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我正在将一个 json 文件导入 SQL 2016，它在嵌套值结构中具有一些嵌套的名称-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：

Declare @JSON varchar(max)
SELECT @JSON = BulkColumn
FROM OPENROWSET (BULK 'C:\temp\fruit.json', SINGLE_CLOB) as j

If (ISJSON(@JSON)=1) BEGIN
    Select * from openjson ( @JSON )
    WITH(
        id int,
        fruit varchar(20),
        Colour varchar(20) '$.values.Colour',
        Weight int '$.values.Weight'
    ) as Orders
END
ELSE
  Select 'JSON is invalid!'

其结果是：

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

完整（测试）数据..

[ { "id": 1, "fruit": "orange", "values": [ { "name": "Color", "value": "Orange" }, { "name": "Weight", "value ": "16" }, { "name": "卡路里", "value": "25" } ] }, { "id": 23, "fruit": "Banana", "values": [ { "name “：“颜色”，“价值”：“黄色”}，{“名称”：“体重”，“价值”：“30”}，{“名称”：“卡路里”，“价值”：“250”} ] } ]

4

1 回答 1

2

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

如果您想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   Colour  Weight
1   orange  Orange  16
23  Banana  Yellow  30
于 2019-02-14T10:50:07.250 回答