SQL LEFT JOIN 到许多类别

Question

假设以下简单的scenario，其中一个产品行连接到一个主要类别、子类别和子子类别。

DECLARE @PRODUCTS TABLE (ID int, DESCRIPTION varchar(50), CAT varchar(30), SUBCAT varchar(30), SUBSUBCAT varchar(30));

INSERT @PRODUCTS (ID, DESCRIPTION, CAT, SUBCAT, SUBSUBCAT) VALUES
(1, 'NIKE MILLENIUM', '1', '10', '100'),
(2, 'NIKE CORTEZ', '1', '12', '104'),
(3, 'ADIDAS PANTS', '2', '27', '238'),
(4, 'PUMA REVOLUTION 5', '3', '35', '374'),
(5, 'SALOMON SHELTER CS', '4', '15', '135'),
(6, 'NIKE EBERNON LOW', '2', '14', '157');

DECLARE @CATS TABLE (ID int, DESCR varchar(100));

INSERT @CATS (ID, DESCR) VALUES
(1, 'MEN'),
(2, 'WOMEN'),
(3, 'UNISEX'),
(4, 'KIDS'),
(5, 'TEENS'),
(6, 'BACK TO SCHOOL');

DECLARE @SUBCATS TABLE (ID int, DESCR varchar(100));

INSERT @SUBCATS (ID, DESCR) VALUES
(10, 'FOOTWEAR'),
(12, 'OUTERWEAR'),
(14, 'SWIMWEAR'),
(15, 'HOODIES'),
(27, 'CLOTHING'),
(35, 'SPORTS');

DECLARE @SUBSUBCATS TABLE (ID int, DESCR varchar(100));

INSERT @SUBSUBCATS (ID, DESCR) VALUES
(100, 'RUNNING'),
(104, 'ZIP TOPS'),
(135, 'FLEECE'),
(157, 'BIKINIS'),
(238, 'PANTS'),
(374, 'JOGGERS');


SELECT prod.ID,
    prod.DESCRIPTION,
    CONCAT(cat1.DESCR, ' > ', cat2.DESCR, ' > ', cat3.DESCR) AS CATEGORIES
FROM @PRODUCTS AS prod
LEFT JOIN @CATS AS cat1 ON cat1.ID = prod.CAT
LEFT JOIN @SUBCATS AS cat2 ON cat2.ID = prod.SUBCAT
LEFT JOIN @SUBSUBCATS AS cat3 ON cat3.ID = prod.SUBSUBCAT;

现在假设@PRODUCTS 表上的外键不仅仅是它们各自表的索引。它们是多个类别、子类别和子子类别（如 here）的逗号分隔索引。

DECLARE @PRODUCTS TABLE (ID int, DESCRIPTION varchar(50), CAT varchar(30), SUBCAT varchar(30), SUBSUBCAT varchar(30));

INSERT @PRODUCTS (ID, DESCRIPTION, CAT, SUBCAT, SUBSUBCAT) VALUES
(1, 'NIKE MILLENIUM', '1, 2', '10, 12', '100, 135'),
(2, 'NIKE CORTEZ', '1, 5', '12, 15', '104, 374'),
(3, 'ADIDAS PANTS', '2, 6', '27, 35', '238, 374');

DECLARE @CATS TABLE (ID int, DESCR varchar(100));

INSERT @CATS (ID, DESCR) VALUES
(1, 'MEN'),
(2, 'WOMEN'),
(3, 'UNISEX'),
(4, 'KIDS'),
(5, 'TEENS'),
(6, 'BACK TO SCHOOL');

DECLARE @SUBCATS TABLE (ID int, DESCR varchar(100));

INSERT @SUBCATS (ID, DESCR) VALUES
(10, 'FOOTWEAR'),
(12, 'OUTERWEAR'),
(14, 'SWIMWEAR'),
(15, 'HOODIES'),
(27, 'CLOTHING'),
(35, 'SPORTS');

DECLARE @SUBSUBCATS TABLE (ID int, DESCR varchar(100));

