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移除嵌套在对象内部的数组项

array-splice javascript-objects arrays javascript

JackJack · 技术社区 · 6 年前

我正在尝试基于数组中的标题属性从对象数组中移除特定项。我一直遇到一个问题,在这个问题中我可以查看数组项,但是我无法根据在remove函数中输入的参数将该项从数组中拼接出来。我只是从我的 else 函数中的语句。

我试过用 find, forEach, findIndex 并匹配该情况,以测试是否根据索引或 key 'text' . 在搜索论坛建议的答案之前,我对所有尝试过的功能进行了评论。我所有的配方功能都在工作,还有我的 createIngredient 函数,它向配方数组添加一个对象。但是 removeIngredient 我一直在努力工作的函数,不是因为上面提到的问题。

let recipes = []

// Read existing recipes from localStorage
const loadRecipes = () => {
    const recipesJSON = localStorage.getItem('recipes')

    try {
        return recipesJSON ? JSON.parse(recipesJSON) : []
    } catch (e) {
        return []
    } 
}

// Expose recipes from module
const getRecipes = () => recipes

const createRecipe = () => {
    const id = uuidv4()
    const timestamp = moment().valueOf()

    recipes.push({
        id: id,
        title: '',
        body: '',
        createdAt: timestamp,
        updatedAt: timestamp,
        ingredient: []
    })
    saveRecipes()

    return id
}

// Save the recipes to localStorage
const saveRecipes = () => {
    localStorage.setItem('recipes', JSON.stringify(recipes))
}

// Remove a recipe from the list
const removeRecipe = (id) => {
    const recipeIndex = recipes.findIndex((recipe) => recipe.id === id)

    if (recipeIndex > -1) {
        recipes.splice(recipeIndex, 1)
        saveRecipes()
    }
}

// Remove all recipes from the recipe array
const cleanSlate = () => {
    recipes = []
    saveRecipes()
}

const updateRecipe = (id, updates) => {
    const recipe = recipes.find((recipe) => recipe.id === id)

    if (!recipe) {
        return
    }

    if (typeof updates.title === 'string') {
        recipe.title = updates.title
        recipe.updatedAt = moment().valueOf()
    }

    if (typeof updates.body === 'string') {
        recipe.body = updates.body
        recipe.updateAt = moment().valueOf()
    }

    saveRecipes()
    return recipe
}

const createIngredient = (id, text) => {
    const recipe = recipes.find((recipe) => recipe.id === id)

    const newItem = {
        text,
        have: false
    }
    recipe.ingredient.push(newItem)
    saveRecipes()
}

const removeIngredient = (id) => {
    const ingredient = recipes.find((recipe) => recipe.id === id)
    console.log(ingredient)
    const allIngredients = ingredient.todo.forEach((ingredient) => console.log(ingredient.text))

    // const recipeIndex = recipes.find((recipe) => recipe.id === id)

    // for (let text of recipeIndex) {
    //     console.log(recipdeIndex[text])
    // }

// Attempt 3
    // if (indexOfIngredient === 0) {
    //     ingredientIndex.splice(index, 1)
    //     saveRecipes()
    // } else {
    //     console.log('error')
    // }
    // Attempt 2
    // const recipe = recipes.find((recipe) => recipe.id === id)
    // const ingredients = recipe.todo 
    // // let newItem = ingredients.forEach((item) => item)

    // if (ingredients.text === 'breadcrumbs') {
    //     ingredients.splice(ingredients, 1)
    //     saveRecipes()
    // }
    // Attempt 1
    // const ingredientName = ingredients.forEach((ingredient, index, array) => console.log(ingredient, index, array))
    // console.log(ingredientName)

    // const recipeIndex = recipes.findIndex((recipe) => recipe.id === id)

    // if (recipeIndex > -1) {
    //     recipes.splice(recipeIndex, 1)
    //     saveRecipes()
    // }
}

recipes = loadRecipes()

产量

{id: "ef88e013-9510-4b0e-927f-b9a8fc623450", title: "Spaghetti", body: "", createdAt: 1546878594784, updatedAt: 1546878608896,Â â¦}
recipes.js:94 breadcrumbs
recipes.js:94 noodles
recipes.js:94 marinara
recipes.js:94 meat
recipes.js:94 ground beef
recipes.js:94 milk

所以我可以查看上面打印的输出并查看 ingredients 数组,但尝试根据 index 数字或 key 对于我来说,还不能使用我已经尝试过的函数,以及我在StackOverflow上找到的关于对象、数组和拼接方法的信息。

1 回复 | 直到 6 年前

Dacre Denny 6 年前

如果我理解正确(在阅读了代码中注释掉的尝试之后),您将尝试从对应于 id 传递给 removeIngredient() 功能。

在这种情况下,也许你可以采取一种稍微不同的方法从食谱中去除成分。 todo 数组,通过 Array#filter 方法?

你可以使用 filter() 用以下方法“过滤”(即除去)面包屑 托多 通过以下筛选逻辑排列:

// Keep any ingredients that do not match ingredient (ie if ingredient
// equals "breadcrumbs")
todo.filter(todoIngredient => todoIngredient !== ingredient)

你可以考虑修改你的 删除redient()。 功能;

添加附加项 ingredient 函数参数的参数。这允许您指定要从对应的配方中删除的成分 recipeId
并且,介绍 过滤器() 描述的想法:

const removeIngredient = (recipeId, ingredient) => {

    const recipe = recipes.find(recipe => recipe.id === recipeId)

    if(recipe) {

        // Filter recipe.todo by ingredients that do not match  
        // ingredient argument, and reassign the filtered array 
        // back to the recipie object we're working with
        recipe.todo = recipe.todo.filter(todoIngredient => 
                                         (todoIngredient !== ingredient));
    }

}

现在,当您为每种配料引入“删除”按钮时,您将调用 删除redient()。 如下:

var recipeId = /* got id from somewhere */
var ingredientText = /* got ingredient from somewhere */

removeIngredient( recipeId, ingredientText );

希望这有帮助!