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Supabase Backend as a Service · 课时

高级 SQL 查询与连接

掌握复杂的 SQL 查询，包括各种连接类型、子查询和窗口函数，以检索复杂数据集。

高级 SQL 查询与连接 是 CoddyKit 上的免费 Supabase Backend as a Service 课时。 这是第 1 节课，共 3 节。 你可以在下方免费阅读本课时的完整内容 — 然后在浏览器中使用内置代码编辑器和全天候 AI 导师进行实践。 这是 Supabase Backend as a Service 学习路径的一部分，你的进度在网页和 CoddyKit 应用中同步。 Supabase Backend as a Service 课程共包含 3 节课。

本课时的部分内容尚未翻译，以英文显示。

Beyond Basic Queries

Welcome to Advanced SQL Queries! So far, you've learned to select, filter, and sort data. But real-world applications often need to combine data from multiple sources or perform complex calculations.

This lesson will equip you with powerful techniques to retrieve sophisticated datasets, making your Supabase applications even smarter.

Joins Recap: Inner Join

Let's quickly refresh our memory on joins. A JOIN combines rows from two or more tables based on a related column between them.

An INNER JOIN returns only the rows where there is a match in both tables. If a row in one table doesn't have a match in the other, it's excluded.

Try running this example to set up our tables and see an INNER JOIN:

CREATE TABLE departments (
  department_id INT PRIMARY KEY,
  department_name VARCHAR(50) NOT NULL
);

CREATE TABLE employees (
  employee_id INT PRIMARY KEY,
  first_name VARCHAR(50) NOT NULL,
  last_name VARCHAR(50) NOT NULL,
  department_id INT REFERENCES departments(department_id)
);

INSERT INTO departments (department_id, department_name) VALUES
(1, 'Sales'),
(2, 'Marketing'),
(3, 'Engineering'),
(4, 'HR');

INSERT INTO employees (employee_id, first_name, last_name, department_id) VALUES
(101, 'Alice', 'Smith', 1),
(102, 'Bob', 'Johnson', 2),
(103, 'Charlie', 'Brown', 1),
(104, 'Diana', 'Prince', 3),
(105, 'Eve', 'Adams', NULL);

SELECT
  e.first_name,
  e.last_name,
  d.department_name
FROM
  employees e
INNER JOIN
  departments d ON e.department_id = d.department_id;

Left Join: All from the Left

What if you want to see all employees, even those not assigned to a department yet? That's where LEFT JOIN (or LEFT OUTER JOIN) comes in handy.

A LEFT JOIN returns all rows from the 'left' table (the first one mentioned) and the matching rows from the 'right' table. If there's no match on the right, NULL values are returned for the right table's columns.

Run this to see employee Eve Adams, who has no department:

SELECT
  e.first_name,
  e.last_name,
  d.department_name
FROM
  employees e
LEFT JOIN
  departments d ON e.department_id = d.department_id;

Right Join: All from the Right

The opposite of a LEFT JOIN is a RIGHT JOIN (or RIGHT OUTER JOIN). It returns all rows from the 'right' table and matching rows from the 'left' table.

If there's no match on the left, NULL values are returned for the left table's columns. This is less common, as you can often rewrite it as a LEFT JOIN by swapping table order.

Let's find all departments, including 'HR' which currently has no employees:

SELECT
  e.first_name,
  e.last_name,
  d.department_name
FROM
  employees e
RIGHT JOIN
  departments d ON e.department_id = d.department_id;

Full Outer Join: Everything!

Want to see everything? A FULL OUTER JOIN (or just FULL JOIN) returns all rows when there is a match in either the left or the right table.

If a row doesn't have a match in the other table, the columns from the non-matching side will have NULL values. It's like combining a LEFT and a RIGHT join.

Observe how both Eve (no department) and HR (no employees) appear:

SELECT
  e.first_name,
  e.last_name,
  d.department_name
FROM
  employees e
FULL OUTER JOIN
  departments d ON e.department_id = d.department_id;

Self Join: Table to Itself

Sometimes, you need to join a table to itself. This is called a SELF JOIN and is useful for finding relationships within the same table, like 'employees and their managers'.

To do this, you use table aliases to treat the same table as two separate entities in your query.

