在 SQL Server 2016 中将 JSON 转换为表

Question

我正在开发一个 Web 项目，其中客户端应用程序通过 JSON 与 DB 进行通信。

最初的实现发生在 SQL Server 2012（不支持 JSON，因此我们实现了一个处理解析的存储函数），现在我们正在转向 2016（支持 JSON）。

到目前为止，我们正在将处理时间大大缩短（在某些情况下，速度提高了 200 倍以上！）。

有些交互包含需要转换成表格的数组。为此，OPENJSON 函数几乎满足了我们的需要。

在其中一些（基于数组的）情况下，数组中的记录具有一个或多个字段，这些字段也是 OBJECTS（在这种特殊情况下，也是数组），例如：

    [{
        "Formal_Round_Method": "Floor",
        "Public_Round_Method": "Closest",
        "Formal_Precision": "3",
        "Public_Precision": "3",
        "Formal_Significant_Digits": "3",
        "Public_Significant_Digits": "3",
        "General_Comment": [{
            "Timestamp": "2018-07-16 09:19",
            "From": "1",
            "Type": "Routine_Report",
            "Body": "[To + Media + What]: Comment 1",
            "$$hashKey": "object:1848"
        }, {
            "Timestamp": "2018-07-16 09:19",
            "From": "1",
            "Type": "User_Comment",
            "Body": "[]: Comment 2",
            "$$hashKey": "object:1857"
        }, {
            "Timestamp": "2018-07-16 09:19",
            "From": "1",
            "Type": "Routine_Report",
            "Body": "[To + Media + What]: Comment 3",
            "$$hashKey": "object:1862"
        }]
    }, {
        "Formal_Round_Method": "Floor",
        "Public_Round_Method": "Closest",
        "Formal_Precision": "3",
        "Public_Precision": "3",
        "Formal_Significant_Digits": "3",
        "Public_Significant_Digits": "3",
        "General_Comment": []

    }]

这里General_Comment也是一个数组。

运行命令时：

SELECT *
  FROM OPENJSON(@_l_Table_Data)
  WITH (    Formal_Round_Method                 NVARCHAR(16)    '$.Formal_Round_Method'               ,
            Public_Round_Method                 NVARCHAR(16)    '$.Public_Round_Method'               ,
            Formal_Precision                    INT             '$.Formal_Precision'                  ,
            Public_Precision                    INT             '$.Public_Precision'                  ,
            Formal_Significant_Digits           INT             '$.Formal_Significant_Digits'         ,
            Public_Significant_Digits           INT             '$.Public_Significant_Digits'         ,
            General_Comment                     NVARCHAR(4000)  '$.General_Comment'                   
        ) ;

[@_l_Table_Data 是一个保存 JSON 字符串的变量]

我们得到了 General_Comment = NULL 列，即使里面有数据（至少在数组的第一个元素中）。

我想我应该对那些可能包含 OBJECTS 而不是 SIMPLE VALUES 的列使用不同的语法，但我不知道该语法应该是什么。

Answer 1

我找到了一个真正解决问题的微软页面。

查询应该是这样的：

SELECT *
  FROM OPENJSON(@_l_Table_Data)
  WITH (    Formal_Round_Method        NVARCHAR(16)    '$.Formal_Round_Method'               ,
            Public_Round_Method        NVARCHAR(16)    '$.Public_Round_Method'               ,
            Formal_Precision           INT             '$.Formal_Precision'                  ,
            Public_Precision           INT             '$.Public_Precision'                  ,
            Formal_Significant_Digits  INT             '$.Formal_Significant_Digits'         ,
            Public_Significant_Digits  INT             '$.Public_Significant_Digits'         ,
            General_Comment            NVARCHAR(MAX)   '$.General_Comment'   AS JSON                
    ) ;

因此，您需要在列定义的末尾添加AS JSON，并且（天知道为什么）类型必须为NVARCHAR(MAX)。

确实很简单！！！

Answer 2

创建函数ParseJson：

Create or Alter FUNCTION [dbo].[ParseJson]  (@JSON NVARCHAR(MAX))
RETURNS @Unwrapped TABLE
  (
  [id] INT IDENTITY, --just used to get a unique reference to each json item
  [level] INT, --the hierarchy level
  [key] NVARCHAR(100), --the key or name of the item
  [Value] NVARCHAR(MAX),--the value, if it is a null, int,binary,numeric or string
  type INT, --0 TO 5, the JSON type, null, numeric, string, binary, array or object
  SQLDatatype sysname, --whatever the datatype can be parsed to
  parent INT, --the ID of the parent
  [path] NVARCHAR(4000) --the path as used by OpenJSON
  )
AS begin
INSERT INTO @Unwrapped ([level], [key], Value, type, SQLDatatype, parent,
[path])
VALUES
  (0, --the level
   NULL, --the key,
   @json, --the value,
   CASE WHEN Left(ltrim(@json),1)='[' THEN 4 ELSE 5 END, --the type
   'json', --SQLDataType,
   0 , --no parent
   '$' --base path
  );
DECLARE @ii INT = 0,--the level
@Rowcount INT = -1; --the number of rows from the previous iteration
WHILE @Rowcount <> 0 --while we are still finding levels
  BEGIN
    INSERT INTO @Unwrapped ([level], [key], Value, type, SQLDatatype, parent,
    [path])
      SELECT [level] + 1 AS [level], new.[Key] AS [key],
        new.[Value] AS [value], new.[Type] AS [type],
-- SQL Prompt formatting off
/* in order to determine the datatype of a json value, the best approach is to a determine
the datatype that can be parsed. It JSON, an array of objects can contain attributes that arent
consistent either in their name or value. */
       CASE
        WHEN new.Type = 0 THEN 'bit null'
        WHEN new.[type] IN (1,2)  then COALESCE(
            CASE WHEN TRY_CONVERT(INT,new.[value]) IS NOT NULL THEN 'int' END,
            CASE WHEN TRY_CONVERT(NUMERIC(14,4),new.[value]) IS NOT NULL THEN 'numeric' END,
            CASE WHEN TRY_CONVERT(FLOAT,new.[value]) IS NOT NULL THEN 'float' END,
          CASE WHEN TRY_CONVERT(MONEY,new.[value]) IS NOT NULL THEN 'money' END,
            CASE WHEN TRY_CONVERT(DateTime,new.[value],126) IS NOT NULL THEN 'Datetime2' END,
          CASE WHEN TRY_CONVERT(Datetime,new.[value],127) IS NOT NULL THEN 'Datetime2' END,
          'nvarchar')
       WHEN new.Type = 3 THEN 'bit'
       WHEN new.Type = 5 THEN 'object' ELSE 'array' END AS SQLDatatype,
        old.[id],
        old.[path] + CASE WHEN old.type = 5 THEN '.' + new.[Key]
                       ELSE '[' + new.[Key] COLLATE DATABASE_DEFAULT + ']' END AS path
-- SQL Prompt formatting on
      FROM @Unwrapped old
        CROSS APPLY OpenJson(old.[Value]) new
          WHERE old.[level] = @ii AND old.type IN (4, 5);
    SELECT @Rowcount = @@RowCount;
    SELECT @ii = @ii + 1;
  END;
  return
END

用途：

select * from ParseJson(jsonString)