将嵌套的名称-值对从json导入SQL Server

我正在将一个json文件导入SQL 2016，该文件在嵌套值结构中有一些嵌套的名称-值对。这是我遇到问题的这些对中的值，例如

{ "name": "Colour", "value": "Orange" }, { "name": "Calories", "value": "25" } 

sql:

其结果是：

id  fruit   Colour  Weight
1   orange  NULL    NULL
23  Banana  NULL    NULL

完整（测试）数据

[ { “id”：1， “水果”：“橙色”， “价值观”：[ {“名称”：“颜色”，“值”：“橙色”}， {“名称”：“重量”，“值”：“16”}， {“名称”：“卡路里”，“值”：“25”} ] }, { “id”：23， “水果”：“香蕉”， “价值观”：[ {“名称”：“颜色”，“值”：“黄色”}， {“名称”：“重量”，“值”：“30”}， {“名称”：“卡路里”，“值”：“250”} ] }

---

]

您可以尝试下一种方法，该方法将返回完整数据：

DECLARE @json nvarchar(max)
SET @json = N'[ { "id": 1, "fruit": "orange", "values": [ { "name": "Colour", "value": "Orange" }, { "name": "Weight", "value": "16" }, { "name": "Calories", "value": "25" } ] }, { "id": 23, "fruit": "Banana", "values": [ { "name": "Colour", "value": "Yellow" }, { "name": "Weight", "value": "30" }, { "name": "Calories", "value": "250" } ] } ]'

SELECT i.id, i.fruit, v.[name], v.[value]
FROM OPENJSON(@json) 
WITH (
   id int '$.id',
   fruit nvarchar(50) '$.fruit',
   [values] nvarchar(max) '$.values' AS JSON
) AS i
CROSS APPLY (
   SELECT *
   FROM OPENJSON(i.[values])
   WITH (
      [name] nvarchar(max) '$.name',
      [value] nvarchar(max) '$.value'
   )
) v

输出：

id  fruit   name     value
1   orange  Colour   Orange
1   orange  Weight   16
1   orange  Calories 25
23  Banana  Colour   Yellow
23  Banana  Weight   30
23  Banana  Calories 250

id  fruit   Colour  Weight
1   orange  Orange  16
23  Banana  Yellow  30

如果希望输出有关颜色和重量的信息，请尝试以下操作：

DECLARE @json nvarchar(max)
SET @json = N'[ { "id": 1, "fruit": "orange", "values": [ { "name": "Colour", "value": "Orange" }, { "name": "Weight", "value": "16" }, { "name": "Calories", "value": "25" } ] }, { "id": 23, "fruit": "Banana", "values": [ { "name": "Colour", "value": "Yellow" }, { "name": "Weight", "value": "30" }, { "name": "Calories", "value": "250" } ] } ]'

SELECT i.id, i.fruit, v1.[value] AS Colour, v2.[value] AS Weight
FROM OPENJSON(@json) 
WITH (
   id int '$.id',
   fruit nvarchar(50) '$.fruit',
   [values] nvarchar(max) '$.values' AS JSON
) AS i
OUTER APPLY (
   SELECT *
   FROM OPENJSON(i.[values])
   WITH (
      [name] nvarchar(max) '$.name',
      [value] nvarchar(max) '$.value'
   )
   WHERE [name] = 'Colour'
) v1 
OUTER APPLY (
   SELECT *
   FROM OPENJSON(i.[values])
   WITH (
      [name] nvarchar(max) '$.name',
      [value] nvarchar(max) '$.value'
   )
   WHERE [name] = 'Weight'
) v2

