Javascript-比較兩個數組並將缺少的項添加到第一個數組中

Question

可以說我有兩個數組details1和details2。 以下是數組的內容

var details1 =[];
details1[0] = {'ActivityName' : 'Act1',
              'Status' : 'Done'};
details1[1] = {'ActivityName' : 'Act2',
              'Status' : 'InProgress'};
details1[2] = {'ActivityName' : 'Act5',
              'Status' : 'Done'};

var details2 =[];
details2[0] = {'ActivityName' : 'Act2',
              'Status' : 'Done'};
details2[1] = {'ActivityName' : 'Act3',
              'Status' : 'Done'};

我需要比較兩個數組並添加缺少的項，並根據數組details1中的名稱更新狀態。 我的輸出應該是

var details1 =[];
details1[0] = {'ActivityName' : 'Act1',
              'Status' : 'Done'};
details1[1] = {'ActivityName' : 'Act2',
              'Status' : 'Done'};
details1[2] = {'ActivityName' : 'Act3',
              'Status' : 'Done'};
details1[3] = {'ActivityName' : 'Act5',
              'Status' : 'Done'};

實現此目標的最佳方法是什么？

Answer 1

將兩個數組都轉換為對象，合並它們，對鍵進行排序並提取值：

 var details1 =[]; details1[0] = {'ActivityName' : 'Act1','Status' : 'Done1'}; details1[1] = {'ActivityName' : 'Act2','Status' : 'InProgress'}; details1[2] = {'ActivityName' : 'Act5','Status' : 'Done1'}; var details2 =[]; details2[0] = {'ActivityName' : 'Act2','Status' : 'Done2'}; details2[1] = {'ActivityName' : 'Act3','Status' : 'Done2'}; let toObject = (a, key) => a.reduce((o, x) => Object.assign(o, {[x[key]]: x}), {} ); let merge = Object.assign( toObject(details1, 'ActivityName'), toObject(details2, 'ActivityName') ); let result = Object.keys(merge).sort().map(k => merge[k]); console.log(result); 

Answer 2

您可以使用哈希表作為對項目的引用。

 var details1 = [{ ActivityName: 'Act1', Status: 'Done' }, { ActivityName: 'Act2', Status: 'InProgress' }, { ActivityName: 'Act5', Status: 'Done' }], details2 = [{ ActivityName: 'Act2', Status: 'Done' }, { ActivityName: 'Act3', Status: 'Done' }], hash = Object.create(null); details1.forEach(function (a) { hash[a.ActivityName] = a; }); details2.forEach(function (a) { if (hash[a.ActivityName]) { hash[a.ActivityName].Status = a.Status; return; } details1.push(hash[a.ActivityName] = a); }); details1.sort(function (a, b) { return a.ActivityName.localeCompare(b.ActivityName); }); console.log(details1); 

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Answer 3

var details1 = [];
details1[0] = {'ActivityName': 'Act1',
    'Status': 'Done'};
details1[1] = {'ActivityName': 'Act2',
    'Status': 'InProgress'};
details1[2] = {'ActivityName': 'Act5',
    'Status': 'Done'};

var details2 = [];
details2[0] = {'ActivityName': 'Act2',
    'Status': 'Done'};
details2[1] = {'ActivityName': 'Act3',
    'Status': 'Done'};

//Merge function
var __merge = function (arr1, arr2) {
    return arr1.concat(arr2).reduce(function (prev, current, index) {

        if (!(current.ActivityName in prev.keys)) {
            prev.keys[current.ActivityName] = index;
            prev.result.push(current);
        } else {
            prev.result[prev.keys[current.ActivityName]] = current;
        }

        return prev;
    }, {result: [], keys: {}}).result;
}; 

details1 = (__merge(details1, details2));

結果就是：

[{ ActivityName="Act1",  Status="Done"}, { ActivityName="Act2",  Status="Done"}, { ActivityName="Act5",  Status="Done"}, { ActivityName="Act3",  Status="Done"}]

Answer 4

使用Map的 ES6 +解決方案。

要點是要遍歷第一個數組，並將所有項目添加到以ActivityName為鍵，而實際對象為值的地圖中。 然后遍歷第二個數組，並與現有ActivityName鍵上的值合並或添加一個新值。

最后，對結果進行排序。

注意：此解決方案不會更改現有陣列

 const details1 = [{ 'ActivityName': 'Act1', 'Status': 'Done' }, { 'ActivityName': 'Act2', 'Status': 'InProgress' }, { 'ActivityName': 'Act5', 'Status': 'Done' }]; const details2 = [{ 'ActivityName': 'Act2', 'Status': 'Done' }, { 'ActivityName': 'Act3', 'Status': 'Done' }]; const { assign } = Object; const map = new Map(); const addToMap = (detail) => { const { ActivityName: name } = detail; if (map.has(name)) { // if detail already exists, // create a new object by merging the current detail and the new detail detail = assign({}, map.get(name), detail); } map.set(name, detail); }; // add first then second details to a map details1.concat(details2).forEach(addToMap); // sort the keys and map them to values const result = [...map.keys()] .sort() .map(key => map.get(key)); // show the new result (NOTE: old arrays are unchanged) console.log(result);