javascript - 如何按特定顺序正确收集 json 数据

Question

如果我的语言不是“ technical”，我真的很抱歉。我会学得很快。我有json对象，我试图找到正确的方法以正确的方式在单独的数组中收集信息

我正在使用map函数将每个键分隔为一个单独的键，因为有些键属于另一个数组，我有点困惑。我很想得到一些解释。

我在我的问题的底部添加了我预期的解决方案，如果我不清楚这有帮助的话。

以下是 json 对象：

var json_museums = {
    "museums_costs": [
        {
            "museumType": "Art",
            "costs": [
                {
                    "date": "2018-10-01",
                    "employees": 1082220.3945979946,
                    "outsource": 9585.8794674401543
                },
                {
                    "date": "2018-11-01",
                    "employees": 1056643.6756958894,
                    "outsource": 11018.898987932616
                },
                {
                    "date": "2018-12-01",
                    "employees": 1022322.1218654147,
                    "outsource": 11954.748339502334
                }
            ]
        },
        {
            "museumType": "History",
            "costs": [
                {
                    "date": "2018-10-01",
                    "employees": 91336.329372028267,
                    "outsource": 0
                },
                {
                    "date": "2018-11-01",
                    "employees": 78130.417193652393,
                    "outsource": 0
                },
                {
                    "date": "2018-12-01",
                    "employees": 100755.62136033915,
                    "outsource": 0
                }
            ]
        },
        {
            "museumType": "Culture",
            "costs": [
                {
                    "date": "2018-10-01",
                    "employees": 66604.417069221658,
                    "outsource": 0
                },
                {
                    "date": "2018-11-01",
                    "employees": 63882.129662868538,
                    "outsource": 0
                },
                {
                    "date": "2018-12-01",
                    "employees": 108781.84734382466,
                    "outsource": 0
                }
            ]
        },
        {
            "museumType": "News",
            "costs": [
                {
                    "date": "2018-10-01",
                    "employees": 0,
                    "outsource": 1452.6775522557543
                },
                {
                    "date": "2018-11-01",
                    "employees": 0,
                    "outsource": 2366.423254655545
                },
                {
                    "date": "2018-12-01",
                    "employees": 0,
                    "outsource": 3485.5840149129986
                }
            ]
        },
        {
            "museumType": "Science",
            "costs": [
                {
                    "date": "2018-10-01",
                    "employees": 3237887.746893588,
                    "outsource": 0
                },
                {
                    "date": "2018-11-01",
                    "employees": 3218724.4935297123,
                    "outsource": 0
                },
                {
                    "date": "2018-12-01",
                    "employees": 2925562.2916941536,
                    "outsource": 0
                }
            ]
        },
        {
            "museumType": "Religious",
            "costs": [
                {
                    "date": "2018-10-01",
                    "employees": 291293.35185208195,
                    "outsource": 0
                },
                {
                    "date": "2018-11-01",
                    "employees": 171962.47103846565,
                    "outsource": 0
                },
                {
                    "date": "2018-12-01",
                    "employees": 95615.439855929668,
                    "outsource": 0
                }
            ]
        }
    ]
}

var results = [];



results = json_museums.museums_costs.map(function (x) {
    return [x.museumType, x. visits.map(function(c) { c.date), x.costs.reduce(function (a, b, c) {
        return a + b + c;
    }, 0)]
});


(18) [Array(3), Array(3), Array(3), Array(3), Array(3), Array(3)]
0: (3) ["Art", "2018-10-01",  outsource + employee]
1: (3) ["Art", "2018-11-01",  outsource +employee]
2: (3) ["Art", "2018-12-01",  outsource +employee]
3: (3) ["History", "2018-10-01",  outsource + employee]
4: (3) ["History", "2018-11-01",  outsource + employee]
5: (3) ["History", "2018-12-01",  outsource + employee]
6: (3) ["Culture", "2018-10-01",  outsource + employee]
7: (3) ["Culture", "2018-11-01",  outsource + employee]
8: (3) ["Culture", "2018-12-01",  outsource + employee]
9: (3) ["News", "2018-10-01",  outsource + employee]
10: (3) ["News", "2018-11-01",  outsource + employee]
11: (3) ["News", "2018-12-01",  outsource + employee]
12: (3) ["Science", "2018-10-01",  outsource + employee]
13: (3) ["Science", "2018-11-01",  outsource + employee]
14: (3) ["Science", "2018-12-01",  outsource + employee]
15: (3) ["Religious", "2018-10-01",  outsource + employee]
16: (3) ["Religious", "2018-11-01",  outsource + employee]
17: (3) ["Religious", "2018-12-01",  outsource + employee]

标签： javascriptjson

Answer 1

您可以reduce从一个空数组开始作为累加器的顶级数组，然后 map超过成本并将结果数组的元素推送到累加器：

var json_museums = {"museums_costs":[{"museumType":"Art","costs":[{"date":"2018-10-01","employees":1082220.3945979946,"outsource":9585.8794674401543},{"date":"2018-11-01","employees":1056643.6756958894,"outsource":11018.898987932616},{"date":"2018-12-01","employees":1022322.1218654147,"outsource":11954.748339502334}]},{"museumType":"History","costs":[{"date":"2018-10-01","employees":91336.329372028267,"outsource":0},{"date":"2018-11-01","employees":78130.417193652393,"outsource":0},{"date":"2018-12-01","employees":100755.62136033915,"outsource":0}]},{"museumType":"Culture","costs":[{"date":"2018-10-01","employees":66604.417069221658,"outsource":0},{"date":"2018-11-01","employees":63882.129662868538,"outsource":0},{"date":"2018-12-01","employees":108781.84734382466,"outsource":0}]},{"museumType":"News","costs":[{"date":"2018-10-01","employees":0,"outsource":1452.6775522557543},{"date":"2018-11-01","employees":0,"outsource":2366.423254655545},{"date":"2018-12-01","employees":0,"outsource":3485.5840149129986}]},{"museumType":"Science","costs":[{"date":"2018-10-01","employees":3237887.746893588,"outsource":0},{"date":"2018-11-01","employees":3218724.4935297123,"outsource":0},{"date":"2018-12-01","employees":2925562.2916941536,"outsource":0}]},{"museumType":"Religious","costs":[{"date":"2018-10-01","employees":291293.35185208195,"outsource":0},{"date":"2018-11-01","employees":171962.47103846565,"outsource":0},{"date":"2018-12-01","employees":95615.439855929668,"outsource":0}]}]}

var results = json_museums.museums_costs.reduce((a, x) => {
  a.push(...x.costs.map(cost => [x.museumType, cost.date, cost.outsource + cost.employees]));
  return a;
}, []);

console.log(results);

另一种方法是使用flatMap，尽管这并未得到普遍支持。