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

基本上保持表evaluation_grid的顺序，但仅根据start_value、end_value、provier、property_1、property_2和property_3对不是逗号分隔值字符串范围的标准进行分组

我试过这样：

 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

但它返回错误的数据，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, 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 也是，但还没有找到解决方案，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！

4

2 回答 2

1

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.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
于 2021-02-04T20:03:45.200 回答
0

遵循要求有点困难。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', 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;
于 2021-02-03T00:23:18.990 回答