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刪除對象項并將其添加到數組中，如果數組中已存在特定項，則更改對象鍵

JavaScript

炎炎設計 2023-03-24 15:39:38

我有這個食譜數組，其中每個對象都是一個特定的食譜，每個食譜都有一個成分數組，每個成分都是由_id name quantity. 如果你們知道為什么會這樣，請在下面添加我的問題，請告訴我。我掙扎了 3 天......任何建議將不勝感激。多謝！(3) [{…}, {…}, {…}]0:ingredients: Array(4)0: {_id: "5f6628d0029e87e02c79ce0a", name: "chia", quantity: 10}1: {_id: "5f6628d0029e87e02c79ce0b", name: "apple", quantity: 15}2: {_id: "5f6628d0029e87e02c79ce0c", name: "honey", quantity: 30}3: {_id: "5f6628d0029e87e02c79ce0d", name: "almond flour", quantity: 35}length: 4__proto__: Array(0)name: "Coconut Chia Pudding"__v: 0_id: "5f6628d0029e87e02c79ce09"__proto__: Object1: {_id: "5f6628d0029e87e02c79ce0e", name: "Peanut Butter Cookies", ingredients: Array(4), __v: 0}2: {_id: "5f6628d0029e87e02c79ce13", name: "Caprese Avocado Bowls", ingredients: Array(3), __v: 0}length: 3__proto__: Array(0)我在 UI 中擁有的是一個包含上述食譜的列表，用戶可以勾選和取消勾選，在勾選食譜后，其成分顯示在列表中。HTML<ion-content> <ion-grid> <ion-row> <ion-col> <ion-list> <ion-item *ngFor="let recipe of loadedRecipes; let lastRecipe = last" [ngClass]="{ 'last-recipe': lastRecipe }" > <ion-checkbox (ionChange)="onCheckRecipe($event)" value="{{recipe.name}}" ></ion-checkbox> <ion-label>{{recipe.name}}</ion-label> <ion-button [routerLink]="['/','recipes','recipe-details', recipe._id]" >></ion-button > </ion-item> </ion-list> </ion-col> </ion-row> <ion-row> <ion-col> <h6 class="ion-padding-start" *ngIf="groceryList.length > 0"> Grocery List </h6> <ion-list *ngIf="groceryList.length > 0"> <ion-item *ngFor="let ingredient of groceryList"> <ion-label>{{ingredient.name}}</ion-label> <ion-note slot="end">{{ingredient.quantity}} g</ion-note> </ion-item> </ion-list> </ion-col> </ion-row> </ion-grid></ion-content>

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2 回答

慕村9548890

TA貢獻1884條經驗獲得超4個贊

問題在于您的 if 語句的流程。在代碼的“onRemove”部分，您說的是“如果成分在列表中，則將其從列表中刪除。如果不在，則減少其數量?！?第二部分沒有任何意義，更重要的是，你永遠也做不到，因為成分應該總是在列表中。

for (let eachIngredient of recipe.ingredients) {

let matched = this.groceryList.find(function(foundIngre) {

return foundIngre.name === eachIngredient.name;

});

if (

matched.name === eachIngredient.name &&

matched._id === eachIngredient._id

) {

let index = this.groceryList.findIndex(

(x) => x._id === matched._id

);

// Problem e ca eachIngredient.quantity se schimba

this.groceryList.splice(index, 1);

} else {

matched.quantity = matched.quantity - eachIngredient.quantity;

}

根據你所說的，你想要做的是：

減去歸因于已刪除配方的數量
如果新數量為零，則從列表中刪除該成分（盡管您也可以保留它并忽略數量為零的成分）

試試這個：

for (let eachIngredient of recipe.ingredients) {

// I am assuming that ids are unique so I am not checking foundIngre.name at all,

// since I assume that ingredients with the same name must also have the same name

// I am also using findIndex first so that you don't need a second find when removing

const matchIndex = this.groceryList.findIndex(

(foundIngre) => foundIngre._id === eachIngredient._id

);

if ( matchIndex ) { // this should always be true

const matched = this.groceryList[matchIndex];

// preserve the entry if there is still some quantity

if ( matched.quantity > eachIngredient.quantity ) {

matched.quantity = matched.quantity - eachIngredient.quantity; // can use -= to shorten

