MySQL多表连接查询高阶技巧和高阶函数示例详解

-- 先通过子查询筛选出需要的订单数据，再与客户表连接
SELECT c.customer_name, o.order_id, o.order_date
FROM customers c
JOIN (SELECT * FROM orders WHERE order_date >= '2024-01-01') o
ON c.customer_id = o.customer_id;

-- 如果只需要订单表中存在的客户信息，使用 INNER JOIN
SELECT c.customer_name, o.order_id, o.order_date
FROM customers c
INNER JOIN orders o
ON c.customer_id = o.customer_id;
-- 如果需要所有客户的信息，即使他们没有订单，使用 LEFT JOIN
SELECT c.customer_name, o.order_id, o.order_date
FROM customers c
LEFT JOIN orders o
ON c.customer_id = o.customer_id;

-- 为 customer_id 字段创建索引
CREATE INDEX idx_customer_id ON orders(customer_id);
-- 查询时利用索引加速连接
SELECT c.customer_name, o.order_id, o.order_date
FROM customers c
JOIN orders o
ON c.customer_id = o.customer_id;

-- 使用 EXISTS 替代 IN
SELECT c.customer_name
FROM customers c
WHERE EXISTS (
    SELECT 1
    FROM orders o
    WHERE o.customer_id = c.customer_id
);
-- 使用 NOT EXISTS 替代 NOT IN
SELECT c.customer_name
FROM customers c
WHERE NOT EXISTS (
    SELECT 1
    FROM orders o
    WHERE o.customer_id = c.customer_id
);

-- 使用 JOIN 替代子查询
SELECT c.customer_name, o.order_id, o.order_date
FROM customers c
JOIN orders o
ON c.customer_id = o.customer_id
WHERE o.order_date >= '2024-01-01';
-- 原子查询版本
SELECT c.customer_name, (SELECT o.order_id FROM orders o WHERE o.customer_id = c.customer_id AND o.order_date >= '2024-01-01')
FROM customers c;

SELECT CHAR_LENGTH(name) AS name_length FROM employees;

SELECT CONCAT(first_name, ' ', last_name) AS full_name FROM employees;

SELECT SUBSTRING(name, 1, 5) AS name_part FROM employees;

SELECT REPLACE(name, 'John', 'Jane') AS updated_name FROM employees;

SELECT TRIM(name) AS trimmed_name FROM employees;

SELECT COUNT(*) AS employee_count FROM employees;

SELECT SUM(salary) AS total_salary FROM employees;

SELECT AVG(salary) AS average_salary FROM employees;

SELECT MAX(salary) AS max_salary, MIN(salary) AS min_salary FROM employees;

SELECT GROUP_CONCAT(name) AS all_names FROM employees;

SELECT name, IF(salary > 50000, 'High', 'Low') AS salary_level FROM employees;

SELECT name,
       CASE
           WHEN salary < 30000 THEN 'Low'
           WHEN salary BETWEEN 30000 AND 50000 THEN 'Medium'
           ELSE 'High'
       END AS salary_level
FROM employees;

SELECT name, COALESCE(department, 'Unknown') AS department_name FROM employees;

SELECT name, NULLIF(salary, 0) AS adjusted_salary FROM employees;

SELECT name, salary, ROW_NUMBER() OVER (ORDER BY salary DESC) AS row_num FROM employees;

SELECT name, salary, RANK() OVER (ORDER BY salary DESC) AS rank FROM employees;
SELECT name, salary, DENSE_RANK() OVER (ORDER BY salary DESC) AS dense_rank FROM employees;

SELECT name, salary, NTILE(4) OVER (ORDER BY salary DESC) AS quartile FROM employees;

SELECT name, salary, LAG(salary, 1) OVER (ORDER BY salary) AS prev_salary FROM employees;
SELECT name, salary, LEAD(salary, 1) OVER (ORDER BY salary) AS next_salary FROM employees;