问题描述
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
基本上保持表evaluation_grid的顺序，但只对不是范围的标准进行分组 在基于 start_value、end_value、provier、property_1、property_2 和 property_3 的逗号分隔值字符串中
我是这样试的：
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('')
),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，如下所示：
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,test2 NULL external 2 8 1
product_category test3,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;”的版本同样，但还没有设法找到解决方案，是的，我已经检查了类似的问题。 :) PS：我不能使用 STRING_AGG，因为我们的 sql Server 版本低于 2017。 任何建议将不胜感激，谢谢！
解决方法
据我所知，此查询返回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.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
按照要求有点困难。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',(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
(1,3,NULL,'internal',100,(5,4,50,(6,'external',8,(7,5,(8,150,2),(9,1)
declare @product_category table (id int,[name] varchar(50))
insert into @product_category
values (1,'test1'),'test2'),'test3'),'test4');
declare @product_subcategory table (id int,[name] varchar(50))
insert into @product_subcategory
values (1,'producttest1'),'producttest2'),'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('')),'') 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,property_3
union all
select c.criteria_name,cast(start_value as varchar(10)),property_3
from @evaluation_grid eg
join @criteria c on eg.criteria_id = c.criteria_id
where c.is_range = 1;