Many to many relationships&Assignment

最新推荐文章于 2023-08-11 02:11:35 发布

原创最新推荐文章于 2023-08-11 02:11:35 发布 · 461 阅读

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本文介绍了一种通过创建中间表来实现表与表之间的多对多关联的方法，并提供了具体的Python代码示例，展示了如何使用SQLite数据库进行操作。

表与表之间多对多关联

import json
import sqlite3

conn = sqlite3.connect('rosterdb.sqlite')
cur = conn.cursor()

# Do some setup
cur.executescript('''
DROP TABLE IF EXISTS User;
DROP TABLE IF EXISTS Member;
DROP TABLE IF EXISTS Course;

CREATE TABLE User (
    id     INTEGER NOT NULL PRIMARY KEY AUTOINCREMENT UNIQUE,
    name   TEXT UNIQUE
);

CREATE TABLE Course (
    id     INTEGER NOT NULL PRIMARY KEY AUTOINCREMENT UNIQUE,
    title  TEXT UNIQUE
);

CREATE TABLE Member (
    user_id     INTEGER,
    course_id   INTEGER,
    role        INTEGER,
    PRIMARY KEY (user_id, course_id)
)
''')

fname = raw_input('Enter file name: ')
if ( len(fname) < 1 ) : fname = 'roster_data.json'

# [
#   [ "Charley", "si110", 1 ],
#   [ "Mea", "si110", 0 ],

str_data = open(fname).read()
json_data = json.loads(str_data)

for entry in json_data:

    name = entry[0];
    title = entry[1];
    role=entry[2];
    cur.execute('''INSERT OR IGNORE INTO User (name) 
        VALUES ( ? )''', ( name, ) )
    cur.execute('SELECT id FROM User WHERE name = ? ', (name, ))
    user_id = cur.fetchone()[0]

    cur.execute('''INSERT OR IGNORE INTO Course (title) 
        VALUES ( ? )''', ( title, ) )
    cur.execute('SELECT id FROM Course WHERE title = ? ', (title, ))
    course_id = cur.fetchone()[0]

    cur.execute('''INSERT OR REPLACE INTO Member
        (user_id, course_id,role) VALUES ( ?, ?, ? )''', 
        ( user_id, course_id ,role) )

    conn.commit()

要点其实是需要再建立一张表用来存入course和users的primary key。
最后在SQL中输入

SELECT hex(User.name || Course.title || Member.role ) AS X FROM 
    User JOIN Member JOIN Course 
    ON User.id = Member.user_id AND Member.course_id = Course.id
    ORDER BY X