Javascript 如何将返回的数据作为未定义的数据处理,这些数据需要添加到apollo和reactjs中的useQuery挂钩中
我有下面的代码,但它在第一次呈现和查询时呈现了未定义的cookieData,因此查询无法获取cookie,并且验证失败。使查询在运行之前等待对CookieAPI的调用返回的任何方法Javascript 如何将返回的数据作为未定义的数据处理,这些数据需要添加到apollo和reactjs中的useQuery挂钩中,javascript,reactjs,graphql,next.js,apollo,Javascript,Reactjs,Graphql,Next.js,Apollo,我有下面的代码,但它在第一次呈现和查询时呈现了未定义的cookieData,因此查询无法获取cookie,并且验证失败。使查询在运行之前等待对CookieAPI的调用返回的任何方法 const { data: cookieData, error: cookieError } = useSWR( "/api/cookie", fetcher ); console.log(cookieData); var test = `Bearer ${coo
const { data: cookieData, error: cookieError } = useSWR(
"/api/cookie",
fetcher
);
console.log(cookieData);
var test = `Bearer ${cookieData}`;
const { loading, error, data } = useQuery(FORMS, {
context: {
headers: {
authorization: test,
},
},
});
更新:我最终使用了上面的惰性查询,但我也会尝试跳过,但我现在一直在尝试在这个变种上实现跳过,它说id是未定义的,它在页面上进行了控制台操作,但有几次没有定义
const addFormClicked = async (data) => {
//console.log(data);
const res = await createForm({
variables: {
name: data.name,
user: user.id,
},
skip: !user.id,
});
console.log(res);
Router.push(`/formBuild/${res.data.createForm._id}`);
};
下面是上下文的全部代码
import { useMutation, gql } from "@apollo/client";
import Layout from "../components/Layout";
import { useForm } from "react-hook-form";
import { useRouter } from "next/router";
import { FORMS } from "../components/Layout";
import useSWR from "swr";
import { useState } from "react";
const ADD_FORM = gql`
mutation AddForm($name: String!, $id: ID!) {
createForm(data: { name: $name, user: { connect: $id } }) {
name
_id
}
}
`;
const fetcher = (url) => fetch(url).then((r) => r.json());
export default function AddForm() {
const { data: user } = useSWR("/api/user"); // add
const { data: cookieData, error: cookieError } = useSWR(
"/api/cookie",
fetcher
);
var test = `Bearer ${cookieData}`;
const Router = useRouter();
const [
createForm,
{
data: createFormData,
error: createFormError,
loading: createFormLoading,
},
] = useMutation(ADD_FORM, {
refetchQueries: [{ query: FORMS }],
context: {
headers: {
authorization: test,
},
},
});
const addFormClicked = async (data) => {
//console.log(data);
const res = await createForm({
variables: {
name: data.name,
user: user.id,
},
skip: !user.id,
});
console.log(res);
Router.push(`/formBuild/${res.data.createForm._id}`);
};
const { register, handleSubmit, errors, reset } = useForm();
if (createFormLoading) return <p>Loading</p>;
if (createFormError) return <p>Error: {createFormError.message}</p>;
//console.log(createFormData);
return (
<Layout>
<form onSubmit={handleSubmit(addFormClicked)}>
<h1>Form Name</h1>
<input type="text" name="name" ref={register()} />
<button type="submit">Add Form</button>
</form>
</Layout>
);
}
当查询处于加载状态时,
user
变量将是undefined
。与cookieData
相同。中没有可用的skip
选项,因为它不会在组件渲染时自动执行变异
一个简单的解决方案是仅当user
和cookieData
存在时才呈现表单。通过这种方式,您可以确定在提交表单时用户id和令牌是否可用
//添加'userError'与'user'结合使用,以检查查询是否正在加载
const{data:user,error:userError}=useSWR('/api/user',userFetcher)
常数[
createForm,
{data:createFormData,错误:createFormError,加载:createFormLoading},
]=使用变异(添加形式{
refetchQueries:[{query:FORMS}],
})
const addFormClicked=async(数据)=>{
const res=wait createForm({
背景:{
标题:{
授权:`Bearer${cookieData}`,
},
},
变量:{
name:data.name,
用户:user.id,
},
})
Router.push(`/formBuild/${res.data.createForm.\u id}`)
}
if(userError | | cookieError){
返回出了问题的地方
}
如果(!user | |!cookieData){
返回加载。。。
}
//渲染形式
skip
optionok如果我通过skip true,那么当cookieData返回时我如何重新运行。。。为什么为true?@AndersKitson如果cookieData
是undefined
-skip:,您只想跳过查询!cookieData
。所以我稍微改变了一下问题,我用lazyquery解决了第一个问题,但我也会尝试跳过,但您能帮助我理解为什么跳过不适用于user.id、user.id控制台,但在前几次它有一些未定义的
const addFormClicked = async (data) => {
//console.log(data);
const res = await createForm({
variables: {
name: data.name,
id: user.id, //NOT user:user.id BUT id:user.id
},
skip: !user.id,
});
console.log(res);
Router.push(`/formBuild/${res.data.createForm._id}`);
};