如何从 JSON_Value 结果中删除重复结果

Question

如果我有两张桌子：

PartDetailsTable

Id   Details                            Price   Brand Id
--------------------------------------------------------
1   {"partNumber":"012345","qty":14}    14.99          1
2   {"partNumber":"012345","qty":11}     4.88          1
3   {"partNumber":"44444","qty":1}       3.24          3
4   {"partNumber":"72431223","qty":9}    3.33          1
5   {"partNumber":"98989","qty":1}       3.33          2

Brands Table
Id   BrandName   
----------
1   BrandOne
2   BrandTwo
3   BrandThree

我想从 PartDetailsTable 返回Details.PartNumber、Details.Qty 和 BrandId，但还包括 BrandName

我正在使用这个：

SELECT DISTINCT i.id,
JSON_VALUE(i.Details, '$.partNumber') AS 'PartNumber', 
i.BrandId, 
b.Name AS 'BrandName',
    JSON_VALUE(i.Details, '$.qty') AS 'Qty', 
FROM
    PartDetailsTable i INNER JOIN Brands b ON (i.BrandId = b.Id)

但我想删除重复的零件编号记录，因此对于上面的 PartDetailsTable 中的示例，我只想返回 1,3,4,5 记录，请提供任何建议（因为 id 1 和 2 具有相同的零件编号 012345 （我不需要对数量或类似内容进行求和，只需选择一个并删除任何重复项

Answer 1

您可以按 JSON_VALUE 将其分组，然后获取 MIN（索引），连接回同一张表以获取数量

SELECT i.id,
    JSON_VALUE(i.Details, '$.partNumber') AS 'PartNumber', 
    i.BrandId, 
    b.Name AS 'BrandName',
    JSON_VALUE(i.Details, '$.qty') AS 'Qty', 
FROM (
    SELECT JSON_VALUE(Details, '$.partNumber') AS 'PartNumber', MIN(Id) [Id]
    FROM PartnerDetailsTable
    GROUP BY JSON_VALUE(Details, '$.partNumber')
) gi
JOIN PartnerDetailsTable i
    ON i.Id = gi.Id
INNER JOIN Brands b 
    ON i.BrandId = b.Id