首頁猿問計算嵌套對象數

計算嵌套對象數

JavaScript

鴻蒙傳說 2023-02-24 16:46:17

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3 回答

幕布斯6054654

TA貢獻1876條經驗獲得超7個贊

使用reduce()

const json = [{"order":1111,"items":[{"colour1":"red","colour2":"yellow"},{"colour1":"red","colour2":"red"},{"colour1":"red","colour2":"red"}]},{"order":2222,"items":[{"colour1":"black","colour2":"blue","colour3":"orange"},{"colour1":"white","colour2":"red","colour3":"green"}]}]

const res = json.reduce((acc, order) => {

let red = order.items.filter(color => color.colour2 === 'red')

return red.length ? [...acc, red.length] : acc

}, [])

console.log(res)

注意：要獲得過濾后的結果而不是計數返回[...acc, red]

反對回復 2023-02-24

瀟湘沐

TA貢獻1816條經驗獲得超6個贊

將有多種方法解決此問題，但如果您只對出現“colour2”：“red”感興趣。您也可以使用類似的東西。

let json = [{"order":1111,"items":[{"colour1":"red","colour2":"yellow"},{"colour1":"red","colour2":"red"},{"colour1":"red","colour2":"red"}]},{"order":2222,"items":[{"colour1":"black","colour2":"blue","colour3":"orange"},{"colour1":"white","colour2":"red","colour3":"green"}]}]

let count=[];

for(let i of json){

count.push(JSON.stringify(i).match(/"colour2":"red"/g).length)

}

console.log(count);

反對回復 2023-02-24

烙印99

TA貢獻1829條經驗獲得超13個贊

也許這里的效率不是最好的，但它是可以理解的代碼：

let count = [];

json.forEach((order, index) => {

count[index] = [];

order.items.forEach((item) => {

if (item.colour2 === "red") {

count[index].push(item.colour2);

}

});

console.log(count);

反對回復 2023-02-24

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計算嵌套對象數

計算嵌套對象數

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