选择嵌套 JSON 数组包含特定值的行答案

【问题标题】：Select row where nested JSON array contains certain value选择嵌套 JSON 数组包含特定值的行
【发布时间】：2020-06-18 09:11:19
【问题描述】：

我在Employee 表中有几个这样的数据库行：

EmpId       Name        Information
1           Eric       {“Experience”: [{“Title”: “Assistant Accountant”, “Company”: “ComA”, “YearsOfExperience”: 3}, {“Title”: “Accountant”, “Company”: “ComB”, “YearsOfExperience”: 2}], “EmployedYear”:2016}
2           John       {“Experience”: [{“Title”: “Tech Engineer”, “Company”: “ComX”, “Years”: 5}, {“Title”: “Senior Tech Engineer”, “Company”: “ComY”, “YearsOfExperience”: 2}], “EmployedYear”:2012}
3           Leonard    {“Experience”: [{“Title”: “Junior Engineer”, “Company”: “ComJ”, “Years”: 2}, {“Title”: “Tech Engineer”, “Company”: “ComB”, “YearsOfExperience”: “7”}], “EmployedYear”:2017}

我如何选择没有在 ComB 工作过的员工？

对此有什么查询？到目前为止，由于这个复杂的嵌套 JSON 数组，我一无所获。

我正在尝试：

SELECT Name, Id 
FROM Employee
OUTER APPLY OPENJSON(Information, '$.Experience') WITH (
   Title nvarchar(max) '$.Title', 
   Company nvarchar(max) '$.Company', 
   YearsOfExperience int '$.YearsOfExperience'
) AS [Info]
WHERE [Info].Company != 'ComB'

【问题讨论】：

使用OPENJSON。这个 JSON 并不是特别复杂，所以至少我们为什么你所拥有的东西不起作用。
仅供参考，您的 JSON 不使用风格化双引号（“ 和 ”），因为某些解析将无法读取它们。使用非样式化双引号 (")。
Eric和Leonard在ComB有工作经验。你确定这个问题（我如何选择没有在 ComB 工作过的员工，在这种情况下他们是 Eric 和 Leonard？）？
谢谢大家，解决了，我应该用Exist check

标签： sql json sql-server select sql-server-json

【解决方案1】：

您可以尝试使用EXISTS() 和OPENJSON()。

表：

CREATE TABLE Employee (
    EmpId int,       
    Name varchar(100),        
    Information varchar(1000)
)
INSERT INTO Employee (EmpId, Name, Information)
VALUES
    (1, 'Eric',    '{"Experience":[{"Title":"Assistant Accountant","Company":"ComA","YearsOfExperience":3},{"Title":"Accountant","Company":"ComB","YearsOfExperience":2}],"EmployedYear":2016}'),
    (2, 'John',    '{"Experience":[{"Title":"Tech Engineer","Company":"ComX","Years":5},{"Title":"Senior Tech Engineer","Company":"ComY","YearsOfExperience":2}],"EmployedYear":2012}'),
    (3, 'Leonard', '{"Experience":[{"Title":"Junior Engineer","Company":"ComJ","Years":2},{"Title":"Tech Engineer","Company":"ComB","YearsOfExperience":"7"}],"EmployedYear":2017}')

声明：

SELECT Name, EmpId 
FROM Employee
WHERE NOT EXISTS (
   SELECT 1
   FROM OPENJSON(Information, '$.Experience') WITH (
     Title nvarchar(max) '$.Title', 
     Company nvarchar(max) '$.Company', 
     YearsOfExperience int '$.YearsOfExperience'
   ) AS [Info]
   WHERE [Info].Company = 'ComB'
)

【讨论】：

【解决方案2】：

一种方法是在HAVING 子句中进行一些条件聚合：

SELECT V.EmpId,
       V.Name
FROM (VALUES(1,'Eric',N'{"Experience": [{"Title": "Assistant Accountant", "Company": "ComA", "YearsOfExperience": 3}, {"Title": "Accountant", "Company": "ComB", "YearsOfExperience": 2}], "EmployedYear":2016}'),
            (2,'John',N'{"Experience": [{"Title": "Tech Engineer", "Company": "ComX", "Years": 5}, {"Title": "Senior Tech Engineer", "Company": "ComY", "YearsOfExperience": 2}], "EmployedYear":2012}'),
            (3,'Leonard',N'{"Experience": [{"Title": "Junior Engineer", "Company": "ComJ", "Years": 2}, {"Title": "Tech Engineer", "Company": "ComB", "YearsOfExperience": "7"}], "EmployedYear":2017}'))V(EmpId,[Name],Information)
     CROSS APPLY OPENJSON(Information,'$.Experience')
     WITH (Title nvarchar(50),
           Company nvarchar(50),
           Years int) OJ
GROUP BY V.EmpId, V.Name
HAVING COUNT(CASE OJ.Company WHEN 'ComB' THEN 1 END) = 0;

【讨论】：

【解决方案3】：

您可以使用openjson() 和横向连接。您不需要从嵌套对象中提取所有属性，因为您只对公司名称感兴趣：

select e.empId, e.Name
from employee e
where exists (
    select 1
    from openjson(information, '$.Experience') with(company nvarchar(max) '$.Company')
    where company = 'ComB'
)

Demo on DB Fiddle：

员工 |姓名 ----: | :------ 1 |埃里克 3 |伦纳德

【讨论】：