Question

当我运行该程序时，它不会更改数据库中的值。是因为选项部分有问题还是存在其他问题？我不知道这是什么，希望有人可以帮助我。

import sqlite3

def product_kopen(crsr):
    print ("Which product would you like to buy?")
    product = str(input(""))
    print ("And how many?")
    wanted_amount = int(input(""))
    crsr.execute("SELECT * FROM emp")
    rows = crsr.fetchall()
    for row in rows:
        if row[0] == product:
            actual_amount = int(row[1])
            fetched_amount = actual_amount - wanted_amount
    kopen = "UPDATE emp SET Amount = {amount} WHERE Product = '{name}' ".format(name=product,
                                                                                amount=fetched_amount)
    crsr.execute(kopen)
    print("You have succesfully bought your product!", "\n")


def product_verkopen(crsr):
    print ("Which product would you like to sell?")
    sold_product = str(input(""))
    print("And how many?")
    wanted_sold_amount = int(input(""))
    crsr.execute("SELECT * FROM emp")
    rows = crsr.fetchall()
    for row in rows:
        if row[0] == sold_product:
            actual_amount = int(row[1])
            fetched_amount = actual_amount + wanted_sold_amount

    kopen = "UPDATE emp SET Amount = {amount} WHERE Product = '{name}' ".format(name=sold_product,
                                                                                amount=fetched_amount)
    crsr.execute(kopen)
    print("You have succesfully sold your product")


connection = sqlite3.connect("Stock.db", timeout=10)
crsr = connection.cursor()
connection.commit()


while True:
    print("Welcome to this shop! choose your option", "\n",
          "1. Buy a product.", "\n", "2. Sell a product")
    option = int(input(""))
    if option == 1:
        product_kopen(crsr)
    elif option == 2:
        product_verkopen(crsr)
    else:
        print("This isn't a valid option", "\n")
        continue

connection.close()

Answer 1

对数据库进行更改后，必须commit。 sqlite3在默认情况下不会commit进行更改。每次您进行cursor.execute(...)时，请紧跟着cursor.commit()

crsr.execute(kopen)
crsr.commit()

有关更多信息，请参见此处https://docs.python.org/3/library/sqlite3.html

连接数据库问题

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