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根据javascript中的键值合并两个json对象

json

我需要使用javascript根据键值合并两个json对象。

我有两个不同的变量g和c。

术语:所有值都需要合并。

var g = [ { id: 36, name: 'AAA', goal: 'yes' },
    { id: 40, name: 'BBB', goal: 'yes' },
    { id: 57, name: 'CCC', goal: 'yes' },
    { id: 4, name: 'DDD', goal: 'yes' },
    { id: 39, name: 'EEE', goal: 'yes' },
    { id: 37, name: 'FFF', goal: 'yes' },
    { id: 59, name: 'GGG', goal: 'yes' },
    { id: 50, name: 'III', goal: 'yes' },
    { id: 43, name: 'HHH', goal: 'yes' },
    { id: 35, name: 'JJJ', goal: 'yes' } ]


 var c = [ { id: 36, name: 'AAA', circle: 'yes' },
    { id: 40, name: 'BBB', circle: 'yes' },
    { id: 57, name: 'CCC', circle: 'yes' },
    { id: 42, name: 'ZZZ', circle: 'yes' },
    { id: 4, name: 'DDD', circle: 'yes' },
    { id: 39, name: 'EEE', circle: 'yes' },
    { id: 37, name: 'FFF', circle: 'yes' },
    { id: 59, name: 'GGG', circle: 'yes' },
    { id: 43, name: 'HHH', circle: 'yes' },
    { id: 35, name: 'JJJ', circle: 'yes' },
    { id: 100, name: 'JJJ', circle: 'yes' } ]

我尝试了以下代码:但我合并在’c’变量中具有相同ID的内容。但是我需要合并比较“ g”和“ c”。

   var arrayList = [];
    for(var i in g) {
        var getid = g[i].id;
        var getname = g[i].name;
        var getgoal = g[i].goal;
        for(var j in c){
            var compareid = c[j].id;
            if(getid == compareid){
                var obj = {};
                obj.id = getid;
                obj.name = getname;
                obj.goal =  'yes';
                obj.circle = 'yes';
                console.log(obj);
                arrayList.push(obj);
            }
         }

    }
    console.log(arrayList);

预期产量:

[ { id: 36, name: 'AAA', goal: 'yes',circle: 'yes' },
    { id: 40, name: 'BBB', goal: 'yes',circle: 'yes' },
    { id: 57, name: 'CCC', goal: 'yes',circle: 'yes' },
    { id: 4, name: 'DDD', goal: 'yes' ,circle: 'yes' },
    { id: 39, name: 'EEE', goal: 'yes' ,circle: 'yes' },
    { id: 37, name: 'FFF', goal: 'yes' ,circle: 'yes'},
    { id: 59, name: 'GGG', goal: 'yes' ,circle: 'yes'},
    { id: 50, name: 'III', goal: 'yes' ,circle: 'no'},
    { id: 43, name: 'HHH', goal: 'yes' ,circle: 'yes'},
    { id: 35, name: 'JJJ', goal: 'yes' ,circle: 'yes'} ,
    { id: 42, name: 'ZZZ', goal: 'no' , circle: 'yes' },
    { id: 100, name: 'JJJ',goal: 'no' , circle: 'yes' }]

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2020-07-27

共1个答案

小编典典

如果objID中不存在ID,您忘了推入第一个循环;如果ID中不存在该对象的一个​​或多个ID,您忘了c循环遍历。c``g

var g = [
        { id: 36, name: 'AAA', goal: 'yes' },
        { id: 40, name: 'BBB', goal: 'yes' },
        { id: 57, name: 'CCC', goal: 'yes' },
        { id: 4, name: 'DDD', goal: 'yes' },
        { id: 39, name: 'EEE', goal: 'yes' },
        { id: 37, name: 'FFF', goal: 'yes' },
        { id: 59, name: 'GGG', goal: 'yes' },
        { id: 50, name: 'III', goal: 'yes' },
        { id: 43, name: 'HHH', goal: 'yes' },
        { id: 35, name: 'JJJ', goal: 'yes' }
    ],
    c = [
        { id: 36, name: 'AAA', circle: 'yes' },
        { id: 40, name: 'BBB', circle: 'yes' },
        { id: 57, name: 'CCC', circle: 'yes' },
        { id: 42, name: 'ZZZ', circle: 'yes' },
        { id: 4, name: 'DDD', circle: 'yes' },
        { id: 39, name: 'EEE', circle: 'yes' },
        { id: 37, name: 'FFF', circle: 'yes' },
        { id: 59, name: 'GGG', circle: 'yes' },
        { id: 43, name: 'HHH', circle: 'yes' },
        { id: 35, name: 'JJJ', circle: 'yes' },
        { id: 100, name: 'JJJ', circle: 'yes' }
    ],
    arrayList = [], obj_c_processed = [];

for (var i in g) {
    var obj = {id: g[i].id, name: g[i].name, goal: g[i].goal};

    for (var j in c) {
        if (g[i].id == c[j].id) {
            obj.circle = c[j].circle;
            obj_c_processed[c[j].id] = true;
        }
    }

    obj.circle = obj.circle || 'no';
    arrayList.push(obj);
}

for (var j in c){
    if (typeof obj_c_processed[c[j].id] == 'undefined') {
        arrayList.push({id: c[j].id, name: c[j].name, goal: 'no', circle: c[j].circle});
    }
}

console.log(arrayList);
2020-07-27