使用来自 SQL 服务器中不同表的数据创建逗号分隔值字符串
Create comma separated value strings using data from different tables in SQL Server
我有以下数据库模型：
criteria table:
criteria_id criteria_name is_range
1 product_category 0
2 product_subcategory 0
3 items 1
4 criteria_4 1
evaluation_grid table:
evaluation_grid_id criteria_id start_value end_value provider property_1 property_2 property_3
1 1 3 NULL internal 1 1 1
2 1 1 NULL internal 1 1 1
3 2 1 NULL internal 1 2 1
4 3 1 100 internal 2 1 1
5 4 1 50 internal 2 2 1
6 1 2 NULL external 2 8 1
7 2 2 NULL external 2 5 1
8 3 1 150 external 2 2 2
9 3 1 100 external 2 3 1
product_category table:
id name
1 test1
2 test2
3 test3
product_subcategory table:
id name
1 producttest1
2 producttest2
3 producttest3
我想要实现的是返回这样的值：
criteria start_value end_value provider property_1 property_2 property_3
product_category test3, test1 NULL internal 1 1 1
product_subcategory producttest1 NULL internal 1 2 1
items 1 100 internal 2 1 1
criteria_4 1 50 internal 2 2 1
product_category test2 NULL external 2 8 1
product_subcategory producttest2 NULL external 2 5 1
items 1 150 external 2 2 2
criteria_4 1 100 external 2 3 1
基本上保持 table evaluation_grid 的顺序，但仅对非范围的标准进行分组
基于 start_value、end_value、provier、property_1、property_2 和 property_3
的逗号分隔值字符串
我这样试过：
SELECT c.criteria_name AS criteria
,CASE WHEN c.criteria_id = 1
THEN
(IsNull(STUFF((SELECT ', ' + RTRIM(LTRIM(pc.name))
FROM product_category pc
INNER JOIN [evaluation_grid] eg ON eg.start_value=pc.id
WHERE srsg.criteria_id=c.criteria_id
FOR XML PATH('')), 1, 2, ''), ''))
WHEN c.criteria_id = 2
THEN (IsNull(STUFF((SELECT ' , ' + RTRIM(LTRIM(psc.name))
FROM product_subcategory psc
INNER JOIN [evaluation_grid] eg ON eg.start_value=psc.id
WHERE srsg.criteria_id=c.criteria_id
FOR XML PATH('')
), 1, 3, ''), ''))
ELSE
CAST(eg.start_value AS VARCHAR)
END AS start_value
,eg.end_value AS end_value
,eg.provider AS provider
,eg.property_1 AS property_1
,eg.property_2 AS property_2
,eg.property_3 AS property_3
FROM [evaluation_grid] eg
INNER JOIN criteria c ON eg.criteria_id = crs.criteria_id
GROUP BY c.criteria_name,c.criteria_id,c.is_range,eg.start_value,eg.end_value,eg.provider,eg.property_1,eg.property_2,eg.property_3
但是它返回了错误的数据，像这样：
criteria start_value end_value provider property_1 property_2 property_3
product_category test3, test1, test2 NULL internal 1 1 1
product_category test3, test1, test2 NULL external 2 8 1
product_category test3, test1, test2 NULL internal 1 1 1
product_subcategory producttest1,producttest2 NULL internal 1 2 1
product_subcategory producttest1,producttest2 NULL external 2 5 1
items 1 100 internal 1 1 1
items 1 150 external 2 2 2
criteria_4 1 50 internal 2 2 1
criteria_4 1 100 external 2 3 1
我尝试了一些带有“with cte;”的版本也没有设法找到解决方案，是的，我已经检查了类似的问题。 :)
PS：我无法使用 STRING_AGG，因为我们的服务器版本低于 2017 Sql。
任何建议将不胜感激，谢谢！
按照要求有点难。您能否查看下面的设置和结果并让我们知道所需的结果集应该是什么？
declare @criteria table (criteria_id int, criteria_name varchar(50), is_range bit)
insert into @criteria
values(1, 'product_category', 0), (2, 'product_subcategory', 0), (3, 'items', 1), (4, 'criteria_4', 1);
declare @evaluation_grid table (evaluation_grid_id int, criteria_id int, start_value int, end_value int, [provider] varchar(50), property_1 int, property_2 int, property_3 int);
insert into @evaluation_grid
values
(1, 1, 3, NULL, 'internal', 1, 1, 1),
(2, 1, 1, NULL, 'internal', 1, 1, 1),
(3, 2, 1, NULL, 'internal', 1, 2, 1),
(4, 3, 1, 100, 'internal', 2, 1, 1),
(5, 4, 1, 50, 'internal', 2, 2, 1),
(6, 1, 2, NULL, 'external', 2, 8, 1),
(7, 2, 2, NULL, 'external', 2, 5, 1),
(8, 3, 1, 150, 'external', 2, 2, 2),
(9, 4, 1, 100, 'external', 2, 3, 1)
declare @product_category table (id int, [name] varchar(50))
insert into @product_category
values (1, 'test1'), (2, 'test2'), (3, 'test3'), (4, 'test4');
declare @product_subcategory table (id int, [name] varchar(50))
insert into @product_subcategory
values (1, 'producttest1'), (2, 'producttest2'), (3, 'producttest3');
select c.criteria_name,
stuff(( select ',' + ipc.[name]
from @evaluation_grid ieg
join @product_category ipc on ieg.start_value = ipc.id
where [provider] = eg.[provider] and property_1 = eg.property_1 and property_2 = eg.property_2 and property_3 = eg.property_3
order by ieg.evaluation_grid_id
for xml path('')), 1 ,1, '') as start_value,
end_value,
[provider],
property_1,
property_2,
property_3
from @evaluation_grid eg
join @criteria c on eg.criteria_id = c.criteria_id
where c.is_range = 0
group
by c.criteria_name, end_value, [provider], property_1, property_2, property_3
union all
select c.criteria_name, cast(start_value as varchar(10)), end_value, [provider], property_1, property_2, property_3
from @evaluation_grid eg
join @criteria c on eg.criteria_id = c.criteria_id
where c.is_range = 1;
据我所知，此查询 returns 您正在寻找的确切输出。
with cte as (
select c.criteria_name,
eg.evaluation_grid_id,
case when c.criteria_id = 1 then pc.[name]
when c.criteria_id = 2 then psc.[name]
else null end pc_cat,
c.criteria_id,c.is_range, eg.start_value, eg.end_value,
eg.[provider], eg.property_1, eg.property_2,eg.property_3
from @evaluation_grid eg
join @criteria c ON eg.criteria_id = c.criteria_id
left join @product_category pc on eg.start_value=pc.id
left join @product_subcategory psc on eg.start_value=psc.id)
select c.criteria_name as criteria,
case when c.is_range=0 then
STUFF((SELECT ', ' + RTRIM(LTRIM(c2.pc_cat))
FROM cte c2
WHERE c2.criteria_id=c.criteria_id
and c2.is_range=c.is_range
and c2.[provider]=c.[provider]
and c2.property_1=c.property_1
and c2.property_2=c.property_2
and c2.property_3=c.property_3
FOR XML PATH('')), 1, 2, '')
else max(cast(c.start_value as varchar(50))) end as start_value,
c.end_value, c.[provider], c.property_1, c.property_2, c.property_3
from cte c
group by c.criteria_name, c.criteria_id, c.is_range, c.end_value,
c.[provider], c.property_1, c.property_2, c.property_3
order by max(c.evaluation_grid_id);
输出
criteria start_value end_value provider property_1 property_2 property_3
product_category test3, test1 NULL internal 1 1 1
product_subcategory producttest1 NULL internal 1 2 1
items 1 100 internal 2 1 1
criteria_4 1 50 internal 2 2 1
product_category test2 NULL external 2 8 1
product_subcategory producttest2 NULL external 2 5 1