Let's add a manager column and find out who manages whom:

ALTER TABLE employees
ADD COLUMN manager_id INT REFERENCES employees(employee_id);

UPDATE employees SET manager_id = 101 WHERE employee_id = 102;
UPDATE employees SET manager_id = 101 WHERE employee_id = 103;
UPDATE employees SET manager_id = 104 WHERE employee_id = 105;

SELECT
  E.first_name AS employee_name,
  M.first_name AS manager_name
FROM
  employees E
INNER JOIN
  employees M ON E.manager_id = M.employee_id;

Introducing Subqueries

Beyond joins, subqueries (also called inner queries or nested queries) are another powerful tool. A subquery is simply a SQL query nested inside a larger query.

They can be used to:

  • Filter data in a WHERE clause.
  • Define columns in a SELECT clause.
  • Create derived tables in a FROM clause.

Subqueries execute first, and their result is then used by the outer query.

Subqueries in WHERE Clause

A common use for subqueries is in the WHERE clause to filter results dynamically. You can use operators like IN, EXISTS, =, <, > with subqueries.

For example, let's find all employees who work in the 'Sales' department without knowing the department_id beforehand:

SELECT
  first_name, last_name
FROM
  employees
WHERE
  department_id IN (
    SELECT department_id
    FROM departments
    WHERE department_name = 'Sales'
  );

Scalar Subqueries in SELECT

A scalar subquery is a subquery that returns a single value (one row, one column). These are often used in the SELECT clause to add a calculated value to each row of the main query.

Let's find each employee's name and also show the total number of employees in their department. This demonstrates how a subquery can compute a value for each row.

SELECT
  e.first_name,
  e.last_name,
  d.department_name,
  (SELECT COUNT(*)
   FROM employees
   WHERE department_id = e.department_id) AS dept_employee_count
FROM
  employees e
LEFT JOIN
  departments d ON e.department_id = d.department_id;

Advanced Queries Challenge

Consider a scenario where you want to list all departments, and for each department, show the names of employees working there. If a department has no employees, it should still appear in the list with NULL for employee names.

Which SQL JOIN type is most appropriate for this task?

Recap: Your SQL Superpowers

You've gained some serious SQL superpowers today!

  • Joins: Beyond INNER JOIN, you learned about LEFT, RIGHT, and FULL OUTER JOINs to handle different data inclusion needs.
  • Self Join: How to join a table to itself for hierarchical data.
  • Subqueries: Nesting queries to filter data (WHERE clause) or compute scalar values (SELECT clause).

These techniques are fundamental for building powerful and flexible data retrieval logic in your Supabase projects. Keep practicing!

常见问题解答

「高级 SQL 查询与连接」课时是免费的吗？

是的 — 「高级 SQL 查询与连接」的完整文本可在网页上免费阅读。要进行交互式练习（内置代码编辑器和全天候 AI 导师）并解锁 Supabase Backend as a Service 课程的其余内容，请升级到 CoddyKit PRO。 Supabase Backend as a Service 课程共包含 3 节课。

「高级 SQL 查询与连接」这节课中我会学到什么？

掌握复杂的 SQL 查询，包括各种连接类型、子查询和窗口函数，以检索复杂数据集。 你通过在浏览器中直接运行的动手代码来练习 Supabase Backend as a Service，全天候 AI 导师会在你学习这节课的过程中回答你的问题。

学习 Supabase Backend as a Service 需要有经验吗？

无需任何先前经验。CoddyKit 上的 Supabase Backend as a Service 课程适合初学者到高级学习者，你可以从这里开始或从头开始，按照自己的节奏学习。 这是第 1 节课，共 3 节。

「高级 SQL 查询与连接」课时需要多长时间？

大多数 CoddyKit 课程大约需要 5–10 分钟。每节课都很精短且互动，所以你能稳步进步，并在网页和应用中从离开的地方继续。

我能在这节 Supabase Backend as a Service 课中编写并运行代码吗？

能。每节 Supabase Backend as a Service 课都包含内置代码编辑器，你可以在浏览器中直接编写并运行真实代码，并获得即时 AI 反馈 — 无需本地设置。

此课程中的所有课时

  1. 高级 SQL 查询与连接
  2. 用于性能优化的数据库索引
  3. 数据库函数与触发器
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