Jor*_*dan 5 javascript svelte sveltekit
我来自 vue,习惯了可组合函数。我正在尝试找出以 svelte 方式做到这一点的方法
因此,我创建了一个 js 文件并导入存储,然后尝试创建一个可以在多个组件上调用并单独操作的函数
swipe.js 文件
import { writable, derived, get } from 'svelte/store';
function createSwipe() {
const dyFromStart = writable(0)
function moveEvent(eventType, val){
console.log('moveEvent', eventType, val, get(dyFromStart))
dyFromStart.update(n => n + 1);
}
const dxScore = derived(dyFromStart, $dyFromStart => $dyFromStart + 3)
const dyScore = derived(dyFromStart, $dyFromStart => Math.round($dyFromStart + 100));
return {
moveEvent,
dxScore,
dyScore,
};
}
export const swipe = createSwipe();
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然后在 .svelte 组件中导入脚本中的函数并分解为子部分
import { writable, derived, get } from 'svelte/store';
function createSwipe() {
const dyFromStart = writable(0)
function moveEvent(eventType, val){
console.log('moveEvent', eventType, val, get(dyFromStart))
dyFromStart.update(n => n + 1);
}
const dxScore = derived(dyFromStart, $dyFromStart => $dyFromStart + 3)
const dyScore = derived(dyFromStart, $dyFromStart => Math.round($dyFromStart + 100));
return {
moveEvent,
dxScore,
dyScore,
};
}
export const swipe = createSwipe();
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好吧,最终我想变成一个滑动组件,因此得名,但试图了解基本原理。因此,我希望能够为每个组件拥有唯一的存储,并且为此,如果我使用多个此 .svelte 组件,则状态将在所有组件之间共享。
不仅仅是像三个 idk modal.svelte 组件一样,我想对一堆 diff 组件使用滑动,也许 photoViewer.svelte 只是通用滑动功能,并对所有组件使用相同的代码。
或者我只需要保持每个 .svelte 组件中的状态const dyFromStart = writable(0)并将let dyFromStart = 0其传递到返回结果并更新本地 .svelte 变量的纯 js 函数中
将其添加为非存储更纯的 js 东西,我正在尝试,但无法做出反应,因此接受下面关于有效且听起来像的存储方法的答案是正确的方法
export function createSwipe() {
let dyFromStart = 0
function moveEvent(eventType, val){
console.log('moveEvent', eventType, val, dyFromStart, dxScore(), dyScore())
dyFromStart++
}
function dxScore(){ return dyFromStart + 3 }
// const dzScore = derived(dyFromStart, $dyFromStart => $dyFromStart + 3)
const dyScore = () => Math.round(dyFromStart + 100)
return {
moveEvent,
dxScore,
dyScore,
dyFromStart
};
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export function createSwipe() {
let dyFromStart = 0
let dxScore = dyFromStart + 3
let dyScore = Math.round(dyFromStart + 100)
function moveEvent(eventType, val){
console.log('moveEvent', eventType, val, dyFromStart, dxScore, dyScore)
dyFromStart++
dxScore = dyFromStart + 3
dyScore = Math.round(dyFromStart + 100)
}
return {
moveEvent,
dxScore,
dyScore,
dyFromStart
};
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我想这工作得很好,只是不与 $ 发生反应,并且需要调用来更新 diff 本地变量(如果这样做)
就可组合函数类型样式而不是存储类型而言,这对我来说似乎是最优雅的或类似的东西
export function createSwipe() {
let dyFromStart = 0
function moveEvent(eventType, val){
console.log('moveEvent', eventType, val)
dyFromStart++
}
$: dxScore = dyFromStart + 3
$: dyScore = Math.round($dyFromStart + 100)
return {
moveEvent,
dxScore,
dyScore,
};
}
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我不完全理解这个问题,所以我尝试首先重申我认为你想要的内容:
如果这是正确的,那么答案很简单:不要这样做export const swipe = createSwipe()。删除该部分并导出创建函数以直接在组件中使用。这样你每次都会创建一个新的独立实例:
<script>
import { createSwipe } from "$lib/swipe";
let { moveEvent, dxScore, dyScore } = createSwipe()
</script>
<p>{$dxScore}{$dyScore}</p>
<button on:click="() => moveEvent">button</button>
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