[英]Flutter Supabase - Join Tables
我有 2 张桌子
Programs中的 dept_id 是Department表中的外键
我想加入Program表中的department_id和Department表中的id我无法在 Flutter supabase supabase_flutter: ^0.2.12
中做到这一点。 请帮帮我谢谢
编辑:我想从前端执行此操作。 (扑)
CREATE TABLE Departments (
id INTEGER PRIMARY KEY,
name TEXT NOT NULL
);
CREATE TABLE Programs (
id INTEGER PRIMARY KEY,
name TEXT NOT NULL,
dept_id INTEGER,
FOREIGN KEY (dept_id) references Departments(id)
);
INSERT INTO Departments VALUES (1, 'DEPT_A');
INSERT INTO Departments VALUES (2, 'DEPT_B');
INSERT INTO Programs VALUES (1, 'PROG_A', 1);
INSERT INTO Programs VALUES (2, 'PROG_B', 1);
INSERT INTO Programs VALUES (3, 'PROG_C', 2);
INSERT INTO Programs VALUES (4, 'PROG_D', 2);
SELECT Departments.name, Departments.id, Programs.name, Programs.id
FROM Departments
INNER JOIN Programs
ON Departments.id = Programs.dept_id;
如果你的数据库有关系,你也可以查询相关的表。
final res = await supabase
.from('countries')
.select('''
name,
cities (
name
)
''')
.execute();
这直接来自文档: Supabase Dart Docs
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