Question

我有两个具有相同字段的JSON数组如下：

var total_90 = [
    { "date": "2011-11-14T17:22:59Z", "quantity": 2, "total": 90, "tip": 0, "type": "tab" },
    { "date": "2011-11-14T17:07:21Z", "quantity": 2, "total": 90, "tip": 1, "type": "tab" },
    { "date": "2012-11-14T16:30:43Z", "quantity": 3, "total": 90, "tip": 0, "type": "tab" }
];

var tip_0 = [
    { "date": "2011-11-14T17:22:59Z", "quantity": 2, "total": 80, "tip": 0, "type": "tab" },
    { "date": "2011-11-14T17:07:21Z", "quantity": 2, "total": 70, "tip": 0, "type": "tab" },
    { "date": "2011-11-14T16:58:03Z", "quantity": 2, "total": 90, "tip": 0, "type": "tab" },
    { "date": "2011-11-14T16:30:43Z", "quantity": 2, "total": 90, "tip": 0, "type": "tab" }
];

我需要第三个JSON文件，它具有上述两个JSON文件的交集。（通过交集，我的意思是来自两个JSON文件的所有行，其中 TOTAL = 90 AND TIP = 0 ）有没有办法做到这一点？

我的预期输出将是具有以下输出的第三个JSON文件

{"date":"2012-11-14T16:30:43Z","quantity":3,"total":90,"tip":0,"type":"tab"},
{"date":"2011-11-14T16:58:03Z","quantity":2,"total":90,"tip":0,"type":"tab"},
{"date":"2011-11-14T16:30:43Z","quantity":2,"total":90,"tip":0,"type":"tab"}

Answer 1

您需要循环2个对象并将内容合并到1个对象中。

例如，检查此主题，因为这是重复的How can I merge properties of two JavaScript objects dynamically?

Answer 2

您可以执行以下操作以从JSON文件中收集所有行，这些文件的 TOTAL = 90 和 TIP = 0 -

var total_90 = [
    { "date": "2011-11-14T17:22:59Z", "quantity": 2, "total": 90, "tip": 0, "type": "tab" },
    { "date": "2011-11-14T17:07:21Z", "quantity": 2, "total": 90, "tip": 1, "type": "tab" },
    { "date": "2012-11-14T16:30:43Z", "quantity": 3, "total": 90, "tip": 0, "type": "tab" }
];

var tip_0 = [
    { "date": "2011-11-14T17:22:59Z", "quantity": 2, "total": 80, "tip": 0, "type": "tab" },
    { "date": "2011-11-14T17:07:21Z", "quantity": 2, "total": 70, "tip": 0, "type": "tab" },
    { "date": "2011-11-14T16:58:03Z", "quantity": 2, "total": 90, "tip": 0, "type": "tab" },
    { "date": "2011-11-14T16:30:43Z", "quantity": 2, "total": 90, "tip": 0, "type": "tab" }
];

// An empty arrays to contain final intersection array
var result = [];

/* now looping over both arrays to traverse all the elements from them */

// iterating over first array
total_90.forEach(x => {
    // iterating over second array
    tip_0.forEach(y => {
        // push into output array only if total is 90 & tip is 0
        if ((x.total == 90 && y.total == 90) && (x.tip == 0 && y.tip == 0)) {
            result.push({
                date: x.date,
                quantity: x.quantity,
                total: x.total,
                tip: x.tip,
                type: x.type
            });
        }
    });
});

console.log(result);

注意- 可以对其进行优化以降低时间复杂度。

Answer 3

var total_90 = [
{"date":"2011-11-14T17:22:59Z","quantity":2,"total":90,"tip":0,"type":"tab"},
{"date":"2011-11-14T17:07:21Z","quantity":2,"total":90,"tip":1,"type":"tab"},
{"date":"2012-11-14T16:30:43Z","quantity":3,"total":90,"tip":0,"type":"tab"}
              ]

var tip_0 = [
{"date":"2011-11-14T17:22:59Z","quantity":2,"total":80,"tip":0,"type":"tab"},
{"date":"2011-11-14T17:07:21Z","quantity":2,"total":70,"tip":0,"type":"tab"},
{"date":"2011-11-14T16:58:03Z","quantity":2,"total":90,"tip":0,"type":"tab"},
{"date":"2011-11-14T16:30:43Z","quantity":2,"total":90,"tip":0,"type":"tab"}
            ] 

allData['total_90'] = total_90;
allData['tip_0'] = tip_0;

使用allData

Answer 4

function intersection(a, b)
{
  var result = [];

       for (var i = 0; i < a.length; i++){
            if (a[i].total == 90 && a[i].tip == 0)
            {

                result.push(a[i]);
            }
        }

       for (var i = 0; i < b.length; i++){
            if (b[i].total == 90 && b[i].tip == 0)
            {
                result.push(b[i]);
            }
        }
    return result;
}

JSFiddle

编辑：使用concat更新该功能，以提供更短的时间。

function intersection(a, b)
{
  var result = [];

    var c = a.concat(b);


       for (var i = 0; i < c.length; i++){
            if (c[i].total == 90 && c[i].tip == 0)
            {

                result.push(c[i]);
            }
        }

    return result;
}

New JSFiddle

两个json文件的交集

4 个答案: