如何过滤字符串数组的嵌套对象

Question

let products = [
{
name: "A",
color: "Blue",
size: {
size1: 1,
size2: 2,
size3: 3,
},
},
{
name: "B",
color: "Blue",
size: {
size1: 5,
size2: 19,
size3: 22,
},
},
{ name: "C", color: "Black", size: 70 },
{ name: "D", color: "Green", size: 50 },
];

filters = ['Blue','2']; 

结果必须是检查数组中所有字符串的对象，例如

{
name: "A",
color: "Blue",
size: {
size1: 1,
size2: 2,
size3: 3,
},
},

无论

中的值是多少，研究必须被接受。

Answer 1

您可以通过某种方式使用堆栈来解析嵌套，通过递归或显式地迭代使用堆栈。这是一个递归的解决方案:

function getFiltered(obj, filters, found = null) {
let outermostCall = (found === null);
if (outermostCall) { //outermost call
found = [];
for (let index = 0; index < filters.length; index++) {
found[index] = false;
}
}
for (let key in obj) {
if (typeof obj[key] === 'object') {
let tempFound = getFiltered(obj[key], filters, found);
for (let index = 0; index < found.length; index++) {
if (tempFound[index]) found[index] = true;
}
} else {
let foundIndex = -1;
for (let index = 0; index < filters.length; index++) {
if (filters[index] == obj[key]) {
foundIndex = index;
index = filters.length;
}
}
if (foundIndex >= 0) {
found[foundIndex] = true;
}
}
}
if (outermostCall) {
return !found.filter(item => !item).length;
}
return found;
}
function getAllFiltered(array, filters) {
let output = [];
for (let obj of array) {
if (getFiltered(obj, filters)) output.push(obj);
}
return output;
}
let products = [
{
name: "A",
color: "Blue",
size: {
size1: 1,
size2: 2,
size3: 3,
},
},
{
name: "B",
color: "Blue",
size: {
size1: 5,
size2: 19,
size3: 22,
},
},
{ name: "C", color: "Black", size: 70 },
{ name: "D", color: "Green", size: 50 },
];
let filters = ['Blue','2']; 
console.log(getAllFiltered(products, filters));

Answer 2

您可以对任何搜索值进行闭包，并检查它们是否都在对象或嵌套对象中以进行过滤。

const
has = f => {
const check = o => o && typeof o === 'object'
? Object.values(o).some(check)
: f === o;
return check;
},
products = [{ name: "A", color: "Blue", size: { size1: 1, size2: 2, size3: 3 } }, { name: "B", color: "Blue", size: { size1: 5, size2: 19, size3: 22 } }, { name: "C", color: "Black", size: 70 }, { name: "D", color: "Green", size: 50 }],
search = ['Blue', 2],
result = products.filter(o => search.every(f => has(f)(o)));
console.log(result);

.as-console-wrapper { max-height: 100% !important; top: 0; }

Answer 3

您可以使用Array.every来检查对象中是否存在所有过滤器，如果这就是您的意思。

const products = [
{ name: "A", color: "Blue", size: { size1:1, size2:2, size3:3 } },
{ name: "B", color: "Blue", size: { size1:5, size2:19, size3:22 } },
{ name: "C", color: "Black", size: 70 },
{ name: "D", color: "Green", size: 50 },
];
const filters = ['Blue','2'];
const filtered = products.filter(product => {
return Object.values(product).every(value => {
return filters.includes(value);
});
});
console.log(filtered);

Answer 4

函数strings返回对象中的所有嵌套字符串，将任何数字转换为字符串。然后，我们只过滤字符串列表中包含所有必要匹配项的乘积。

const products = [{"name":"A","color":"Blue","size":{"size1":1,"size2":2,"size3":3}},{"name":"B","color":"Blue","size":{"size1":5,"size2":19,"size3":22}},{"name":"C","color":"Black","size":70},{"name":"D","color":"Green","size":50}];
const filters = ['Blue','2'];
const subsetMatch=(a,s)=>s.every(i=>a.includes(i));
const strings=i=>typeof i==='object'?Object.values(i).flatMap(strings):i+'';
console.log(products.filter(i=>subsetMatch(strings(i),filters)));