8

是否可以一次一天地迭代间隔的开始日期和结束日期之间的时间?使用clj-timeClojure 库也可以!

4

4 回答 4

14

是的。

像这样的东西:

DateTime now = DateTime.now();
DateTime start = now;
DateTime stop = now.plusDays(10);
DateTime inter = start;
// Loop through each day in the span
while (inter.compareTo(stop) < 0) {
    System.out.println(inter);
    // Go to next
    inter = inter.plusDays(1);
}

此外,这里是 Clojure 的 clj-time 的实现:

(defn date-interval
  ([start end] (date-interval start end []))
  ([start end interval]
   (if (time/after? start end)
     interval
     (recur (time/plus start (time/days 1)) end (concat interval [start])))))
于 2012-08-19T22:01:42.737 回答
8

这应该有效。

(take-while (fn [t] (cljt/before? t to)) (iterate (fn [t] (cljt/plus t period)) from))
于 2012-08-20T00:12:28.130 回答
2

使用 clj-time,the-interval 是 Joda Interval :

(use '[clj-time.core :only (days plus start in-days)])
(defn each-day [the-interval f]
 (let [days-diff (in-days the-interval)]
    (for [x (range 0 (inc days-diff))] (f (plus (start the-interval) (days x))))))
于 2012-08-19T22:15:50.033 回答
1

Clj-time 具有periodic-seq 功能。它看起来非常像 Hendekagon 的解决方案。这可以与开始/结束功能相结合,以创建类似:

(defn interval-seq 
  "Returns sequence of period over interval. 
   For example, the individual days in an interval using (clj-time.core/days 1) as the period."
  [interval period]
  (let [end (clj-time.core/end interval)]
      (take-while #(clj-time.core/before? % end) (clj-time.periodic/periodic-seq  (clj-time.core/start interval) period ))))
于 2014-05-29T01:28:49.287 回答