我如何破坏质查询异步返回变量，以使其中一个不是log dog nog undefined？

Question

Axios 对象没有 winnersData 属性，因此解构它会在原始 ocde 块中返回 undefined。我需要解构来自两个 Axios 请求的数据，并且我稍后对它们的称呼需要不同。我只是不确定如何实现这一点并且已经挣扎了一段时间。我下面的尝试也失败了。

我怎样才能实现这个？

原始代码块：

async function fetchUploads(){
        const headers = {
            "Accept": 'application/json',
            "Authorization": `Bearer ${authToken}`
        };

        const {data} = await axios.get('http://localhost/api/get-user-uploads-data', {headers});
        return data;
    }

async function fetchWinners(){
        const headers = {
            "Accept": 'application/json',
            "Authorization": `Bearer ${authToken}`
        };

        const {winnersData} =  axios.get('http://localhost/api/choose-winners', {headers});
        return winnersData
    }

    const { data }        = useQuery('uploads', fetchUploads)
    const { winnersData } = useQuery('winners', fetchWinners)
    console.log(data); // logs data correctly
    console.log(winnersData); // logs undefined

    return(....);

尝试修复 fetchWinners() 我已经尝试过，但失败了：

async function fetchWinners(){
    const headers = {
        "Accept": 'application/json',
        "Authorization": `Bearer ${authToken}`
    };

    const { winnersData } =  axios.get('http://localhost/api/choose-winners', {headers});
    return { winnersData };
}



const { data }        = useQuery('uploads', fetchUploads)
const { data: { winnersData } } = useQuery('winners', fetchWinners)
console.log(data.winnersData); // still logs undefined

Answer 1

看起来USEQUERY总是返回一个名为数据的属性，因此您只需要重命名它（对于Axios返回而言相同）。而且不要忘记等待Axio的电话。

const { data } = await axios.get('http://localhost/api/choose-winners', {headers});
return data;
...
const { data } = useQuery('uploads', fetchUploads);
const { data: winnersData } = useQuery('winners', fetchWinners);
console.log('winnersData', winnersData);