dispatch 不是函数，异步

Question

我在 react-native 应用程序中有以下操作。我收到错误消息“dispatch 不是函数。我是 redux 新手，并且在处理异步操作时遇到了困难。有没有办法在不使用 async/await 的情况下将 Asyncstorage 设置为状态？

export const balanceCall =async (dispatch) => {

let token = await getAsyncStoreT();
let mobile= await getAsyncStoreM();


balanceURL = `https://localhost/v1/user/find?token=${token}&mobile=${mobile}`;

console.log(balanceURL);
axios({
    method: 'get',
    url: checkURL,
}).then(function (response) {
    let balance =response.data.map(user=>user.balance);
    dispatch(balanceReceived(balance));
})
    .catch(function (error) {
        console.log(error);
        console.log("NOT AGAIN");
    });
};

Answer 1

balanceCall 应该返回一个以 dispatch 作为参数的函数。相应地更改代码。

export const balanceCall = () => async dispatch => {

let token = await getAsyncStoreT();
let mobile= await getAsyncStoreM();


balanceURL = `https://localhost/v1/user/find?token=${token}&mobile=${mobile}`;

console.log(balanceURL);
axios({
    method: 'get',
    url: checkURL,
}).then(function (response) {
    let balance =response.data.map(user=>user.balance);
    dispatch(balanceReceived(balance));
})
    .catch(function (error) {
        console.log(error);
        console.log("NOT AGAIN");
    });
};