我在Postgresql 13的2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,我得到了一个语法错误。
错误:第6行“DepartmentId”或其附近出现语法错误:... class ='class1 '部门ID'类似于' %e.数据库::'部门ID '类似于'部门ID '类似于'%e.数据库::'部门ID'类似于'部门ID'类似于' % e.数据库::'部门ID '类似于'部门ID '类似于' % e.数据库::'部门ID'类似于'部门ID'类似于' % e.数据库::'部门ID '类似于'部门ID '类似于'%e.数据库::'部门ID'类似于'部门ID'类似于'%e.数据库::'部门ID'类似于'部门ID'类似于' % e.数据库::'部门ID'类似于'部门ID'类似于' % e.数据库::'部门ID'类似于'部门ID'类似于'
SQL FIDDLE for reference
下面是我的疑问:
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
1条答案
按热度按时间bmvo0sr51#
您可以通过将
'%'
连接到like
语句来稍微改变like
语句的构造,这样它就可以工作了。dbfiddle