不能持之以恒地使用简单易感的反应

Question

当将AsyncStorage作为存储引擎传递时，我得到以下错误：

WARN  Possible Unhandled Promise Rejection (id: 0):
Error: [AsyncStorage] Passing null/undefined as value is not supported. If you want to remove value, Use .removeItem method instead.
Passed value: undefined
Passed key: [EasyPeasyStore][0]

这是我的商店声明：

import { action, createStore, persist } from "easy-peasy";
import AsyncStorage from "@react-native-async-storage/async-storage";

const store = createStore(
    persist({
        ...my state goes here
    }, {
        storage: AsyncStorage
    })
);

Answer 1

尝试添加“允许”属性

const store = createStore(
persist({
    ...my state goes here
}, {
    allow: ['state attributes here ex. user'],
    storage: AsyncStorage
})

)；

Answer 2

在调试和搜索GitHub问题上花费了大量时间之后，我找到了这个解决方案。在这里很有魅力。另外，您需要允许持久化哪个状态变量。

import { action, createStore, persist } from "easy-peasy";
import AsyncStorage from "@react-native-async-storage/async-storage";

const store = createStore(
    persist({
        ...my state goes here
    }, {
        storage: {
            getItem: async function (key){
                console.log("GET_ITEM: ", key);
                const value = await AsyncStorage.getItem(key);
                return JSON.parse(value);
            },
            setItem: function (key, value){
                console.log("SET_ITEM: ", key, value);
                AsyncStorage.setItem(key, JSON.stringify(value))
            }
        },
        allow: ['Names of the state variables to be persisted goes here']
    })
);