4 回答
TA贡献1900条经验 获得超5个赞
正如这里的其他人建议使用 modernreduce或数组方法......以防万一您需要支持旧浏览器或更“经典”的实现,您可以使用includes带有数组方法的简单循环。forEachforindexOf
const Ingris = [{ val: "onion,", amount: "1", },
{ val: "paprika", amount: "½ tsp", },
{ val: "yogurt", amount: "1/2 Cup", },
{ val: "fine sea salt", amount: "½ tsp ", },
];
var spices = ["paprika", "parsley", "peppermint", "poppy seed", "rosemary"];
var meats = ["steak", "ground beef", "stewing beef", "roast beef", "ribs"];
var dairy = ["milk", "eggs", "cheese", "yogurt"];
var produce = ["peppers", "radishes", "onions", "tomatoes"];
ShoppingList = {
produceOutput: [],
spicesOutput: [],
meatsOutput: [],
dairyOutput: [],
NoCategoryOutput: [],
};
for (var i = 0; i < Ingris.length; i++) {
var ingredient = Ingris[i];
if (spices.indexOf(ingredient.val) >= 0) {
ShoppingList.spicesOutput.push(ingredient);
} else if (meats.indexOf(ingredient.val) >= 0) {
ShoppingList.meatsOutput.push(ingredient);
} else if (dairy.indexOf(ingredient.val) >= 0) {
ShoppingList.dairyOutput.push(ingredient);
} else if (produce.indexOf(ingredient.val) >= 0) {
ShoppingList.produceOutput.push(ingredient);
} else {
ShoppingList.NoCategoryOutput.push(ingredient);
}
}
console.log(ShoppingList)
还要记住,如果在 Ingris 中有一些重复的成分 - 它们不会加总它们的数量。为此,您需要以某种数字格式(例如十进制数)提供或转换每个金额。此外,如果当前成分已经在列表中,并且在这种情况下 - 添加它们的数量,而不是简单的push会有一些额外的逻辑检查。
TA贡献1951条经验 获得超3个赞
您可以使用基本的减速器来进行此类操作。
const categorizedOutput = Ingris.reduce((acc, cur) => {
if (spices.includes(cur.val)) {
acc.spices.push(cur);
} else if (meats.includes(cur.val)) {
acc.meats.push(cur);
} else if (dairy.includes(cur.val)) {
acc.dairy.push(cur);
} else if (produce.includes(cur.val)) {
acc.produce.push(cur);
} else {
acc.other.push(cur);
}
return acc;
}, {
spices: [],
meats: [],
dairy: [],
produce: [],
other: []
})
TA贡献1895条经验 获得超7个赞
这是一种稍微更加参数化的方法。我将不同的食物类别组合到一个对象中cat,并允许成分的部分匹配(单数匹配复数):
const cat={
spices: ["paprika", "parsley", "peppermint", "poppy seed", "rosemary","fine sea salt"],
meats: ["steak", "ground beef", "stewing beef", "roast beef", "ribs"],
dairy: ["milk", "eggs", "cheese", "yogurt"],
produce: ["peppers", "radishes", "onions", "tomatoes"]};
var ingredients=[
{"val":"onion" , "amount":"1"},
{"val":"paprika" , "amount":"½ tsp"},
{"val":"yogurt" , "amount":"1/2 Cup"},
{"val":"fine sea salt", "amount":"½ tsp"}
];
const shop=ingredients.reduce((acc,ing)=>{
Object.entries(cat).some(([ca,pr])=>
pr.find(p=>p.indexOf(ing.val)>-1) &&
(acc[ca]=acc[ca]||[]).push(ing) )
|| (acc.other=acc.other||[]).push(ing);
return acc;
}, {});
console.log(shop);
TA贡献1786条经验 获得超11个赞
var spices = ["paprika", "parsley", "peppermint", "poppy seed", "rosemary"];
var meats = ["steak", "ground beef", "stewing beef", "roast beef", "ribs"];
var dairy = ["milk", "eggs", "cheese", "yogurt"];
var produce = ["peppers", "radishes", "onions", "tomatoes"];
var ingredients=[
{"val":"onion," , "amount":"1"},
{"val":"paprika" , "amount":"½ tsp"},
{"val":"yogurt" , "amount":"1/2 Cup"},
{"val":"fine sea salt", "amount":"½ tsp"}
];
var shoppingList={spices:[], meats:[], dairy:[], produce:[], other:[]};
ingredients.forEach(ingredient=>{
if(spices.includes(ingredient.val)) shoppingList.spices.push(ingredient);
else if(meats.includes(ingredient.val)) shoppingList.meats.push(ingredient);
else if(dairy.includes(ingredient.val)) shoppingList.dairy.push(ingredient);
else if(produce.includes(ingredient.val)) shoppingList.produce.push(ingredient);
else shoppingList.other.push(ingredient);
});
console.log(shoppingList);
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