嵌套數組 object 與另一個元素數組比較並使用 Javascript 或 ES6 創建新數組

Question

我有一個復雜的數組，如下所示

sectionDetail = [{id: 1, name:'ma'}, {id: 2, name:'na'}, {id: 3, name:'ra'}, {id: 4, name:'ka'}, {id: 5, name:'pa'}];

abc = [{id:'1', name:'zam', sections:['1',4]}, {id:'2', name:'dam', sections:['3']}, {id:'3', name:'nam', sections:['2','4']}];

現在我必須遍歷abc關於節以用它們各自的sectionDetail值替換數組元素

我嘗試將它循環到一個新變量，但我的部分每次都被替換。 下面是我試過的代碼。

const matchingBoost = [];
const getCategoryBasedBoostList = [];
abc.forEach((item, i) => {
    sectionDetail.forEach((val, index) => {
      item.section.forEach((value, x) => {
        if (value == val.Id) {
          matchingBoost.push(val);
        }
      });
    });
    getCategoryBasedBoostList.push({
      Name: item.Name,
      Boost: matchingBoost
    });
  });

所以基本上我正在尋找一個像這樣的新數組

xyz = [{name:'zam',  sections:[{id: 1, name:'ma'}, {id: 4, name:'ka'}]},
{name:'dam',  sections:[{id: 3, name:'ra'}]}, {name:'nam',  sections:[{id: 2, name:'na'}, {id: 4, name:'ka'}]}];

希望我說得通，並希望得到一些回應。

Answer 1

您可以使用Map和 map 獲取帶有sectionDetail項目的數據。

 var sectionDetail = [{ id: 1, name: 'ma' }, { id: 2, name: 'na' }, { id: 3, name: 'ra' }, { id: 4, name: 'ka' }, { id: 5, name: 'pa' }], data = [{ id: '1', name: 'zam', sections: ['1', 4] }, { id: '2', name: 'dam', sections: ['3'] }, { id: '3', name: 'nam', sections: ['2', '4'] }], map = new Map(sectionDetail.map(o => [o.id, o])), result = data.map(({ name, sections }) => ({ name, sections: sections.map(id => map.get(+id)) }) ); console.log(result);

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

Answer 2

您基本上可以根據里面的sectionDetail是否包含在abc的sections中來過濾object.id中的部分。 在這兩種情況下，我都將索引映射到數字，因為一種是字符串，另一種是 integer。

 sectionDetail = [{id: 1, name:'ma'}, {id: 2, name:'na'}, {id: 3, name:'ra'}, {id: 4, name:'ka'}, {id: 5, name:'pa'}]; abc = [{id:'1', name:'zam', sections:['1',4]}, {id:'2', name:'dam', sections:['3']}, {id:'3', name:'nam', sections:['2','4']}]; xyz = abc.map(item => ({...item, sections: sectionDetail.filter(sect => item.sections.map(id => parseInt(id)).includes(parseInt(sect.id)))})); console.log(xyz);

Answer 3

所以你想從 abc 對象中刪除 id 並將 section 數組元素替換為相應的 details 對象？ 這看起來像是 forEach 和 map 的工作。 我即將展示的代碼還對sections 數組進行了一些預處理，以使整體代碼更加高效。

const sections = sectionDetail.reduce((result, section) => {
    result[section.id] = section;
    return result;
}, {});
abc.forEach(item => {
    delete item.id;
    item.sections = item.sections.map(id => sections[id]);
});

Answer 4

試試這樣：

const sectionDetail = [
    { id: 1, name: 'ma' },
    { id: 2, name: 'na' },
    { id: 3, name: 'ra' },
    { id: 4, name: 'ka' },
    { id: 5, name: 'pa' }];

const abc = [
    { id: '1', name: 'zam', sections: ['1', 4] },
    { id: '2', name: 'dam', sections: ['3'] },
    { id: '3', name: 'nam', sections: ['2', '4'] }
];

const desired = abc.map(({id, name, sections}) => {
    return {id, name, sections : sectionDetail.filter(f => {
        return sections.map(s => +s).includes(f.id)
    })};

})

console.log(desired);

其中+s正在轉換為Number類型。