根据最小和最大日期对数组进行排序？

Question

基本上我有一个日期数组：

var array = ["2017-01-22 00:21:17.0",
           "2017-01-27 11:30:23.0",
           "2017-01-24 15:53:21.0",
           "2017-01-27 11:34:18.0",
           "2017-01-26 16:55:48.0",
           "2017-01-22 11:57:12.0",
           "2017-01-27 11:35:43.0"];

我需要根据最大和最小日期对其进行排序；我试过了：

var lowest = _.max(array, function(o){return o.val;});
console.log(lowest);

但是这个returns-Infinity和_.minreturnsInfinity

此致:)

Answer 1

您可以在没有回调的情况下使用 Array#sort。这按字符串排序，它在索引零处获取最小值，在最后一个索引处获取最大日期。

var array = ["2017-01-22 00:21:17.0", "2017-01-27 11:30:23.0", "2017-01-24 15:53:21.0", "2017-01-27 11:34:18.0", "2017-01-26 16:55:48.0", "2017-01-22 11:57:12.0", "2017-01-27 11:35:43.0"];

array.sort();
console.log(array);
console.log('min', array[0]);
console.log('max', array[array.length - 1]);

.as-console-wrapper { max-height: 100% !important; top: 0; }

另一种解决方案是使用 Array#reduce。

var array = ["2017-01-22 00:21:17.0", "2017-01-27 11:30:23.0", "2017-01-24 15:53:21.0", "2017-01-27 11:34:18.0", "2017-01-26 16:55:48.0", "2017-01-22 11:57:12.0", "2017-01-27 11:35:43.0"],
    result = array.reduce(function (r, a, i) {
        if (!i) {
            return { min: a, max: a};
        }
        if (a < r.min) {
            r.min = a;
        }
        if (a > r.max) {
            r.max = a;
        }
        return r;
    }, undefined);

console.log(result);

Answer 2

var array = ["2017-01-22 00:21:17.0",
           "2017-01-27 11:30:23.0",
           "2017-01-24 15:53:21.0",
           "2017-01-27 11:34:18.0",
           "2017-01-26 16:55:48.0",
           "2017-01-22 11:57:12.0",
           "2017-01-27 11:35:43.0"];

var timeStamped = array.map((item) => new Date(item).getTime());

console.log('Max is ', new Date(Math.max.apply(null,timeStamped)). toUTCString());

console.log('Min is ', new Date(Math.min.apply(null,timeStamped)). toUTCString());

根据最小和最大日期对数组进行排序？

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