}

// remove from the list only if there is no quantity remaining

else {

this.groceryList.splice(matchIndex, 1);

}

編輯：嘗試更新和刪除數組中的項目是一種不必要的痛苦。修改后的代碼版本將 _groceryList 存儲在鍵控字典中。我最初打算按成分 ID 鍵，但在查看您的演示后，我發現我的假設是錯誤的，即多個食譜中的相同成分將共享相同的 ID。所以我改為按成分名稱鍵入。這樣你就可以寫入 _groceryList[name] 并且它以前是否存在并不重要。

該類有一個公共的 getter groceryList，它將私有的 _groceryList 字典轉換為一個數組。

我還嘗試通過使用一個通用toggleIngredient函數來消除場景分支中不必要的代碼重復，該函數使用布爾值checked來控制它是通過乘以加號還是減號來控制加法或減法。

import { Component } from "@angular/core";

import { Platform } from "@ionic/angular";

import { SplashScreen } from "@ionic-native/splash-screen/ngx";

import { StatusBar } from "@ionic-native/status-bar/ngx";

import { Subscription } from "rxjs";

export interface Ingredient {

_id: string;

name: string;

quantity: number;

}

export interface Recipe {

_id: string;

name: string;

ingredients: Ingredient[];

}

@Component({

selector: "app-root",

templateUrl: "app.component.html"

})

export class AppComponent {

private _recipesSub: Subscription;

constructor(

private platform: Platform,

private splashScreen: SplashScreen,

private statusBar: StatusBar,

) {

this.initializeApp();

}

initializeApp() {

this.platform.ready().then(() => {

this.statusBar.styleDefault();

this.splashScreen.hide();

});

}

private loadedRecipes: Recipe[] = [/*...*/]

// store the groceryList in a dictionary keyed by name

private _groceryList: Record<string, Ingredient> = {};

// getter returns the groceryList in array format, ignoring 0 quantities

get groceryList(): Ingredient[] {

return Object.values(this._groceryList).filter( ing => ing.quantity > 0 );

}

// get the current quantity for an ingredient by name, or 0 if not listed

currentQuantity( name: string ): number {

const ingredient = this._groceryList[name];

return ingredient ? ingredient.quantity : 0;

}

// update the quantity for an ingredient when checked or unchecked

// will add new ingredients, but never removes old ones

toggleIngredient( ingredient: Ingredient, checked: boolean ): void {

// add to or remove from quantity depending on the value of checked

const quantity = this.currentQuantity(ingredient.name) + (checked ? 1 : -1 ) * ingredient.quantity;

// replace the object in the grocery list dictionary

this._groceryList[ingredient.name] = {

...ingredient,

quantity

}

onCheckRecipe(e) { // you'll want to add a type for e here

for (let recipe of this.loadedRecipes) {

// find the matching recipe

if (recipe.name === e.detail.value) {

// loop through the recipe ingredients

for (let eachIngredient of recipe.ingredients) {

this.toggleIngredient(eachIngredient, e.detail.checked)

}

反對回復 2023-03-24

汪汪一只貓

TA貢獻1898條經驗獲得超8個贊

我認為問題似乎出在代碼的這一部分

if (

matched.name === eachIngredient.name &&

matched._id === eachIngredient._id

) {

let index = this.groceryList.findIndex(

(x) => x._id === matched._id

);

// Problem e ca eachIngredient.quantity se schimba

this.groceryList.splice(index, 1);

} else {

matched.quantity = matched.quantity - eachIngredient.quantity;

}

if 語句應該檢查數量而不是再次驗證名稱和 id ，比如
if(matched.quantity <= eachIngredient.quantity){ // 拼接項目并刪除它。} else { // 減少數量 }
找到匹配成分的小建議。先用findIndex()檢索matchedIndex，再用grocerylist[matchedIndex]檢索item，避免再次遍歷grocerylist尋找索引進行拼接。

反對回復 2023-03-24

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刪除對象項并將其添加到數組中，如果數組中已存在特定項，則更改對象鍵

刪除對象項并將其添加到數組中，如果數組中已存在特定項，則更改對象鍵