我正在尝试通过使用一些虚拟数据制作一个简单的 Node 应用程序来学习 Graphql。
当我尝试运行 Node 时出现以下错误
TypeError: Cannot set property 'name' of undefined
at GraphQLObjectType (/Users/BorisGrunwald/Desktop/programmering/Javascript/Node/graphQLServer/server/node_modules/graphql/type/definition.js:480:15)
at Object.<anonymous> (/Users/BorisGrunwald/Desktop/programmering/Javascript/Node/graphQLServer/server/schema/schema.js:22:23)
at Module._compile (internal/modules/cjs/loader.js:721:30)
at Object.Module._extensions..js (internal/modules/cjs/loader.js:732:10)
at Module.load (internal/modules/cjs/loader.js:620:32)
at tryModuleLoad (internal/modules/cjs/loader.js:560:12)
at Function.Module._load (internal/modules/cjs/loader.js:552:3)
at Module.require (internal/modules/cjs/loader.js:657:17)
at require (internal/modules/cjs/helpers.js:20:18)
at Object.<anonymous> (/Users/BorisGrunwald/Desktop/programmering/Javascript/Node/graphQLServer/server/app.js:6:20)
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这些是我项目中仅有的两个文件:
应用程序.js
const express = require('express');
const graphqlHTTP = require('express-graphql');
const app = express();
const schema = require('./schema/schema');
app.use('/graphql',graphqlHTTP({
schema
}));
app.listen(4000,() => {
console.log('listening on port 4000');
});
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架构.js
const graphql = require('graphql');
const { …Run Code Online (Sandbox Code Playgroud) 我有以下反应组件
import React from "react"
import styled from "styled-components"
import { moveInLeft } from '../../styles/Animations'
const StyledHeadingOne = styled.h1`
font-size: 6rem;
text-transform: uppercase;
color: #fff;
font-weight: 300;
letter-spacing: 0.5rem;
animation: ${moveInLeft} 1s ease-in-out .3s both;
`
export default function HeadingOne({ children }) {
return <StyledHeadingOne>{children}</StyledHeadingOne>
}
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moveInLeft我添加从文件导入的动画:
import { keyframes} from "styled-components"
export const moveInLeft = keyframes`
@keyframes moveInLeft {
0% {
opacity: 0;
transform: translateX(-10rem);
}
100% {
opacity: 1;
transform: translate(0);
}
}
`
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但由于某种原因,动画不适用。谁能弄清楚为什么吗?
我有一个反应应用程序,在初始加载时会执行大量 fetch 调用来检索一些数据。
我使用 redux 和 redux 工具包来管理状态
正如您所看到的,在我的 redux 切片中,有很多链接的 addCase 调用,它们只是处理设置加载状态以获取loading或failed
有没有办法避免这种情况并使我的代码更简洁?一种制作单个 addCase 的方法,该 addCase 处理将加载状态设置loading为所有获取以及同样的情况failed?
export const dashboardSlice = createSlice({
name: "dashboard",
initialState,
reducers: {},
extraReducers(builder) {
builder
//Co2Emission
.addCase(fetchCO2EmissionData.pending, (state, action) => {
state.co2.status = "loading";
})
.addCase(fetchCO2EmissionData.fulfilled, (state, action) => {
state.co2.status = "succeeded";
//We are just using the CPU data from first pod in the array. When when KG-121 it should just be state.Co2DiagramData = action.payload.values
console.log(action.payload[0]); …Run Code Online (Sandbox Code Playgroud) 我试图理解为什么这段代码不像我期望的那样表现:
async function test() {
await setTimeout(() => {
console.log('done')
}, 1000)
console.log('it finished');
}
test();Run Code Online (Sandbox Code Playgroud)
这首先打印it finished,然后打印done。这段代码不应该在执行之前等待超时完成console.log('it finished');还是我误解了什么?
假设我有一个字符串__f__g_f_
我怎样才能将它分割成一个像这样的数组['__','f','__','g','_','f','_']