沒有思路 試了用幾種方法，都沒成功

請把你用過的幾種方法貼出來，讓大家知道你曾經(jīng)努力過

以下是我的思路

const arr1 = [
    {active:true, id:1,name: "業(yè)務(wù)類型"},
    {active:true, id:2,name: "報(bào)表類型"},
    {active:true, id:3,name: "資產(chǎn)類型"},
];
const arr2 = [
    {active:false, id:1,name: "業(yè)務(wù)類型"},
    {active:false, id:2,name: "報(bào)表類型"},
    {active:false, id:3,name: "資產(chǎn)類型"},
    {active:false, id:4,name: "商業(yè)類型"},
    {active:false, id:5,name: "報(bào)銷類型"}
];
const set = new Set();
arr1.forEach(item=>set.add(item.id));
arr2.forEach(item=>{
    if(set.has(item.id)){
        item.active = true;
    }
});

console.table(arr2);

以下是運(yùn)行結(jié)果

// 創(chuàng)建映射
const map = arr1.reduce((p, c) => [p[c.name] = p[c.name] || true, p][1], {})
// 獲取映射結(jié)果
const result = arr2.map(item => [map[item.name] && (item.active = true), item][1]])

話說這兩行代碼我已經(jīng)用來回答三四個(gè)類似的問題了，都是這兩天的，中午還回答了一個(gè)，你這個(gè)問題好像之前就有個(gè)幾乎一樣的啊...

arr1 = [
{active:true, id:1,name:'業(yè)務(wù)類型'},
{active:true, id:2,name:'報(bào)表類型'},
{active:true, id:3,name:'資產(chǎn)類型'}
]

arr2 = [
{active:false, id:1,name: '業(yè)務(wù)類型'},
{active:false, id:2,name: '報(bào)表類型'},
{active:false, id:3,name: '資產(chǎn)類型'},
{active:false, id:4,name: '商業(yè)類型'},
{active:false, id:5,name: '報(bào)銷類型'}
]

let obj = {};
arr1.forEach(item=>{
    obj[item.id] = item;
});

arr2.forEach(item=>{
    if(obj[item.id]){
        item.active = true;
    }
})

const arr1 = [
    {active:true, id:1,name: "業(yè)務(wù)類型"},
    {active:true, id:2,name: "報(bào)表類型"},
    {active:true, id:3,name: "資產(chǎn)類型"},
];
const arr2 = [
    {active:false, id:1,name: "業(yè)務(wù)類型"},
    {active:false, id:2,name: "報(bào)表類型"},
    {active:false, id:3,name: "資產(chǎn)類型"},
    {active:false, id:4,name: "商業(yè)類型"},
    {active:false, id:5,name: "報(bào)銷類型"}
];
arr2.map(v2 => {
    const member = arr1.find(v1 => v1.id === v2.id);
    return member && member.active ? {...v2, active: member.active} : v2
})