选择嵌套 JSON 数组包含特定值的行
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【中文标题】选择嵌套 JSON 数组包含特定值的行【英文标题】:Select row where nested JSON array contains certain value 【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
【参考方案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】:
【参考方案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,'$.Experience')
WITH (Title nvarchar(50),
Company nvarchar(50),
Years int) OJ
GROUP BY V.EmpId, V.Name
HAVING COUNT(CASE OJ.Company WHEN 'ComB' THEN 1 END) = 0;
【讨论】:
【参考方案3】:您可以使用openjson()
和横向连接。您不需要从嵌套对象中提取所有属性,因为您只对公司名称感兴趣:
select e.empId, e.Name
from employee e
where exists (
select 1
from openjson(information, '$.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】:
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