通过pickle存储和调用程序文件数据

Question

我正在尝试通过 pickle 将我的数据保存到文件中。 每当我运行我的程序时，数据似乎永远不会加载到“thelist”。

import pickle

thelist = []

thelist.append(input('Input: '))

def save_data():
    with open('thelist.data', 'wb') as filehandle:
        pickle.dump(thelist, filehandle)
        print('Data saved.')

def load_prev_data():
    try:
        with open('thelist.data', 'rb') as filehandle:
            thelist = pickle.load(filehandle)
    except FileNotFoundError:
        print('No data found. Starting new session.')

load_prev_data()
save_data()

print(thelist)

我还没有尝试过 JSON，但我想知道这是否可行。 我知道答案可能正盯着我看，但我希望有人能提供帮助。 如果有人能指出我正确的方向，将不胜感激。 谢谢你。

Answer 1

您创建局部变量thelist ，您应该使用return将其发送到 function 之外

def load_data():
    try:
        with open('thelist.data', 'rb') as filehandle:
            data = pickle.load(filehandle)
    except FileNotFoundError:
        print('No data found. Starting new session.')
        data = []

    return data

和

thelist = load_data()

类似的方式，您可以在保存时将列表作为参数

def save_data(data):
    with open('thelist.data', 'wb') as filehandle:
        pickle.dump(data, filehandle)
        print('Data saved.')

和

save_data(thelist)

---

完整代码：

import pickle

# --- functions ---

def load_data():
    try:
        with open('thelist.data', 'rb') as filehandle:
            data = pickle.load(filehandle)
    except FileNotFoundError:
        print('No data found. Starting new session.')
        data = []

    return data

def save_data(data):
    with open('thelist.data', 'wb') as filehandle:
        pickle.dump(data, filehandle)
        print('Data saved.')

# --- main ---

thelist = load_data()

thelist.append(input('Input: '))

print(thelist)

save_data(thelist)