我尝试重构以下代码以使用.filter()和.any()。我哪里错了?
目标是归还所有不含坚果或蘑菇的比萨饼。在这两种情况下,productsICanEat.length都应该=== 1.
原始代码:
products = [
{ name: "Sonoma", ingredients: ["artichoke", "sundried tomatoes", "mushrooms"], containsNuts: false },
{ name: "Pizza Primavera", ingredients: ["roma", "sundried tomatoes", "goats cheese", "rosemary"], containsNuts: false },
{ name: "South Of The Border", ingredients: ["black beans", "jalapenos", "mushrooms"], containsNuts: false },
{ name: "Blue Moon", ingredients: ["blue cheese", "garlic", "walnuts"], containsNuts: true },
{ name: "Taste Of Athens", ingredients: ["spinach", "kalamata olives", "sesame seeds"], containsNuts: true }
];
var i,j,hasMushrooms, productsICanEat = [];
for (i = 0; i < products.length; i+=1) {
if (products[i].containsNuts === false) {
hasMushrooms = false;
for (j = 0; j < products[i].ingredients.length; j+=1) {
if (products[i].ingredients[j] === "mushrooms") {
hasMushrooms = true;
}
}
if (!hasMushrooms) productsICanEat.push(products[i]);
}
}
重构代码:
var productsICanEat = [];
/* solve using filter() & all() / any() */
productsICanEat = _(products).chain()
.filter(function(product) {
if(product.containsNuts === false)
return product;
})
.any(function(product) {
if(product.ingredients.indexOf('mushrooms') === -1)
return true;
else
return false;
}).value();
答案 0 :(得分:2)
您可以使用filter执行此操作。 filter
迭代器使用的函数应根据它是否应包含在productsICanEat
中返回一个布尔值:
productsICanEat = _(products).chain().filter(function(product) {
return !product.containsNuts && (product.ingredients.indexOf('mushrooms') === -1);
});
答案 1 :(得分:1)
你过度复杂了。
var productsICanEat = _(products).filter(function(product) {
return !product.containsNuts && product.ingredients.indexOf("mushrooms") === -1;
});