从 SQL Server 2016+ 创建 JSON（用于 JSON 路径命令）

Question

我正在尝试以特定 JSON 格式创建一些输出，这是输出示例（将重复多次 - 每所大学一次）：

sample json image

{
    "id": "37e556ae6b9f620d2b7262d3de971c40",
    "recordType": "school",
    "created": 1532531378,
    "updated": 1532531378,
    "published": 1532531378,
    "name": "ABC University – ABC School of Management",
    "schoolName": "ABC School of Management",
    "universityName": "ABC University",
    "sortableName": "ABC School of Management"
}

我一直在想我会为此使用 SQL，因为它有 FOR JSON 命令来处理 JSON 格式并将处理嵌套。当我 hard-code 值时，我可以使用以下代码 re-create 以上内容：

SELECT 
     HASHBYTES('md5','something') AS id
     ,'school' AS recordType
     ,getdate() AS created
     ,getdate() AS updated
     ,getdate() AS published
     ,'ABC University – ABC School of Management' AS name
     ,'ABC School of Management' as schoolname 
     ,'ABC University' as universityName
     , 'ABC School of Management' as sortableName
FOR JSON PATH, WITHOUT_ARRAY_WRAPPER

我遇到的问题是我需要为不同的大学创建此输出 100 次。我想在 SQL 中创建一个数据库 table(s) 并在其上创建运行某些内容。这是 Excel 中的数据示例，我已经过滤了列表以仅显示上述所需的数据以及稍后单独输出中所需的其他几个数据点（我会担心关于其他输出稍后）：

IndiUID SchoolCode  DataValue   Ranking Band    RegionalRank
OVERALL_SCORE   XYZ     105 D   23
SCHOOL_NAME_FORMAL  XYZ XYZ Institute of Management         
SCHOOL_NAME_SORTABLE    XYZ XYZ Institute of Management         
SCHOOL_NAME_AND_BUSINESS_SCHOOL_NAME    XYZ XYZ Institute of Management, XYZ School of Business         
STUDENTS_MEDIAN_AGE XYZ 26          
OVERALL_SCORE   ABC     35  B   3
SCHOOL_NAME_FORMAL  ABC ABC School of Management            
SCHOOL_NAME_SORTABLE    ABC ABC School of Management            
SCHOOL_NAME_AND_BUSINESS_SCHOOL_NAME    ABC ABC University – ABC School of Management           
STUDENTS_MEDIAN_AGE ABC 26

data sample image

我只是不太明白我需要如何构建我的 table(s)，以便我可以使用 SQL 提取相关部分。我想知道我是否应该旋转 table 以便“IndiUID”的值成为数据库中的字段 table 或为每个当前 header 创建一个新的 table ] 项目（数据值、排名、波段等）。

我想我可以 select 来自 table 其中 indiUID = OVERALL_SCORE, SCHOOL_NAME_FORMAL 等，但我想知道是不是太乱了。

这就是我卡住的地方——我如何将这些数据组织成 tables 以便能够提取它？这主要是关于创建 JSON 输出，所以数据库部分是一个特别优雅的解决方案并不重要（它不会 - 我对 SQL 非常基础！）。

Answer 1

您可以尝试以下方法

示例数据:

CREATE TABLE [#school]
([school_name_and_business_school_name] VARCHAR(255), 
 [school_name]                          VARCHAR(255), 
 [university_name]                      VARCHAR(255), 
 [school_sortable]                      VARCHAR(255)
);

INSERT INTO [#school]
VALUES
('ABC University – ABC School of Management', 
 'ABC School of Management', 
 'ABC University', 
 'ABC School of Management'
),
('XYZ University – XYZ School of Management', 
 'XYZ School of Management', 
 'XYZ University', 
 'XYZ School of Management'
),
('QWE University – QWEC School of Management', 
 'QWE School of Management', 
 'QWE University', 
 'QWE School of Management'
);

然后使用这个查询：

SELECT HASHBYTES('md5', 'something') AS [id], 
  'school' AS [recordType], 
  GETDATE() AS [created], 
  GETDATE() AS [updated], 
  GETDATE() AS [published], 
  [school_name_and_business_school_name] AS [name], 
  [school_name] AS [schoolName], 
  [university_name] AS [universityName], 
  [school_sortable] AS [sortableName]
FROM   [#school] 
FOR JSON PATH, WITHOUT_ARRAY_WRAPPER;

给出以下结果：

{"id":"Q3uTDbhLgHnC3YBKcZNrXw==",
"recordType":"school",
"created":"2018-09-07T10:14:44.130",
"updated":"2018-09-07T10:14:44.130",
"published":"2018-09-07T10:14:44.130",
"name":"ABC University – ABC School of Management",
"schoolName":"ABC School of Management",
"universityName":"ABC University",
"sortableName":"ABC School of Management"
},

{"id":"Q3uTDbhLgHnC3YBKcZNrXw==",
"recordType":"school",
"created":"2018-09-07T10:14:44.130",
"updated":"2018-09-07T10:14:44.130",
"published":"2018-09-07T10:14:44.130",
"name":"XYZ University – XYZ School of Management",
"schoolName":"XYZ School of Management",
"universityName":"XYZ University",
"sortableName":"XYZ School of Management"
},

{"id":"Q3uTDbhLgHnC3YBKcZNrXw==",
"recordType":"school",
"created":"2018-09-07T10:14:44.130",
"updated":"2018-09-07T10:14:44.130",
"published":"2018-09-07T10:14:44.130",
"name":"QWE University – QWEC School of Management",
"schoolName":"QWE School of Management",
"universityName":"QWE University",
"sortableName":"QWE School of Management"
}

这有帮助吗？

从 SQL Server 2016+ 创建 JSON（用于 JSON 路径命令）

Create JSON from SQL Server 2016+ (using for JSON path command)

sql

json

data-structures

for-json