如何解决从文本字段中找到关系级别
我有一张表格，其中包含单位的地理结构。有父子关系列，但我想使用现有的文本字段（而不是递归）来查找项目之间的关系级别。
（这是一个表创建脚本）
drop table if exists #temp_structure
create table #temp_structure
(org_id int,parent_org_id int,org_name nvarchar(255),search_tree nvarchar(255))
insert into #temp_structure
values
(1,null,'World','| 1 |'),(2,1,'Europe','| 1 | 2 |'),(3,'North America','| 1 | 3 |'),(4,'South America','| 1 | 4 |'),(5,'Asia','| 1 | 5 |'),(6,'Africa','| 1 | 6 |'),(7,'Australia','| 1 | 7 |'),(8,2,'Spain','| 1 | 2 | 8 |'),(9,'Germany','| 1 | 2 | 9 |'),(10,'Italy','| 1 | 2 | 10 |'),(11,'France','| 1 | 2 | 11 |'),(12,8,'Madrid ','| 1 | 2 | 8 | 12 |'),(13,'Barcelona ','| 1 | 2 | 8 | 13 |'),(14,9,'Berlin','| 1 | 2 | 9 | 14 |'),(15,'Munich','| 1 | 2 | 9 | 15 |'),(16,10,'Rome','| 1 | 2 | 10 | 16 |'),(17,'Milano','| 1 | 2 | 10 | 17 |'),(18,11,'Paris','| 1 | 2 | 11 | 18 |'),(19,'Marseille','| 1 | 2 | 11 | 19 |')
我希望达到的预期结果显示如下（我仅列出了一个4级示例）：
+--------+-------------+------------+
| org_id | search_item | nest_level |
+--------+-------------+------------+
| 1 | 1 | 1 |
| 2 | 2 | 1 |
| 2 | 1 | 2 |
| 3 | 3 | 1 |
| 3 | 1 | 2 |
| 4 | 4 | 1 |
| 4 | 1 | 2 |
| 5 | 5 | 1 |
| 5 | 1 | 2 |
| 6 | 6 | 1 |
| 6 | 1 | 2 |
| 7 | 7 | 1 |
| 7 | 1 | 2 |
| 8 | 8 | 1 |
| 8 | 2 | 2 |
| 8 | 1 | 3 |
| 9 | 9 | 1 |
| 9 | 2 | 2 |
| 9 | 1 | 3 |
| 10 | 10 | 1 |
| 10 | 2 | 2 |
| 10 | 1 | 3 |
| 11 | 11 | 1 |
| 11 | 2 | 2 |
| 11 | 1 | 3 |
| 12 | 12 | 1 |
| 12 | 8 | 2 |
| 12 | 2 | 3 |
| 12 | 1 | 4 |
.....................................
+--------+-------------+------------+
我能够使用STRING_SPLIT拉出org_id-search_item关系，但我仍然错过了棘手的级别（我想知道枚举'|'字符）
SELECT t.org_id
--,substring(replace(search_tree,' ',''),len(replace(search_tree,'')) - 2),ss.value as search_item
FROM #temp_structure t
CROSS APPLY string_split(substring(replace(search_tree,'|') ss
解决方法
我尚未对此进行彻底测试，但是您可以尝试执行以下操作：
-- Table mock-up.
DECLARE @temp TABLE ( org_id int,parent_org_id int,org_name nvarchar(255),search_tree nvarchar(255) )
-- Insert sample data...
INSERT INTO @temp VALUES
(1,null,'World','| 1 |'),(2,1,'Europe','| 1 | 2 |'),(3,'North America','| 1 | 3 |'),(4,'South America','| 1 | 4 |'),(5,'Asia','| 1 | 5 |'),(6,'Africa','| 1 | 6 |'),(7,'Australia','| 1 | 7 |'),(8,2,'Spain','| 1 | 2 | 8 |'),(9,'Germany','| 1 | 2 | 9 |'),(10,'Italy','| 1 | 2 | 10 |'),(11,'France','| 1 | 2 | 11 |'),(12,8,'Madrid ','| 1 | 2 | 8 | 12 |');
-- Select data in a nested level...
SELECT
org_id,search_item,ROW_NUMBER() OVER ( PARTITION BY org_id ORDER BY org_id,parent_org_id,search_item DESC ) AS nest_level
FROM @temp AS tmp
CROSS APPLY (
SELECT CAST ( [value] AS INT ) AS search_item FROM STRING_SPLIT ( tmp.search_tree,'|' )
WHERE NULLIF ( [value],'' ) IS NOT NULL
) AS tree
ORDER BY
org_id,search_item DESC;
返回
+--------+-------------+------------+
| org_id | search_item | nest_level |
+--------+-------------+------------+
| 1 | 1 | 1 |
| 2 | 2 | 1 |
| 2 | 1 | 2 |
| 3 | 3 | 1 |
| 3 | 1 | 2 |
| 4 | 4 | 1 |
| 4 | 1 | 2 |
| 5 | 5 | 1 |
| 5 | 1 | 2 |
| 6 | 6 | 1 |
| 6 | 1 | 2 |
| 7 | 7 | 1 |
| 7 | 1 | 2 |
| 8 | 8 | 1 |
| 8 | 2 | 2 |
| 8 | 1 | 3 |
| 9 | 9 | 1 |
| 9 | 2 | 2 |
| 9 | 1 | 3 |
| 10 | 10 | 1 |
| 10 | 2 | 2 |
| 10 | 1 | 3 |
| 11 | 11 | 1 |
| 11 | 2 | 2 |
| 11 | 1 | 3 |
| 12 | 12 | 1 |
| 12 | 8 | 2 |
| 12 | 2 | 3 |
| 12 | 1 | 4 |
+--------+-------------+------------+
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