SQLite：仅返回每组中的前 2 个结果

Posted 2023-02-16

【中文标题】SQLite：仅返回每组中的前 2 个结果

【英文标题】：SQLite: return only top 2 results within each group

【发布时间】：2018-05-16 03:18:02

【问题描述】：

我检查了类似问题的其他解决方案，但 sqlite 不支持 row_number() 和 rank() 函数，或者没有涉及连接多个表、按多列分组并仅返回的示例每个组同时获得前 N 个结果。

这是我运行的代码

db = sqlite3.connect('mydb')

cursor = db.cursor()

cursor.execute(
    '''
    CREATE TABLE orders(
        id INTEGER PRIMARY KEY, product_id INTEGER,
        client_id INTEGER
        )
    '''
)

cursor.execute(
    '''
    CREATE TABLE clients(
        id INTEGER PRIMARY KEY, gender TEXT,
        city TEXT
        )
    '''
)

cursor.execute(
    '''
    CREATE TABLE products(
        id INTEGER PRIMARY KEY, category_name TEXT
        )
    '''
)

orders = [
    (9, 6), (3, 10), (8, 6), (4, 8),
    (5, 6), (7, 4), (9, 2), (10, 8),
    (4, 6), (3, 1), (10, 2), (9, 8),
    (9, 7), (4, 9), (7, 10), (2, 7),
    (4, 7), (6, 2), (6, 2), (9, 3),
    (10, 6), (4, 4), (2, 6), (3, 8),
    (9, 2), (1, 9), (3, 9), (9, 4),
    (5, 5), (7, 1), (8, 7), (7, 8),
    (6, 3), (9, 6), (8, 3), (7, 1),
    (10, 5), (7, 10), (8, 1), (7, 9),
    (4, 4), (3, 8), (5, 2), (5, 8),
    (6, 10), (9, 7), (2, 2), (4, 10),
    (5, 10), (3, 9)
]

clients = [
    ('Male', 'NY'),
    ('Female', 'NY'),
    ('Male', 'London'),
    ('Male', 'London'),
    ('Male', 'NY'),
    ('Female', 'NY'),
    ('Female', 'London'),
    ('Male', 'London'),
    ('Male', 'NY'),
    ('Female', 'London')
]

products = [
    ('Kitchen', ),
    ('Sport', ),
    ('Furniture', ),
    ('Furniture', ),
    ('Furniture', ),
    ('Sport', ),
    ('Sport', ),
    ('Kitchen', ),
    ('Kitchen', ),
    ('Kitchen', )
]

cursor.executemany("INSERT INTO orders(product_id, client_id) VALUES(?,?)", orders)
cursor.executemany("INSERT INTO clients(gender, city) VALUES(?,?)", clients)
cursor.executemany("INSERT INTO products(category_name) VALUES(?)", (products))

db.commit()

cursor.execute(
    '''
    SELECT
        category_name,
        city, gender,
        product_id, COUNT(product_id)
    FROM orders
    LEFT JOIN products ON product_id = products.id
    LEFT JOIN clients ON client_id = clients.id
    GROUP BY product_id, category_name, city, gender
    ORDER BY category_name, city, gender, COUNT(product_id) DESC
    '''
)

print('''category_name, city, gender, product_id, COUNT(product_id)''')

all_rows = cursor.fetchall()
for a, b, c, d, e in all_rows:
    print(a, b, c, d, e)

db.close()

现在的问题是如何在单个查询中获得这样的输出？我不需要用红线交叉的行，因为我只需要前 2 个。

【参考方案1】：

这可以通过使用 WITH 将现有查询嵌入到 CTE 中来实现，然后在 WHERE ... IN 子查询中使用它。子查询从 CTE 中选择与 category_name、城市和性别匹配的 LIMIT 2 产品 ID，按产品计数排序。

WITH order_groups AS (
  SELECT
    category_name,
    city, gender,
    product_id,
    COUNT(product_id) AS product_count
  FROM orders OO
  LEFT JOIN products ON product_id = products.id
  LEFT JOIN clients ON client_id = clients.id
  GROUP BY product_id, category_name, city, gender
  ORDER BY category_name, city, gender, COUNT(product_id) DESC
)
SELECT * FROM order_groups OG_outer
WHERE OG_outer.product_id IN (
  SELECT product_id
  FROM order_groups OG_inner
  WHERE 
    OG_outer.category_name = OG_inner.category_name AND
    OG_outer.city = OG_inner.city AND
    OG_outer.gender = OG_inner.gender
  ORDER BY OG_inner.product_count DESC LIMIT 2
)
ORDER BY category_name, city, gender, product_count DESC

这会根据要求输出以下行：

Furniture|London|Female|4|2
Furniture|London|Female|3|1
Furniture|London|Male|4|3
Furniture|London|Male|3|2
Furniture|NY|Female|5|2
Furniture|NY|Female|4|1
Furniture|NY|Male|3|3
Furniture|NY|Male|4|1
Kitchen|London|Female|9|2
Kitchen|London|Female|8|1
Kitchen|London|Male|9|3
Kitchen|London|Male|8|1
Kitchen|NY|Female|9|4
Kitchen|NY|Female|10|2
Kitchen|NY|Male|1|1
Kitchen|NY|Male|8|1
Sport|London|Female|7|2
Sport|London|Female|2|1
Sport|London|Male|7|2
Sport|London|Male|6|1
Sport|NY|Female|2|2
Sport|NY|Female|6|2
Sport|NY|Male|7|3