Question

我有一个JSON对象，它返回一个包含日期和时间的事件列表，我想在下一个事件时间添加一个计数器，但我不知道如何处理它？</ p>

基本上，我的JSON就像：

[{"name": "BAKER-CROSSROADS", "country": "USA", "lon": "165.25", "yield": "21", "lat": "11.58", "depth": ".027", "date": "1946/07/24 21:35:00", "id": "1"}, {"name": "VENUS", "country": "USA", "lon": "-116.2", "yield": "10", "lat": "37.19", "depth": ".03", "date": "1958/02/22 01:00:00", "id": "10", "fill": "green"}, {"name": "HATCHIE", "country": "USA", "lon": "-116.03", "yield": "20", "lat": "37.058", "depth": ".061", "date": "1963/02/08 16:00:01", "id": "100", "fill": "green"}, {"name": "CLIMAX:Upshot-Knothole", "country": "USA", "lon": "-116.0183", "yield": "61", "lat": "37.0875", "depth": "-.41", "date": "1953/06/04 11:14:57", "id": "1000", "fill": "red"}, {"name": "BRAVO:Castle", "country": "USA", "lon": "165.274", "yield": "15000", "lat": "11.698", "depth": "0", "date": "1954/02/28 18:45:00", "id": "1001", "fill": "red"}, ....

所以每个对象：

{＆＃34; name＆＃34;：＆＃34; BAKER-CROSSROADS＆＃34;，＆＃34; country＆＃34;：＆＃34; USA＆＃34;，＆＃34; lon＆＃34; ：＆＃34; 165.25＆＃34;，＆＃34; yield＆＃34;：＆＃34; 21＆＃34;，＆＃34; lat＆＃34;：＆＃34; 11.58＆＃34;，＆＃ 34;深度＆＃34;：＆＃34; .027＆＃34;，＆＃34;日期＆＃34;：＆＃34; 1946/07/24 21：35：00＆＃34;，＆＃34; id＆＃ 34;：＆＃34; 1＆＃34;}

{＆＃34; name＆＃34;：＆＃34; VENUS＆＃34;，＆＃34; country＆＃34;：＆＃34; USA＆＃34;，＆＃34; lon＆＃34;：＆＃34; -116.2＆＃34;，＆＃34;收益＆＃34;：＆＃34; 10＆＃34;，＆＃34; lat＆＃34;：＆＃34; 37.19＆＃34;，＆＃34 ;＆＃34;：＆＃34; .03＆＃34;，＆＃34; date＆＃34;：＆＃34; 1958/02/22 01：00：00＆＃34;，＆＃34; id＆＃34 ;：＆＃34; 10＆＃34;，＆＃34;填充＆＃34;：＆＃34;绿色＆＃34;}

...

这是读取json文件的代码：

    this.loadNext = function () {
    d3.json("./data/detonations.json", function(datum) {
        for(var i = datum.length - 1; i >= 0; --i) {
            var o = datum[i];
            //console.log(o);
            message = {
                country: o.country
                ,date: o.date
                ,depth: o.depth
                ,lon: o.lon
                ,lat: o.lat
                ,type: o.type
                ,yield: o.yield
            };
            self.doSomething(message)
        }
    });
}

我想计算日期/时间差异，所以在这种情况下，差异是＆＃34; 1958/02/22 01：00：00＆＃34;和＆＃34; 1946/07/24 21：35：00＆＃34;并将其传递给消息数组？

Answer 1

要在毫秒级别中有这种差异，您可以使用

var diff = (new Date(s2)).getTime() - (new Date(s1)).getTime();

那是因为您拥有的格式可以由Date类直接解析。

从JSON数组中计算javascript中的日期差异

1 个答案: