利用Python优化数据库N+1查询问题的方法详解

import sqlite3
​
# 建立数据库连接
conn = sqlite3.connect("example.db")
​
# 查询分类
categories_query = "SELECT * FROM categories;"
categories_cursor = conn.execute(categories_query)
​
for category in categories_cursor.fetchall():
    category_id = category[0]
    category_name = category[1]
​
    # 显示分类名称
    print(f"Category: {category_name}")
    
    # 查询该分类的商品
    items_query = "SELECT id, name FROM items WHERE category_id = ? ORDER BY name;"
    items_cursor = conn.execute(items_query, (category_id,))
    
    for item in items_cursor.fetchall():
        # 显示商品 ID 和名称
        print(f"  Item ID: {item[0]}, Item Name: {item[1]}")
​
conn.close()

import sqlite3
​
# 建立数据库连接
conn = sqlite3.connect("example.db")
​
# 使用 JOIN 查询分类和商品
query = """
    SELECT 
        c.id AS category_id, 
        c.name AS category_name, 
        i.id AS item_id, 
        i.name AS item_name
    FROM categories c
    LEFT JOIN items i ON c.id = i.category_id
    ORDER BY c.name, i.name;
"""
cursor = conn.execute(query)
​
last_category_id = None
​
for row in cursor.fetchall():
    category_id = row[0]
    category_name = row[1]
    item_id = row[2]
    item_name = row[3]
​
    # 如果分类发生变化，渲染分类名称
    if category_id != last_category_id:
        print(f"Category: {category_name}")
​
    # 显示商品信息（如果存在）
    if item_id is not None:
        print(f"  Item ID: {item_id}, Item Name: {item_name}")
​
    last_category_id = category_id
​
conn.close()

query = """
    SELECT 
        c.id AS category_id, 
        c.name AS category_name, 
        COUNT(i.id) AS item_count
    FROM categories c
    LEFT JOIN items i ON c.id = i.category_id
    GROUP BY c.id, c.name
    ORDER BY c.name;
"""

import sqlite3
​
# 建立数据库连接
conn = sqlite3.connect("example.db")
​
# 查询分类及商品数据
query = """
    SELECT 
        c.id AS category_id, 
        c.name AS category_name, 
        i.id AS item_id, 
        i.name AS item_name
    FROM categories c
    LEFT JOIN items i ON c.id = i.category_id
    ORDER BY c.name, i.name;
"""
cursor = conn.execute(query)
​
categories = {}
category_items = []
​
last_category_id = None
last_category_name = None
​
for row in cursor.fetchall():
    category_id = row[0]
    category_name = row[1]
    item_id = row[2]
    item_name = row[3]
​
    # 在分类发生变化时，将当前分类的商品数据存储
    if last_category_id is not None and category_id != last_category_id:
        categories[last_category_name] = category_items
        category_items = []
​
    # 添加商品信息到当前分类
    if item_id is not None:
        category_items.append({"item_id": item_id, "item_name": item_name})
​
    last_category_id = category_id
    last_category_name = category_name
​
categories[last_category_name] = category_items
​
# 渲染数据
for category_name, items in categories.items():
    print(f"Category: {category_name}")
    print(f"{len(items)} items")
​
    for item in items:
        print(f"  Item ID: {item['item_id']}, Item Name: {item['item_name']}")
​
conn.close()