items 1 150 external 2 2 2
criteria_4 1 100 external 2 3 1
我有以下数据库模型：
criteria table:
criteria_id criteria_name is_range
1 product_category 0
2 product_subcategory 0
3 items 1
4 criteria_4 1
evaluation_grid table:
evaluation_grid_id criteria_id start_value end_value provider property_1 property_2 property_3
1 1 3 NULL internal 1 1 1
2 1 1 NULL internal 1 1 1
3 2 1 NULL internal 1 2 1
4 3 1 100 internal 2 1 1
5 4 1 50 internal 2 2 1
6 1 2 NULL external 2 8 1
7 2 2 NULL external 2 5 1
8 3 1 150 external 2 2 2
9 3 1 100 external 2 3 1
product_category table:
id name
1 test1
2 test2
3 test3
product_subcategory table:
id name
1 producttest1
2 producttest2
3 producttest3
我想要实现的是返回这样的值：
criteria start_value end_value provider property_1 property_2 property_3
product_category test3, test1 NULL internal 1 1 1
product_subcategory producttest1 NULL internal 1 2 1
items 1 100 internal 2 1 1
criteria_4 1 50 internal 2 2 1
product_category test2 NULL external 2 8 1
product_subcategory producttest2 NULL external 2 5 1
items 1 150 external 2 2 2
criteria_4 1 100 external 2 3 1
基本上保持 table evaluation_grid 的顺序，但仅对非范围的标准进行分组 基于 start_value、end_value、provier、property_1、property_2 和 property_3
的逗号分隔值字符串我这样试过：
SELECT c.criteria_name AS criteria
,CASE WHEN c.criteria_id = 1
THEN
(IsNull(STUFF((SELECT ', ' + RTRIM(LTRIM(pc.name))
FROM product_category pc
INNER JOIN [evaluation_grid] eg ON eg.start_value=pc.id
WHERE srsg.criteria_id=c.criteria_id
FOR XML PATH('')), 1, 2, ''), ''))
WHEN c.criteria_id = 2
THEN (IsNull(STUFF((SELECT ' , ' + RTRIM(LTRIM(psc.name))
FROM product_subcategory psc
INNER JOIN [evaluation_grid] eg ON eg.start_value=psc.id
WHERE srsg.criteria_id=c.criteria_id
FOR XML PATH('')
), 1, 3, ''), ''))
ELSE
CAST(eg.start_value AS VARCHAR)
END AS start_value
,eg.end_value AS end_value
,eg.provider AS provider
,eg.property_1 AS property_1
,eg.property_2 AS property_2
,eg.property_3 AS property_3
FROM [evaluation_grid] eg
INNER JOIN criteria c ON eg.criteria_id = crs.criteria_id
GROUP BY c.criteria_name,c.criteria_id,c.is_range,eg.start_value,eg.end_value,eg.provider,eg.property_1,eg.property_2,eg.property_3
但是它返回了错误的数据，像这样：
criteria start_value end_value provider property_1 property_2 property_3
product_category test3, test1, test2 NULL internal 1 1 1
product_category test3, test1, test2 NULL external 2 8 1
product_category test3, test1, test2 NULL internal 1 1 1
product_subcategory producttest1,producttest2 NULL internal 1 2 1
product_subcategory producttest1,producttest2 NULL external 2 5 1
items 1 100 internal 1 1 1
items 1 150 external 2 2 2
criteria_4 1 50 internal 2 2 1
criteria_4 1 100 external 2 3 1
我尝试了一些带有“with cte;”的版本也没有设法找到解决方案，是的，我已经检查了类似的问题。 :) PS：我无法使用 STRING_AGG，因为我们的服务器版本低于 2017 Sql。 任何建议将不胜感激，谢谢！
按照要求有点难。您能否查看下面的设置和结果并让我们知道所需的结果集应该是什么？
declare @criteria table (criteria_id int, criteria_name varchar(50), is_range bit)
insert into @criteria
values(1, 'product_category', 0), (2, 'product_subcategory', 0), (3, 'items', 1), (4, 'criteria_4', 1);
declare @evaluation_grid table (evaluation_grid_id int, criteria_id int, start_value int, end_value int, [provider] varchar(50), property_1 int, property_2 int, property_3 int);
