Javascript Redux操作返回未定义的
我有以下减速器:Javascript Redux操作返回未定义的,javascript,reactjs,ecmascript-6,redux,Javascript,Reactjs,Ecmascript 6,Redux,我有以下减速器: export const showResult = (state = [], action) => { switch (action.type) { case 'GET_NUMBER': return [ ...state, action.n ] case 'CLEAR_NUMBER': return [ ...state
export const showResult = (state = [], action) => {
switch (action.type) {
case 'GET_NUMBER':
return [
...state,
action.n
]
case 'CLEAR_NUMBER':
return [
...state
].pop()
case 'RESET_RESULT':
return []
default:
return state
}
}
单击“deleteNumber”时满足“CLEAR_NUMBER”情况:
import React, { Component } from 'react'
import Display from './Container'
import Operator from './Container'
import { selectOperator } from './../util'
class Calculator extends Component {
constructor(props) {
super(props)
this.state = {
val1: '',
val2: ''
}
}
displayNumber = e => {
const { getNumber, resetResult, getOperator } = this.props
getNumber(parseInt(e.target.dataset.n))
}
getOperator = e => {
const { getOperator, n, resetResult, resetOperator } = this.props
resetOperator()
getOperator(e.target.dataset.sign)
if(this.state.val1 == "") {
this.setState({
val1: n
})
resetResult()
}
}
getSum = val => {
val.reduce((acc, val) => acc + val)
}
deleteNumber = () => {
const { clearNumber } = this.props
clearNumber()
}
getTotal = () => {
const { n, operator, resetResult } = this.props,
{ val1, val2 } = this.state,
value1 = this.getSum(val1),
value2 = val2 != "" ? this.getSum(val2) : null;
let operatorVal = operator[0]
this.setState({ val2: n })
resetResult()
selectOperator(operatorVal, value1, value2)
}
render() {
const { n, operator, getNumber } = this.props
return (
<div>
<Display val={n == "" ? null : n} />
<Operator val={operator} />
{/* NUMBERS */}
<p data-n="1" onClick={this.displayNumber}>1</p>
<p data-n="2" onClick={this.displayNumber}>2</p>
<p data-n="3" onClick={this.displayNumber}>3</p>
{/* OPERATORS */}
<p data-sign="+" onClick={this.getOperator}>+</p>
<p data-sign="-" onClick={this.getOperator}>-</p>
<p data-sign="/" onClick={this.getOperator}>/</p>
<p data-sign="*" onClick={this.getOperator}>*</p>
<p onClick={this.getTotal}>equal</p>
<p onClick={this.deleteNumber}>clear</p>
</div>
)
}
}
export default Calculator
import React,{Component}来自“React”
从“./Container”导入显示
从“./Container”导入运算符
从“/../util”导入{selectOperator}
类计算器扩展组件{
建造师(道具){
超级(道具)
此.state={
值1:“”,
值2:“”
}
}
displayNumber=e=>{
const{getNumber,resetResult,getOperator}=this.props
getNumber(parseInt(e.target.dataset.n))
}
getOperator=e=>{
const{getOperator,n,resetResult,resetOperator}=this.props
重置运算符()
getOperator(e.target.dataset.sign)
如果(this.state.val1==“”){
这是我的国家({
瓦尔1:n
})
resetResult()
}
}
getSum=val=>{
val.reduce((acc,val)=>acc+val)
}
deleteNumber=()=>{
const{clearNumber}=this.props
clearNumber()
}
getTotal=()=>{
常量{n,运算符,resetResult}=this.props,
{val1,val2}=this.state,
value1=此.getSum(val1),
value2=val2!=“this.getSum(val2):null;
let operatorVal=运算符[0]
this.setState({val2:n})
resetResult()
选择Operator(operatorVal、value1、value2)
}
render(){
const{n,operator,getNumber}=this.props
返回(
{/*数字*/}
1
2
3
{/*运算符*/}
+
-
/
*
相等
清除
)
}
}
导出默认计算器
它实际上是从数组中删除最后一个元素(就像在计算器上有一个clear函数一样)
我得到的错误是当数组中还有一个项目时:
这很好,但是如果数组是空的呢
pop
将返回undefined
从而返回错误
case 'CLEAR_NUMBER':
return [
...state
].pop()
编辑:
case 'CLEAR_NUMBER': {
if(state.length > 0) {
return [
...state
].pop()
}
return state;
}
pop
返回删除的元素,而不是修改的数组
你需要做什么
const newState = [...state];
newState.pop();
return newState;
或者,使用
返回状态.slice(0,-1)
您应该这样使用以避免此类错误:
case 'CLEAR_NUMBER':
if ([...state].length > 1) {
return [...state].pop()
}
return [] // or return null, but not undefined ie. no explicit return
不完全是这样,我得到的错误,即使在有一个数字在左边的array@Alex你能在编辑中做一个简单的检查吗?顺便说一句,pop返回数组中的最后一个元素,不确定您是否知道。@Alex这对state有效。长度>=2?它会将您的状态从数组更改为元素。。。