INSERT @SUBSUBCATS (ID, DESCR) VALUES
(100, 'RUNNING'),
(104, 'ZIP TOPS'),
(135, 'FLEECE'),
(157, 'BIKINIS'),
(238, 'PANTS'),
(374, 'JOGGERS');


SELECT prod.ID,
    prod.DESCRIPTION
    --CONCAT(cat1.DESCR, ' > ', cat2.DESCR, ' > ', cat3.DESCR) AS CATEGORIES
FROM @PRODUCTS AS prod
--LEFT JOIN @CATS AS cat1 ON cat1.ID = prod.CAT
--LEFT JOIN @SUBCATS AS cat2 ON cat2.ID = prod.SUBCAT
--LEFT JOIN @SUBSUBCATS AS cat3 ON cat3.ID = prod.SUBSUBCAT;

在这种情况下，我想实现以下目标：

1. 能够检索猫、子猫、子子猫的各自名称，即。对于猫 '1, 2' 能够检索他们的名字（我试过 LEFT JOIN @CATS AS cat1 ON cat1.ID IN prod.CAT 但它不起作用）
2. 创建相应猫、子猫、子子猫的三元组，即。对于

• 猫'1、2'
• subcats '12, 17'
• sub-subcats '239, 372'

（在检索到适当的名称后）创建管道分隔的类别路由，例如name of cat 1 > name of subcat 12 > name of sub-subcat 239 | name of cat 2 > name of subcat 17 > name of sub-subcat 372

所以，对于像(1, 'NIKE MILLENIUM', '1, 2', '10, 12', '100, 135'), 这样的一行

我想得到以下结果

ID	DESCRIPTION	CATEGORIES
1	NIKE MILLENIUM	MEN > FOOTWEAR > RUNNING @ WOMEN > OUTERWEAR > FLEECE (I had to use @ as the delimiter of the two triplets because pipe messed with the table's columns)

如果用户愚蠢地存储了比子猫 ID 或子子猫 ID 更多的猫 ID，则查询应该只匹配具有相应位置匹配的那些，即 for

• 猫'1、2'
• 子目录“12”
• sub-subcats '239, 372'

它应该只创建一个三元组，例如 name of 1 > name of 12 > name of 239

Answer 1

STRING_SPLIT() 不承诺按特定顺序返回值，因此在这种情况下它不起作用，因为序数位置很重要。

使用OPENJSON() 将字符串拆分为单独的行，以确保以相同的顺序返回值。
OPENJSON() 还返回一个key 字段，因此您可以加入每个分组中的行号。您需要INNER JOIN，因为您的要求是该“列”中的所有值都必须存在。
使用STUFF() 组装各种 cat>subcat>subsubcat 值。

DECLARE @PRODUCTS TABLE (ID int, DESCRIPTION varchar(50), CAT varchar(30), SUBCAT varchar(30), SUBSUBCAT varchar(30));

INSERT @PRODUCTS (ID, DESCRIPTION, CAT, SUBCAT, SUBSUBCAT) VALUES
(1, 'NIKE MILLENIUM', '1, 2', '10, 12', '100, 135'),
(2, 'NIKE CORTEZ', '1, 5', '12, 15', '104, 374'),
(3, 'ADIDAS PANTS', '2, 6, 1', '27, 35, 10', '238, 374, 100'),
(4, 'JOE THE PLUMBER JEANS', '1, 5', '27', '238, 374');

DECLARE @CATS TABLE (ID int, DESCR varchar(100));

INSERT @CATS (ID, DESCR) VALUES
(1, 'MEN'),
(2, 'WOMEN'),
(3, 'UNISEX'),
(4, 'KIDS'),
(5, 'TEENS'),
(6, 'BACK TO SCHOOL');

DECLARE @SUBCATS TABLE (ID int, DESCR varchar(100));

INSERT @SUBCATS (ID, DESCR) VALUES
(10, 'FOOTWEAR'),
(12, 'OUTERWEAR'),
(14, 'SWIMWEAR'),
(15, 'HOODIES'),
(27, 'CLOTHING'),
(35, 'SPORTS');

DECLARE @SUBSUBCATS TABLE (ID int, DESCR varchar(100));

