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85 linhas
3.7 KiB
85 linhas
3.7 KiB
%%% -------------------------------------------------------
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%%% @author huangyongxing@yeah.net
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%%% @doc
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%%% 测试计算日期跨度的两种接口效率
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%%% @end
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%%% -------------------------------------------------------
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-module(test_day_span).
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-export([
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test/1
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]).
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-include("common.hrl").
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%% 测试运行100w次，
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%% 结果表明，使用格里高利历天数对比效率更高，大约耗时是另一种方法的70%
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%% test_day_span:test(300000).
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%% "my calc day_span1" [total: 266(266)ms avg: 0.887(0.887)us]
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%% "use gregorian_days" [total: 172(187)ms avg: 0.573(0.623)us]
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%% ========================================================================
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%% my calc day_span1 = 266.00ms [ 100.00%] 266.00ms [ 100.00%]
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%% use gregorian_days = 172.00ms [ 64.66%] 187.00ms [ 70.30%]
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%% ========================================================================
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test(N) ->
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T1 = utime:unixtime(),
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T2 = utime:unixtime() - urand:rand(1000000, 9000000),
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Fun1 = fun(_) -> day_span1(T1, T2) end,
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Fun2 = fun(_) -> day_span2(T1, T2) end,
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ptester:run(N, [
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{"my calc day_span1", Fun1},
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{"use gregorian_days", Fun2}
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]).
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day_span1(Unixtime1, Unixtime2) ->
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get_diff_day(Unixtime1, Unixtime2) + 1.
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day_span2(Unixtime1, Unixtime2) ->
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{D1, _} = utime:unixtime_to_localtime(Unixtime1),
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{D2, _} = utime:unixtime_to_localtime(Unixtime2),
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calendar:date_to_gregorian_days(D1) - calendar:date_to_gregorian_days(D2).
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%% 原utime中的get_diff_day
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%% 早期不兼容夏令时的时候，少一次unixdate及相关调用，测试过比用格里高利历的稍快，
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%% 加上针对夏令时的逻辑处理，重新测试结果已经不及格里高利历天数的计算方式）
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get_diff_day(Seconds, Seconds) -> 0;
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get_diff_day(Seconds1, Seconds2) ->
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Midnight1 = utime:unixdate(Seconds1),
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Midnight2 = utime:unixdate(Seconds2),
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DiffSecs = Midnight1 - Midnight2,
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DiffDays = abs(DiffSecs div ?ONE_DAY_SECONDS),
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Remainder = DiffSecs rem ?ONE_DAY_SECONDS,
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case Remainder =:= 0 of
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true -> % 整除意味着两个时刻同处于夏令时、非夏令时或地区无夏令时制
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DiffDays;
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false -> % 两个时刻中有一个处于夏令时中
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%% 由于各地夏令时规定的调整时长不明确，
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%% 有半小时有一小时，总之不可能太长，所以以3小时为界。
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%% 余数少于3小时说明跨越“结束夏令时”的时间点，应向下取整
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%% 余数多于3小时说明跨越“进入夏令时”的时间点，应向上取整
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case abs(Remainder) < 10800 of
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true -> DiffDays;
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false -> DiffDays + 1
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end
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end.
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% %% 此方法存在夏令时问题
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% %% 返回结果为两个UNIX时间戳的日期跨度，例如：同一天和相邻2天，返回结果为1和2（而非0和1）
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% %% 经测试，与分别使用calendar:date_to_gregorian_days/1转换为两个格里高利历天数进行比较，
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% %% 运行30w次，这个方法耗时约0.175秒，转换为格里高历利天数计算约需0.198秒
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% get_diff_day(Seconds, Seconds) -> 0;
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% get_diff_day(Seconds1, Seconds2) ->
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% {_Date, Time} = unixtime_to_localtime(Seconds1),
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% DiffSecs = Seconds1 - calendar:time_to_seconds(Time) - Seconds2,
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% DiffDays = DiffSecs div ?ONE_DAY_SECONDS,
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% Remainder = DiffSecs rem ?ONE_DAY_SECONDS,
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% case DiffSecs > 0 of
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% true -> % 第一个时间的0点比第二个时间大
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% case Remainder =:= 0 of
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% true -> 0;
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% false -> 1 % 并且不整除(第二个时间非0点)，diff days少算了一天，要加1
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% end
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% + DiffDays;
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% false ->
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% - DiffDays
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% end.
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