我正试图在Typescript中使用useQuery

Question

在typescript/react中使用useQuery钩子时，我得到一个错误，说Argument of type '(param: Param) => Promise<any>'不能分配给类型为'QueryFunction<IResponsePlanets, QueryKey>'的参数。

但是记住onSuccess函数在这里工作。即使我得到一个错误

是参数

的接口

type Param = {
pageParam?: undefined;
queryKey: [string, { page: number }];
}

这是async函数

const fetchFunction = async (param: Param) => {
const [key, { page }] = param.queryKey;
const response = await fetch(`https://swapi.dev/api/planet/?page=${page}`);
return response.json();
};

是useQuery钩子

const { status, error, data } = useQuery<IResponsePlanets, Error>(["planets", { page: 1 }], fetchFunction, {
staleTime: 5000,
onSuccess: () => {
console.log('working👍')
}
};

Answer 1

问题在你的fetchFunction签名

const fetchFunction = async (param: Param) => {
^^^^^
const [key, { page }] = param.queryKey;
const response = await fetch(`https://swapi.dev/api/planet/?page=${page}`);
return response.json();
};

参数类型应为QueryFunctionContext<Param>