INSERT @SUBSUBCATS (ID, DESCR) VALUES
(100, 'RUNNING'),
(104, 'ZIP TOPS'),
(135, 'FLEECE'),
(157, 'BIKINIS'),
(238, 'PANTS'),
(374, 'JOGGERS');

;
with prod as (
    SELECT p.ID,
        p.DESCRIPTION
        --CONCAT(cat1.DESCR, ' > ', cat2.DESCR, ' > ', cat3.DESCR) AS CATEGORIES
        , c.value as CatId
        , c.[key] as CatKey
        , sc.value as SubCatId
        , sc.[key] as SubCatKey
        , ssc.value as SubSubCatId
        , ssc.[key] as SubSubCatKey
    FROM @PRODUCTS p
      cross apply OPENJSON(CONCAT('["', REPLACE(cat, ', ', '","'), '"]')) c
      cross apply OPENJSON(CONCAT('["', REPLACE(subcat, ', ', '","'), '"]')) sc
      cross apply OPENJSON(CONCAT('["', REPLACE(subsubcat, ', ', '","'), '"]')) ssc
    where c.[key] = sc.[key]
      and c.[key] = ssc.[key]
)
, a as (
    select p.ID
    , p.DESCRIPTION
    , c.DESCR + ' > ' + sc.DESCR + ' > ' + ssc.DESCR as CATEGORIES
    , p.CatKey
    from prod p
      inner join @CATS c on c.ID = p.CatId
      inner join @SUBCATS sc on sc.ID = p.SubCatId
      inner join @SUBSUBCATS ssc on ssc.ID = p.SubSubCatId
)

select DISTINCT ID
, DESCRIPTION
, replace(STUFF((SELECT distinct ' | ' + a2.CATEGORIES
            from a a2
            where a.ID = a2.ID
            FOR XML PATH(''))
        ,1,2,''), '>', '>') CATEGORIES
from a

Answer 2

由于旧技术的变化，答案完全不同。我认为我的原始答案对于使用当前 SQL Server 版本的人来说仍然很好，所以我不想删除它。

我不记得我是从哪里得到这个函数的。当我今天找到它时，它被命名为 split_delimiter。我更改了名称，添加了一些 cmets，并加入了具有超过一个字符长度的分隔符的功能。

CREATE FUNCTION [dbo].[udf_split_string](@delimited_string VARCHAR(8000), @delimiter varchar(10))
RETURNS TABLE AS
RETURN
WITH cte10(num) AS (    --  10 rows
    SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL
    SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL
    SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1
)                         
, cte100(num) AS (  --  100 rows
    SELECT 1 
    FROM   cte10 t1, cte10 t2
)
, cte10000(num) AS (    --  10000 rows
    SELECT 1 
    FROM   cte100 t1, cte100 t2
)
, cte1(num) AS (    --  1 row per character
    SELECT TOP (ISNULL(DATALENGTH(@delimited_string), 0)) ROW_NUMBER() OVER (ORDER BY (SELECT NULL)) 
    FROM   cte10000
)
, cte2(num) AS (    --  locations of strings
    SELECT  1
    UNION ALL
    SELECT t.num + len(replace(@delimiter, ' ', '_'))
    FROM   cte1 t
    WHERE  SUBSTRING(@delimited_string, t.num, len(replace(@delimiter, ' ', '_'))) = @delimiter
)
, cte3(num, [len]) AS (
    SELECT t.num
         , ISNULL(NULLIF(CHARINDEX(@delimiter, @delimited_string, t.num), 0) - t.num, 8000)
    FROM  cte2 t
)

SELECT [Key]   = ROW_NUMBER() OVER (ORDER BY t.num)
     , [Value] = SUBSTRING(@delimited_string, t.num, t.[len])
FROM cte3 t;
GO



DECLARE @PRODUCTS TABLE (ID int, DESCRIPTION varchar(50), CAT varchar(30), SUBCAT varchar(30), SUBSUBCAT varchar(30));

