHEN d.lvl + 1 = 2 THEN mt.[desc] ELSE d.desc2 END as nvarchar(512)) as desc2
    , CAST(CASE WHEN d.lvl + 1 = 3 THEN mt.[desc] ELSE d.desc3 END as nvarchar(512)) as desc3
    , CAST(CASE WHEN d.lvl + 1 = 4 THEN mt.[desc] ELSE d.desc4 END as nvarchar(512)) as desc4
    , CAST(CASE WHEN d.lvl + 1 = 5 THEN mt.[desc] ELSE d.desc5 END as nvarchar(512)) as desc5
    , mt.[path]
    , mt.lvl
  FROM destruct as d
    INNER JOIN mytable as mt
      ON mt.lvl = d.lvl + 1
      AND mt.[path] like d.[path] + '%'
  AND d.lvl < 20  --Safety for debugging. Remove in Production.
)
SELECT *
FROM destruct as d
WHERE d.lvl = 5

Results:

| code1 | code2 | code3 | code4 | code5 |   desc1 |       desc2 | desc3 |   desc4 |         desc5 |          path | lvl |
|-------|-------|-------|-------|-------|---------|-------------|-------|---------|---------------|---------------|-----|
|     D |   ADB |   761 |   G93 |   276 | Chicago | Los Angeles | Miami | Detroit | San Francisco | DADB761G93276 |   5 |
|     D |   ADB |   761 |   G93 |   277 | Chicago | Los Angeles | Miami | Detroit |       Seattle | DADB761G93277 |   5 |

注意：所有CAST 表达式都是为了确保根查询和子查询具有相同的数据类型。如果没有它们，SQL Fiddle 会抛出有关类型不匹配的错误。

【讨论】：