输出：

id  fruit   name     value
1   orange  Colour   Orange
1   orange  Weight   16
1   orange  Calories 25
23  Banana  Colour   Yellow
23  Banana  Weight   30
23  Banana  Calories 250

id  fruit   Colour  Weight
1   orange  Orange  16
23  Banana  Yellow  30

---

您可以尝试下一种方法，该方法将返回完整数据：

DECLARE @json nvarchar(max)
SET @json = N'[ { "id": 1, "fruit": "orange", "values": [ { "name": "Colour", "value": "Orange" }, { "name": "Weight", "value": "16" }, { "name": "Calories", "value": "25" } ] }, { "id": 23, "fruit": "Banana", "values": [ { "name": "Colour", "value": "Yellow" }, { "name": "Weight", "value": "30" }, { "name": "Calories", "value": "250" } ] } ]'

SELECT i.id, i.fruit, v.[name], v.[value]
FROM OPENJSON(@json) 
WITH (
   id int '$.id',
   fruit nvarchar(50) '$.fruit',
   [values] nvarchar(max) '$.values' AS JSON
) AS i
CROSS APPLY (
   SELECT *
   FROM OPENJSON(i.[values])
   WITH (
      [name] nvarchar(max) '$.name',
      [value] nvarchar(max) '$.value'
   )
) v

输出：

id  fruit   name     value
1   orange  Colour   Orange
1   orange  Weight   16
1   orange  Calories 25
23  Banana  Colour   Yellow
23  Banana  Weight   30
23  Banana  Calories 250

id  fruit   Colour  Weight
1   orange  Orange  16
23  Banana  Yellow  30

如果希望输出有关颜色和重量的信息，请尝试以下操作：

DECLARE @json nvarchar(max)
SET @json = N'[ { "id": 1, "fruit": "orange", "values": [ { "name": "Colour", "value": "Orange" }, { "name": "Weight", "value": "16" }, { "name": "Calories", "value": "25" } ] }, { "id": 23, "fruit": "Banana", "values": [ { "name": "Colour", "value": "Yellow" }, { "name": "Weight", "value": "30" }, { "name": "Calories", "value": "250" } ] } ]'

SELECT i.id, i.fruit, v1.[value] AS Colour, v2.[value] AS Weight
FROM OPENJSON(@json) 
WITH (
   id int '$.id',
   fruit nvarchar(50) '$.fruit',
   [values] nvarchar(max) '$.values' AS JSON
) AS i
OUTER APPLY (
   SELECT *
   FROM OPENJSON(i.[values])
   WITH (
      [name] nvarchar(max) '$.name',
      [value] nvarchar(max) '$.value'
   )
   WHERE [name] = 'Colour'
) v1 
OUTER APPLY (
   SELECT *
   FROM OPENJSON(i.[values])
   WITH (
      [name] nvarchar(max) '$.name',
      [value] nvarchar(max) '$.value'
   )
   WHERE [name] = 'Weight'
) v2

输出：

id  fruit   name     value
1   orange  Colour   Orange
1   orange  Weight   16
1   orange  Calories 25
23  Banana  Colour   Yellow
23  Banana  Weight   30
23  Banana  Calories 250

id  fruit   Colour  Weight
1   orange  Orange  16
23  Banana  Yellow  30

---

非常感谢Zhorov，他正在阅读数据，但不是在我要找的结构中。有趣的方法-我会看看我是否能得到我想要的，直到a能提取出所需的结构。@Fetchezlavache更新了答案。非常感谢。我仍然不相信这是最好的方法，但它是有效的…@Fetchezlavache我已将

交叉应用

更改为

外部应用

——如果你的JSON中没有

颜色

或

权重

。谢谢，这很有道理，谢谢。仍然在寻找替代方案，因为事实上，数据与水果无关，我有20多个名称/值对..：DMany谢谢Zhorov，这是在读取数据，但不是在我要找的结构中。有趣的方法-我会看看我是否能得到我想要的，直到a能提取出所需的结构。@Fetchezlavache更新了答案。非常感谢。我仍然不相信这是最好的方法，但它是有效的…@Fetchezlavache我已将

交叉应用

更改为

外部应用

——如果你的JSON中没有

颜色

或

权重

。谢谢，这很有道理，谢谢。仍然在寻找替代方案，因为事实上，数据与水果无关，我有20多个名称/值对..：D