INSERT @PRODUCTS (ID, DESCRIPTION, CAT, SUBCAT, SUBSUBCAT) VALUES
(1, 'NIKE MILLENIUM', '1, 2', '10, 12', '100, 135'),
(2, 'NIKE CORTEZ', '1, 5', '12, 15', '104, 374'),
(3, 'ADIDAS PANTS', '2, 6, 1', '27, 35, 10', '238, 374, 100'),
(4, 'JOE THE PLUMBER JEANS', '1, 5', '27', '238, 374');

DECLARE @CATS TABLE (ID int, DESCR varchar(100));

INSERT @CATS (ID, DESCR) VALUES
(1, 'MEN'),
(2, 'WOMEN'),
(3, 'UNISEX'),
(4, 'KIDS'),
(5, 'TEENS'),
(6, 'BACK TO SCHOOL');

DECLARE @SUBCATS TABLE (ID int, DESCR varchar(100));

INSERT @SUBCATS (ID, DESCR) VALUES
(10, 'FOOTWEAR'),
(12, 'OUTERWEAR'),
(14, 'SWIMWEAR'),
(15, 'HOODIES'),
(27, 'CLOTHING'),
(35, 'SPORTS');

DECLARE @SUBSUBCATS TABLE (ID int, DESCR varchar(100));

INSERT @SUBSUBCATS (ID, DESCR) VALUES
(100, 'RUNNING'),
(104, 'ZIP TOPS'),
(135, 'FLEECE'),
(157, 'BIKINIS'),
(238, 'PANTS'),
(374, 'JOGGERS');

;
with prod as (
    SELECT p.ID,
        p.DESCRIPTION
        , c.value as CatId
        , c.[key] as CatKey
        , sc.value as SubCatId
        , sc.[key] as SubCatKey
        , ssc.value as SubSubCatId
        , ssc.[key] as SubSubCatKey
    FROM @PRODUCTS p
      cross apply dbo.udf_split_string(cat, ', ') c
      cross apply dbo.udf_split_string(subcat, ', ') sc
      cross apply dbo.udf_split_string(subsubcat, ', ') ssc
    where c.[key] = sc.[key]
      and c.[key] = ssc.[key]
)
, a as (
    select p.ID
    , p.DESCRIPTION
    , c.DESCR + ' > ' + sc.DESCR + ' > ' + ssc.DESCR as CATEGORIES
    , p.CatKey
    from prod p
      inner join @CATS c on c.ID = p.CatId
      inner join @SUBCATS sc on sc.ID = p.SubCatId
      inner join @SUBSUBCATS ssc on ssc.ID = p.SubSubCatId
)

select DISTINCT ID
, DESCRIPTION
, replace(STUFF((SELECT distinct ' | ' + a2.CATEGORIES
            from a a2
            where a.ID = a2.ID
            FOR XML PATH(''))
        ,1,2,''), '>', '>') CATEGORIES
from a

Answer 3

那应该可以了，我把你的字符“>”改成了“-”，只是为了让数据更简单。

您的桌子设计并不完美，但第一次尝试几乎从未如此。

select mainp.ID, mainp.DESCRIPTION, stuff(ppaths.metapaths, len(ppaths.metapaths),1,'') metalinks
from @PRODUCTS mainp
cross apply(
select
(select 
  c.DESCR + '-' + sc.DESCR + '-' + sbc.DESCR + '|'
from @PRODUCTS p    
    cross apply (select row_number() over(order by Value) id, Value from split(p.CAT, ','))cat_ids
    inner join @cats c on c.ID = cat_ids.Value
    cross apply (select row_number() over(order by Value) id, Value from split(p.SUBCAT, ','))subcat_ids
    inner join @SUBCATS sc on sc.ID = subcat_ids.Value
    and subcat_ids.id = subcat_ids.id
    cross apply (select row_number() over(order by Value) id, Value  from split(p.SUBSUBCAT, ','))subsubcat_ids
    inner join @SUBSUBCATS sbc on sbc.ID = subsubcat_ids.Value
    and subsubcat_ids.id = subcat_ids.id
where p.id = mainp.ID
for xml path('')) metapaths
) ppaths

拆分功能的链接 https://desarrolladores.me/2014/03/sql-server-funcion-split-para-dividir-un-string/