我正在使用以下数据结构：
"data": {
"products": [
[
{
"category": "A",
"items": [
{
"name": "Aloe",
"price": 10
},
{
"name": "Apples",
"price": 5
}
]
},
{
"category": "B",
"items": [
{
"name": "Bread",
"price": 5
}
]
}
],
[
{
"category": "C",
"items": [
{
"name": "Candy",
"price": 5
},
{
"name": "Crayon",
"price": 5
}
]
},
{
"category": "D",
"items": [
{
"name": "Dice",
"price": 5
},
{
"name": "Doll",
"price": 10
}
]
}
]
]
}
我想提取它的一部分来展平，结果如下：
[
{
"name": "Aloe",
"price": 10
},
{
"name": "Apples",
"price": 5
},
{
"name": "Bread",
"price": 5
},
{
"name": "Candy",
"price": 5
},
{
"name": "Crayon",
"price": 5
},
{
"name": "Dice",
"price": 5
},
{
"name": "Doll",
"price": 10
}
]
我该如何做到这一点？
我试过这个：
for (var sets in data.products) {
for (var categories in sets) {
for (var items in categories) {
for (var item in items) {
// assemble new array
}
}
}
}
...但在循环子对象时遇到问题。我找到了其他几个similar questions，但它们似乎解决了更简单的数据结构，并且整理了整个对象而不是它的一部分。
任何指针都会受到赞赏。
答案 0 :(得分：2)
def sumofsquares(x1):
result = [] # list to store squares of each number
for i in range(len(x1)): # iterating through the list
result.append(x1[i] ** 2) # squaring each number and appending to the list
return (sum(result)) # returning the sum of the list
product = [1, 2, 3] # declaring the array to find sum of sqaure
print (sumofsquares(product)) # calling the function and displaying the return value
# Output
14
maube第1行的更改取决于您的变量名称
input.data.products. ....
答案 1 :(得分：0)
你必须尝试这样的事情：
data.products以获取单个产品详细信息。items。诀窍是，你必须使用Array.prototype.concat来合并返回的数组，以展平它们。
var data = {
"products": [
[{
"category": "A",
"items": [{
"name": "Aloe",
"price": 10
},
{
"name": "Apples",
"price": 5
}
]
},
{
"category": "B",
"items": [{
"name": "Bread",
"price": 5
}]
}
],
[{
"category": "C",
"items": [{
"name": "Candy",
"price": 5
},
{
"name": "Crayon",
"price": 5
}
]
},
{
"category": "D",
"items": [{
"name": "Dice",
"price": 5
},
{
"name": "Doll",
"price": 10
}
]
}
]
]
}
var result = [].concat.apply([], data.products.map(function(product){
return [].concat.apply([], product.map(function(p){
return p.items
}))
}))
console.log(result)
答案 2 :(得分：0)
我先将数据的JSON对象分配给变量。请参阅以下代码：
var pro={
"products": [
[
{
"category": "A",
"items": [
{
"name": "Aloe",
"price": 10
},
{
"name": "Apples",
"price": 5
}
]
},
{
"category": "B",
"items": [
{
"name": "Bread",
"price": 5
}
]
}
],
[
{
"category": "C",
"items": [
{
"name": "Candy",
"price": 5
},
{
"name": "Crayon",
"price": 5
}
]
},
{
"category": "D",
"items": [
{
"name": "Dice",
"price": 5
},
{
"name": "Doll",
"price": 10
}
]
}
]
]
}
var flatArray=[];
for(var i=0; i<pro.products.length; i++){
for(var j=0; j<pro.products[i].length;j++){
for(var k=0; k<pro.products[i][j].items.length;k++){
flatArray.push(pro.products[i][j].items[k]);
}
}
}
flatArray将是您所需的扁平对象数组。希望这很清楚。
答案 3 :(得分：0)
您可以使用array#reduce，array#forEach和array#concat。
var input = { data: { products: [[{ category: "A", items: [{ name: "Aloe", price: 10 }, { name: "Apples", price: 5 }] }, { category: "B", items: [{ name: "Bread", price: 5 }] }], [{ category: "C", items: [{ name: "Candy", price: 5 }, { name: "Crayon", price: 5 }] }, { category: "D", items: [{ name: "Dice", price: 5 }, { name: "Doll", price: 10 }] }]] } };
const result = input.data.products.reduce((res, arr) => {
arr.forEach( o => res = res.concat(o.items));
return res;
},[]);
console.log(result);.as-console-wrapper { max-height: 100% !important; top: 0; }
答案 4 :(得分：0)
您可以通过检查items属性或返回简化属性的结果来使用递归方法。此方法适用于任何深度，最后具有预期的items属性。
function flat(object) {
return object.items || Object.keys(object).reduce(function (r, k) {
return r.concat(flat(object[k]));
}, []);
}
var object = { data: { products: [[{ category: "A", items: [{ name: "Aloe", price: 10 }, { name: "Apples", price: 5 }] }, { category: "B", items: [{ name: "Bread", price: 5 }] }], [{ category: "C", items: [{ name: "Candy", price: 5 }, { name: "Crayon", price: 5 }] }, { category: "D", items: [{ name: "Dice", price: 5 }, { name: "Doll", price: 10 }] }]] } }
console.log(flat(object));.as-console-wrapper { max-height: 100% !important; top: 0; }