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我正在尝试根据数组中的标题属性从对象数组中删除特定项目。我一直遇到一个问题，我可以查看数组项，但我无法根据删除函数中输入的参数将项从数组中拼接出来。我只是从else函数中的语句中得到错误消息。

我尝试使用find, forEach, findIndex并匹配这种情况，以测试根据索引或key 'text'. 在论坛推荐中搜索答案之前，我将我尝试过的所有功能都注释掉了。我所有的配方函数都在工作，我的createIngredient函数将一个对象添加到配方数组中。但是removeIngredient我一直试图开始工作的功能并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()

输出

{id: "ef88e013-9510-4b0e-927f-b9a8fc623450", title: "Spaghetti", body: "", createdAt: 1546878594784, updatedAt: 1546878608896, …}
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

因此，我可以查看上面打印的输出并查看ingredients数组中的每个项目，但是尝试根据index数字拼接项目，或者key使用我已经尝试过的功能和我找到的信息对我不起作用到目前为止有关对象、数组和拼接方法的 Stackoverflow。

4

1 回答 1

1

如果我理解正确（在阅读了代码中注释掉的尝试之后），您正在尝试从与id传递给removeIngredient()函数的配方中删除“面包屑”成分。

在这种情况下，也许您可​​以采取稍微不同的方法来todo通过方法从食谱数组中删除成分Array#filter？

您可以使用filter()以下方式todo通过以下过滤器逻辑从数组中“过滤掉”（即删除）“面包屑”成分：

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

您可以考虑通过以下方式修改您的removeIngredient()功能；

  • 向函数参数添加一个附加ingredient参数。这允许您指定要从对应于配方中删除的成分recipeId

  • 并且，介绍filter()所描述的想法：

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，当您为每种成分引入“删除”按钮时，您将调用removeIngredient()如下：

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

removeIngredient( recipeId, ingredientText );

希望这可以帮助！

于 2019-01-07T20:32:44.443 回答