I want to:
- Find the cheapest and most expensive food and drink.
- Find the id and name of drinks and foods if their price is higher than 10.
My attempt:
let menu = [
{ id: 1, name: "Soda",price: 3.12,size: "4oz",type: "Drink" },
{ id: 2, name: "Beer", price: 6.50, size: "8oz", type: "Drink" },
{ id: 3, name: "Margarita", price: 12.99, size: "12oz", type: "Drink" },
{ id: 4, name: "Pizza", price: 25.10, size: "60oz", type: "Food" },
{ id: 5, name: "Kebab", price: 31.48, size: "42oz", type: "Food" },
{ id: 6, name: "Berger", price: 23.83, size: "99oz", type: "Food" }
]
I would be happy and thankful if anybody could help.
Thanks in Advanced.
Shakiba Moshiri
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2 Answers 2
here is an example :
let menu = [
{ id: 1, name: "Soda",price: 3.12,size: "4oz",type: "Drink" },
{ id: 2, name: "Beer", price: 6.50, size: "8oz", type: "Drink" },
{ id: 3, name: "Margarita", price: 12.99, size: "12oz", type: "Drink" },
{ id: 4, name: "Pizza", price: 25.10, size: "60oz", type: "Food" },
{ id: 5, name: "Kebab", price: 31.48, size: "42oz", type: "Food" },
{ id: 6, name: "Berger", price: 23.83, size: "99oz", type: "Food" }
];
function getCheapest(array) {
return Math.min(...array.map(item => item.price));
}
function getExpensive(array) {
return Math.max(...array.map(item => item.price));
}
function getFoodsAndDrinksMoreThan(array, minVal = 10){
return array.filter(item => item.price > minVal).map(item => ({id: item.id, name: item.name}));
}
console.log(getCheapest(menu));
console.log(getExpensive(menu));
console.log(getFoodsAndDrinksMoreThan(menu));
answered Feb 28, 2021 at 20:14
sohaieb azaiez
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The first one could be achieved with Array.prototype.reduce by comparing the first and second argument of the function you pass into it and returning the smaller one.
For 2. that sounds like a case for Array.prototype.filter to me.
Let me know if you need more guidance.
answered Feb 28, 2021 at 20:13
Ryuno-Ki
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