Question

我想减少以下代码。这个问题可能非常la脚，对此感到抱歉。我想用for循环中的一个变量替换顿饭1..10，但是我不确定是否可以在节点js中完成。

function hasPortion(meals) {
const portions = ["4", "3", "2", "1", "1/8", "1/4", "1/2"];
if (meals.meal1 != undefined && meals.meal1.activado == "on" && portions.indexOf(meals.meal1.porcion) < 0) { return false; }
if (meals.meal2 != undefined && meals.meal2.activado == "on" && portions.indexOf(meals.meal2.porcion) < 0) { return false; }
if (meals.meal3 != undefined && meals.meal3.activado == "on" && portions.indexOf(meals.meal3.porcion) < 0) { return false; }
if (meals.meal4 != undefined && meals.meal4.activado == "on" && portions.indexOf(meals.meal4.porcion) < 0) { return false; }
if (meals.meal5 != undefined && meals.meal5.activado == "on" && portions.indexOf(meals.meal5.porcion) < 0) { return false; }
if (meals.meal6 != undefined && meals.meal6.activado == "on" && portions.indexOf(meals.meal6.porcion) < 0) { return false; }
if (meals.meal7 != undefined && meals.meal7.activado == "on" && portions.indexOf(meals.meal7.porcion) < 0) { return false; }
if (meals.meal8 != undefined && meals.meal8.activado == "on" && portions.indexOf(meals.meal8.porcion) < 0) { return false; }
if (meals.meal9 != undefined && meals.meal9.activado == "on" && portions.indexOf(meals.meal9.porcion) < 0) { return false; }
if (meals.meal10 != undefined && meals.meal10.activado == "on" && portions.indexOf(meals.meal10.porcion) < 0) { return false; } 
return true;

}

Answer 1

如果膳食仅具有这10种特性，您可能会喜欢

function hasPortion(meals) {
    const portions = ["4", "3", "2", "1", "1/8", "1/4", "1/2"];
    for (const prop in meals) {
        const meal = meals[prop];
        if (meal != undefined && meal.activado == "on" && portions.indexOf(meal.porcion) < 0) {
            return false;
        }
    }
    return true;
}

如果您有更多属性，但只想检查这10个属性，您可能会喜欢，

function hasPortion(meals) {
    const portions = ["4", "3", "2", "1", "1/8", "1/4", "1/2"];
    for (let i = 1; i <= 10; i++) {
        const prop = `meal${i}`;
        const meal = meals[prop];
        if (meal != undefined && meal.activado == "on" && portions.indexOf(meal.porcion) < 0) {
            return false;
        }
    }
    return true;
}

追加：新解决方案

正如我们提到的，您应该使用数组存储所有餐食，也可以使用Set存储部分，因为在Set中搜索比Array快。

const meals = [
    { activado: "on", porcion: "4" },
    { activado: "on", porcion: "1/8" },
    { activado: "on", porcion: "3" },
    { activado: "on", porcion: "1/2" },
];

function hasPortion(meals) {
    const portions = new Set(["4", "3", "2", "1", "1/8", "1/4", "1/2"]);
    return !meals.some((meal) => meal.activado === "on" && !portions.has(meal.porcion));
}

// Good case
console.log(hasPortion(meals));                // Output: true
// Add bad item
meals.push({ activado: "on", porcion: "1/3" })
console.log(hasPortion(meals));                // Output: false

Nodejs Express-有什么方法可以将以下代码简化为更简单的代码？

1 个答案: