CodePattern/A 代码模式/13 多计算单元/数据共享/3 数据库/数据库.py

import sqlite3, os.path
from cppy.cp_util import testfilepath,db_filename,extract_file_words


# 数据库表结构
TABLES = {
    'words': '''CREATE TABLE IF NOT EXISTS words (
                    doc_name INTEGER NOT NULL,
                    value TEXT NOT NULL
                )''',    
}


# 创建数据库表
def create_db_schema(connection):
    for table, sql in TABLES.items():
        c = connection.cursor()
        c.execute(sql)
        connection.commit()
        c.close()


def load_file_into_database(path_to_file, connection):        
    words = extract_file_words( path_to_file )

    doc_name = os.path.basename(testfilepath).split('.')[0]     
    c = connection.cursor()        
    for w in words:
        c.execute("INSERT INTO words (doc_name, value) VALUES (?, ?)", (doc_name, w))        
    connection.commit()
    c.close()

#######################################################
# 建数据库，处理数据入库
#######################################################

# 构造数据库文件的完整路径  
current_dir = os.path.dirname(os.path.abspath(__file__))   
db_file_path = os.path.join(current_dir, db_filename)    
if os.path.exists(db_file_path):          
    os.remove(db_file_path)   

if not os.path.isfile(db_file_path):
    with sqlite3.connect(db_file_path) as connection:
        create_db_schema(connection)
        load_file_into_database(testfilepath, connection)


# 查询输出
with sqlite3.connect(db_file_path) as connection:
    c = connection.cursor()
    c.execute("SELECT value, COUNT(*) as C FROM words GROUP BY value ORDER BY C DESC LIMIT 10")
    for row in c.fetchall():
        print(row[0], '-', row[1])


'''
也可以把数据库看做解决共享数据的竞争死锁的办法
不过本例中的计算太快
用数据库共享数据成本太高
'''