insert into @evaluation_grid
values
(1, 1, 3, NULL, 'internal', 1, 1, 1),
(2, 1, 1, NULL, 'internal', 1, 1, 1),
(3, 2, 1, NULL, 'internal', 1, 2, 1),
(4, 3, 1, 100, 'internal', 2, 1, 1),
(5, 4, 1, 50, 'internal', 2, 2, 1),
(6, 1, 2, NULL, 'external', 2, 8, 1),
(7, 2, 2, NULL, 'external', 2, 5, 1),
(8, 3, 1, 150, 'external', 2, 2, 2),
(9, 4, 1, 100, 'external', 2, 3, 1)
declare @product_category table (id int, [name] varchar(50))
insert into @product_category
values (1, 'test1'), (2, 'test2'), (3, 'test3'), (4, 'test4');
declare @product_subcategory table (id int, [name] varchar(50))
insert into @product_subcategory
values (1, 'producttest1'), (2, 'producttest2'), (3, 'producttest3');
select c.criteria_name,
stuff(( select ',' + ipc.[name]
from @evaluation_grid ieg
join @product_category ipc on ieg.start_value = ipc.id
where [provider] = eg.[provider] and property_1 = eg.property_1 and property_2 = eg.property_2 and property_3 = eg.property_3
order by ieg.evaluation_grid_id
for xml path('')), 1 ,1, '') as start_value,
end_value,
[provider],
property_1,
property_2,
property_3
from @evaluation_grid eg
join @criteria c on eg.criteria_id = c.criteria_id
where c.is_range = 0
group
by c.criteria_name, end_value, [provider], property_1, property_2, property_3
union all
select c.criteria_name, cast(start_value as varchar(10)), end_value, [provider], property_1, property_2, property_3
from @evaluation_grid eg
join @criteria c on eg.criteria_id = c.criteria_id
where c.is_range = 1;
据我所知，此查询 returns 您正在寻找的确切输出。
with cte as (
select c.criteria_name,
eg.evaluation_grid_id,
case when c.criteria_id = 1 then pc.[name]
when c.criteria_id = 2 then psc.[name]
else null end pc_cat,
c.criteria_id,c.is_range, eg.start_value, eg.end_value,
eg.[provider], eg.property_1, eg.property_2,eg.property_3
from @evaluation_grid eg
join @criteria c ON eg.criteria_id = c.criteria_id
left join @product_category pc on eg.start_value=pc.id
left join @product_subcategory psc on eg.start_value=psc.id)
select c.criteria_name as criteria,
case when c.is_range=0 then
STUFF((SELECT ', ' + RTRIM(LTRIM(c2.pc_cat))
FROM cte c2
WHERE c2.criteria_id=c.criteria_id
and c2.is_range=c.is_range
and c2.[provider]=c.[provider]
and c2.property_1=c.property_1
and c2.property_2=c.property_2
and c2.property_3=c.property_3
FOR XML PATH('')), 1, 2, '')
else max(cast(c.start_value as varchar(50))) end as start_value,
c.end_value, c.[provider], c.property_1, c.property_2, c.property_3
from cte c
group by c.criteria_name, c.criteria_id, c.is_range, c.end_value,
c.[provider], c.property_1, c.property_2, c.property_3
order by max(c.evaluation_grid_id);
输出
criteria start_value end_value provider property_1 property_2 property_3
product_category test3, test1 NULL internal 1 1 1
product_subcategory producttest1 NULL internal 1 2 1
items 1 100 internal 2 1 1
criteria_4 1 50 internal 2 2 1
product_category test2 NULL external 2 8 1
product_subcategory producttest2 NULL external 2 5 1
items 1 150 external 2 2 2
criteria_4 1 100 external 2 3 1