我在Postgresql 13的两个表中有JSON数据,我试图使用like操作符连接它们。但是由于字段中的数据是JSON,所以我得到了一个语法错误。
错误:语法错误在"DepartmentId“第6行或附近:.ta::json->‘dv_DeptId’类‘%e.json_data::json->’Department.
以下是我的查询:
CREATE SCHEMA dbo;
CREATE TABLE dbo.Emp
(
EmpId varchar(50),
json_data varchar
);
INSERT INTO dbo.Emp (EmpId,json_data) values ('E1', '{"EmpId":{"value":"E1","display_value":"E1"},"DepartmentId":{"value":"D1","display_value":"D1"}}')
,('E2', '{"EmpId":{"value":"E2","display_value":"E2"},"DepartmentId":{"value":"D2","display_value":"D2"}}');
select * from dbo.Emp;
CREATE TABLE dbo.Dept
(
DeptId varchar(50),
json_data varchar
);
INSERT INTO dbo.Dept (DeptId, json_data) values ('D1', '{"DeptId":"D1","dv_DeptId":"D1","DeptName":"IT","dv_DeptName":"IT"}}')
,('D2', '{"DeptId":"D2","dv_DeptId":"D2","DeptName":"Marketing","dv_DeptName":"Marketing"}}')
,('D3', '{"DeptId":"D3","dv_DeptId":"D3","DeptName":"HR","dv_DeptName":"HR"}}');
SELECT * FROM dbo.Dept;
select
e.json_data::json->'DepartmentId' ->> 'value' as "Emp_DepartmentId"
, d.json_data::json->>'dv_DeptId' as "Dept_DepartmentId"
from dbo.Emp e
LEFT JOIN dbo.Dept as d
ON d.json_data::json->>'dv_DeptId' like '%e.json_data::json->'DepartmentId' ->> 'value'%' -- throws syntax error预期输出
Emp_DepartmentId Dept_DepartmentId
D1 D1
D2 D2发布于 2022-11-18 15:07:32
通过将like连接到'%',您可以构造稍微不同的'%'语句,它将工作。
select
e.json_data::json->'DepartmentId' ->> 'value' as "Emp_DepartmentId"
, d.json_data::json ->> 'dv_DeptId' as "Dept_DepartmentId"
from dbo.Emp e
LEFT JOIN dbo.Dept as d
ON d.json_data::json ->> 'dv_DeptId' like '%' || cast(e.json_data::json->'DepartmentId' ->> 'value' as varchar) || '%'https://stackoverflow.com/questions/74491471
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