Question

我试图创建一个键/值表，该表将获取老师的姓名和超过80000的薪水。我无法在此处进行SELECT语句。在这里使用Json。

CREATE TABLE instructortest (
 ID  INT PRIMARY KEY,
 info VARCHAR(max) NOT NULL
);

ALTER TABLE instructortest
ADD CONSTRAINT "valid JSON"
CHECK (ISJSON (info) = 1);

INSERT INTO instructortest 
VALUES (78699,'{"name":"Pingr","Department":"Statistics","salary":"59303.62"}' )

select JSON_VALUE(info, '$.name') from instructortest 
where ('$.salary') = 59303.62

Answer 1

您只需要在where子句中再次使用JSON_VALUE：

select JSON_VALUE(info, '$.name') from instructortest 
where cast(JSON_VALUE(info, '$.salary') as decimal(18,2)) > 80000

JSON创建，插入和选择查询

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