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javascript - 使用虚拟键盘时 useState 的初始值未更新

转载 作者:行者123 更新时间:2023-12-05 03:51:19 24 4
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这里是初学者。试图获得 react-simple-keyboard使用 Gatsby 和 React。

我用某种状态 (firstName: "Johnn") 初始化我的表单。这应该是初始状态。我希望用户能够修改这个名称,并将修改后的版本保存在状态中。

我在这里初始化我的状态:

 const [inputs, setInputs] = useState({
firstName: "Johnn"
})

当我单击该字段并按下虚拟键盘上的一个按钮(比如一个字母)时,它会删除整个字段的内容并将字母放在那里,而不是将字母添加到已经存在的内容中。另外:单击该字段并按退格键(在 react-simple-keyboard 上)不会执行任何操作。这是为什么?

import React, { useRef, useState, useContext, useEffect } from "react"
import styled from "styled-components"
import ReactDOM from "react-dom"
import Keyboard from "react-simple-keyboard"
import "react-simple-keyboard/build/css/index.css"
import Layout from "@components/layout"
import { useForm } from "react-hook-form"
import { Flex, Box } from "rebass/styled-components"
import Input from "@atoms/Input"

import {
GlobalDispatchContext,
GlobalStateContext,
} from "../context/GlobalContextProvider"

function App() {
const dispatch = useContext(GlobalDispatchContext)
const state = useContext(GlobalStateContext)

const [inputs, setInputs] = useState({
firstName: "Johnn",
// firstName: state.customers[state.currentCustomer].firstName,
})
const [layoutName, setLayoutName] = useState("default")
const [inputName, setInputName] = useState("default")

const [isShiftPressed, setShiftPressed] = useState(false)
const [isCaps, setCaps] = useState(false)
const [isKeyboardVisible, setKeyboardVisible] = useState(false)

const { register, handleSubmit, errors } = useForm()
const keyboard = useRef()

const onChangeAll = newInputs => {
/**
* Here we spread the inputs into a new object
* If we modify the same object, react will not trigger a re-render
*/
setInputs({ ...newInputs })
}

const handleShift = () => {
const newLayoutName = layoutName === "default" ? "shift" : "default"
setLayoutName(newLayoutName)
}

const onKeyPress = button => {
if (isShiftPressed === true && !isCaps) {
setShiftPressed(false)
handleShift()
}
if (button === "{lock}") {
setCaps(true)
}
if (button === "{shift}" || button === "{lock}") {
setShiftPressed(true)
handleShift()
}
}

const onChangeInput = event => {
const inputVal = event.target.value

setInputs({
...inputs,
[inputName]: inputVal,
})

keyboard.current.setInput(inputVal)
}

const getInputValue = inputName => {
return inputs[inputName] || ""
}

return (
<Layout>
<Flex flexDirection="column" style={{ height: "100%" }}>
<form onSubmit={handleSubmit(onSubmit)}>
<Input
id="firstName"
name="firstName"
value={getInputValue("firstName")}
onFocus={() => {
setInputName("firstName")
}}
placeholder={"First Name"}
onChange={onChangeInput}
/>
</form>

<Keyboard
keyboardRef={r => (keyboard.current = r)}
inputName={inputName}
layoutName={layoutName}
onChangeAll={onChangeAll}
onKeyPress={onKeyPress}
/>
</Flex>
</Layout>
)
}

export default App

最佳答案

您可能需要使用 useEffect Hook 设置初始键盘值,并在后续更改时删除 keyboard.current.setInput(inputVal)

const {firstName} = input;
useEffect(() => {
keyboard.current.setInput(firstName);
}, [firstName]);

这将确保在键盘实例中设置 firstName 的初始和后续更改。

代码沙箱:https://codesandbox.io/s/distracted-aryabhata-j3whs?file=/src/index.js

关于javascript - 使用虚拟键盘时 useState 的初始值未更新,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/62999497/

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