Question

我无法让它正常工作。我有4个表：产品，供应商，X_Product_供应商和评论。我想查询它们并使用以下查询将它们放入JSON：

WITH Products (Id, Name, Price) As (
    SELECT 1, 'First Product', 10
), Suppliers (Id, Name) As (
    SELECT 1, 'Factory1' UNION ALL
    SELECT 2, 'Factory2'
), Comments (Id, [Text], ProductId) As (
    SELECT 1, 'Good Product', 1 UNION ALL
    SELECT 2, 'Fantastic!'  , 1
), X_Product_Supplier (ProductId, SupplierId) As (
    SELECT 1, 1 UNION ALL
    SELECT 1, 2
)
SELECT Products.*, Suppliers.*, Comments.* FROM Products
LEFT OUTER JOIN X_Product_Supplier ON X_Product_Supplier.ProductId = Products.Id
LEFT OUTER JOIN Suppliers ON X_Product_Supplier.SupplierId = Suppliers.Id
LEFT OUTER JOIN Comments ON Comments.ProductId = Products.Id
FOR JSON AUTO

出于某种原因，sql-server会将注释嵌套在供应商下而不是产品下：

   {  
      "Id":1,
      "Name":"First Product",
      "Price":"10",
      "Suppliers":[  
         {  
            "Id":1,
            "Name":"Factory1",
            "Comments":[  //THIS SHOULD BE UNDER PRODUCT, NOT SUPPLIER
               {  
                  "Id":1,
                  "Text":"Good Product",
                  "ProductId":1
               },
               {  
                  "Id":2,
                  "Text":"Fantastic!",
                  "ProductId":1
               }
            ]
         },
         {  
            "Id":2,
            "Name":"Factory2",
            "Comments":[  //THIS IS NOW DUPLICATE
               {  
                  "Id":1,
                  "Text":"Good Product",
                  "ProductId":1
               },
               {  
                  "Id":2,
                  "Text":"Fantastic!",
                  "ProductId":1
               }
            ]
         }
      ]
   }

我真正想要的是：

   {  
      "Id":1,
      "Name":"First Product",
      "Price":"10",
      "Suppliers":[  
         {  
            "Id":1,
            "Name":"Factory1"
         },
         {  
            "Id":2,
            "Name":"Factory2"
         }
      ],
      "Comments":[  
               {  
                  "Id":1,
                  "Text":"Good Product",
                  "ProductId":1
               },
               {  
                  "Id":2,
                  "Text":"Fantastic!",
                  "ProductId":1
               }
            ]
   }

我该怎么做？

Answer 1

如果有人在寻找答案，这是我写的简单查询。根据您的架构更改查询，它应该提供适当的结构化结果。

SELECT Products.*,
(SELECT Suppliers.*
 FROM Suppliers
 WHERE Suppliers.Id = Products.SuppliersId
 FOR JSON AUTO) As Suppliers,
(SELECT Comments.*
 FROM Comments 
 WHERE Comments.ProductId = Products.Id
 FOR JSON AUTO) As Comments
FROM Products
FOR JSON AUTO

Answer 2

SQL Server会将结果表转换为JSON，其列的顺序与它们在SELECT语句中的顺序相同。它不会对列分组做出任何假设。为了帮助它，可以为每个产品的JSON对象预先安排供应商和评论：

WITH Products (Id, Name, Price) As (
    SELECT 1, 'First Product', 10
), Suppliers (Id, Name) As (
    SELECT 1, 'Factory1' UNION ALL
    SELECT 2, 'Factory2'
), Comments (Id, [Text], ProductId) As (
    SELECT 1, 'Good Product', 1 UNION ALL
    SELECT 2, 'Fantastic!'  , 1
), X_Product_Supplier (ProductId, SupplierId) As (
    SELECT 1, 1 UNION ALL
    SELECT 1, 2
)
SELECT Products.*,
    (SELECT Suppliers.*
     FROM Suppliers
     INNER JOIN X_Product_Supplier ON X_Product_Supplier.SupplierId = Suppliers.Id
     WHERE X_Product_Supplier.ProductId = Products.Id
     FOR JSON AUTO, TYPE) As Suppliers,
    (SELECT Comments.*
     FROM Comments
     WHERE Comments.ProductId = Products.Id
     FOR JSON AUTO, TYPE) As Comments
FROM Products
FOR JSON AUTO

在XML上测试，因为我目前无权访问SQL Server 2016。如果这不起作用，请告诉我。

Answer 3

我知道它来晚了，但是我遇到了同样的问题，但是答案并不能帮助我过滤下面的解决方案where子句中的空值

SELECT  Products.*, 
        Suppliers.Id as [Suppliers.Id], 
        Suppliers.Name as [Suppliers.Name],
        Comments.Id as [Comments.Id],
        Comments.Text as [Comments.Text],
        Comments.ProductId as [Comments.ProductId] 
FROM Products
LEFT OUTER JOIN X_Product_Supplier ON X_Product_Supplier.ProductId = Products.Id
LEFT OUTER JOIN Suppliers ON X_Product_Supplier.SupplierId = Suppliers.Id
LEFT OUTER JOIN Comments ON Comments.ProductId = Products.Id
FOR JSON PATH

经过测试并在SQL SERVER 2017上工作

我怎样才能做一个＆＃34; FOR JSON＆＃34;有2个联接

3 个答案: