從現有數組創建一個新數組並將每個值連接成一個唯一值並檢查是否已在新數組中設置唯一連接值

Question

我正在嘗試創建一個用於上傳帶有變體的產品的表單。

我將 originalArray 中的每個值連接到 newArray 中的唯一值。 將值合並為一個唯一值的示例：red/small/modern

但是，我想檢查 newArray 中是否已經有一個唯一的（連接的）值。 如果為真，則在 newArray 中保留現有價值和價格。

因此，如果值“red/small/modern”已經存在且價格為 888，則不應將價格為 null 的數組項返回到 newArray 中。

 let originalArray = [ [ { value: 'red', id: 99, price: null }, { value: 'blue', id: 100, price: null }, ], [ { value: 'small', id: 101, price: null }, { value: 'medium', id: 102, price: null }, ], [ { value: 'modern', id: 103, price: null }, { value: 'classic', id: 104, price: null }, ], ]; // // existing array item let newArray = [ { value: 'red/small/modern', id: 1, price: 888 }, { value: 'blue/medium/modern', id: 2, price: 100 }, ]; // console.log('example 1:', newArray); // [{…}, {…}] // newArray = originalArray.map((elem) => { return elem.map(({ value }) => value); }).reduce((acc, cur) => { return acc.flatMap((seq) => { return cur.map((part) => `${seq}/${part}`); }); }).map((elem) => { return { value: elem, price: null }; }); // // // price for value "red/small/modern" and "blue/medium/modern" should not be null, as they are already set in the newArray console.log('example 2:', newArray);

希望問題有意義。

Answer 1

您可以filter尚未在newArray中的元素，並將新元素推入其中：

 let originalArray=[[{value:"red",id:99,price:null},{value:"blue",id:100,price:null}],[{value:"small",id:101,price:null},{value:"medium",id:102,price:null}],[{value:"modern",id:103,price:null},{value:"classic",id:104,price:null}]]; let newArray=[{value:"red/small/modern",id:1,price:888},{value:"blue/medium/modern",id:2,price:100}]; const allCombinations = originalArray.map((elem) => { return elem.map(({ value }) => value); }).reduce((acc, cur) => { return acc.flatMap((seq) => { return cur.map((part) => `${seq}/${part}`); }); }).map((elem) => ({ value: elem, price: null })); newArray.push(...allCombinations.filter(({ value }) => { return newArray.every(existing => existing.value;== value); }) ). console:log('example 2,'; newArray);

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Answer 2

只需添加一個測試？：
if (-1 === newArray.findIndex(row=>row.value===value))

 const originalArray = [ [ { value: 'red', id: 99, price: null }, { value: 'blue', id: 100, price: null } ], [ { value: 'small', id: 101, price: null }, { value: 'medium', id: 102, price: null } ], [ { value: 'modern', id: 103, price: null }, { value: 'classic', id: 104, price: null } ] ], newArray = // existing array item [ { value: 'red/small/modern', id: 1, price: 888 }, { value: 'blue/medium/modern', id: 2, price: 100 } ] originalArray.map((elem) => elem.map(({ value }) => value) ).reduce((acc, cur) => acc.flatMap(seq => cur.map(part => `${seq}/${part}`))).forEach (value => { if (-1 === newArray.findIndex(row=>row.value===value)) newArray.push({value, price: null}) }) console.log( newArray )

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否則，如果您更喜歡創建other newArray

 const originalArray = [ [ { value: 'red', id: 99, price: null }, { value: 'blue', id: 100, price: null } ], [ { value: 'small', id: 101, price: null }, { value: 'medium', id: 102, price: null } ], [ { value: 'modern', id: 103, price: null }, { value: 'classic', id: 104, price: null } ] ], newArray = // existing array item [ { value: 'red/small/modern', id: 1, price: 888 }, { value: 'blue/medium/modern', id: 2, price: 100 } ], otherNewArray = originalArray.map(elem => elem.map(({ value }) => value) ).reduce((acc, cur) => acc.flatMap(seq => cur.map(part => `${seq}/${part}`))).map(value => { let price = newArray.find(row=>row.value===value)?.price || null return({value, price}) }) console.log( otherNewArray )

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