C#中的随机日期
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Random date in C#
提问by Judah Gabriel Himango
I'm looking for some succinct, modern C# code to generate a random date between Jan 1 1995 and the current date.
我正在寻找一些简洁的现代 C# 代码来生成 1995 年 1 月 1 日和当前日期之间的随机日期。
I'm thinking some solution that utilizes Enumerable.Range somehow may make this more succinct.
我在想一些利用 Enumerable.Range 的解决方案可能会使这更简洁。
采纳答案by Joel Coehoorn
private Random gen = new Random();
DateTime RandomDay()
{
DateTime start = new DateTime(1995, 1, 1);
int range = (DateTime.Today - start).Days;
return start.AddDays(gen.Next(range));
}
For better performance if this will be called repeatedly, create the startand gen(and maybe even range) variables outsideof the function.
如果这将被重复调用，为了获得更好的性能，请在函数外部创建start和gen（甚至可能是range）变量。
回答by Gabriele D'Antona
Start with a fixed date object (Jan 1, 1995), and add a random number of days with AddDays (obviusly, pay attention not surpassing the current date).
从固定日期对象（1995 年 1 月 1 日）开始，并使用 AddDays 添加随机天数（显然，注意不要超过当前日期）。
回答by JaredPar
This is in slight response to Joel's comment about making a slighly more optimized version. Instead of returning a random date directly, why not return a generator function which can be called repeatedly to create a random date.
这是对 Joel 关于制作稍微更优化版本的评论的轻微回应。与其直接返回一个随机日期，不如返回一个可以重复调用的生成器函数来创建一个随机日期。
Func<DateTime> RandomDayFunc()
{
DateTime start = new DateTime(1995, 1, 1);
Random gen = new Random();
int range = ((TimeSpan)(DateTime.Today - start)).Days;
return () => start.AddDays(gen.Next(range));
}
回答by James Curran
Well, if you gonna present alternate optimization, we can also go for an iterator:
好吧，如果您要展示替代优化，我们也可以使用迭代器：
static IEnumerable<DateTime> RandomDay()
{
DateTime start = new DateTime(1995, 1, 1);
Random gen = new Random();
int range = ((TimeSpan)(DateTime.Today - start)).Days;
while (true)
yield return start.AddDays(gen.Next(range));
}
you could use it like this:
你可以这样使用它：
int i=0;
foreach(DateTime dt in RandomDay())
{
Console.WriteLine(dt);
if (++i == 10)
break;
}
回答by prespic
I have taken @Joel Coehoorn answer and made the changes he adviced - put the variable out of the method and put all in class. Plus now the time is random too. Here is the result.
我已经接受了@Joel Coehoorn 的回答并进行了他建议的更改 - 将变量从方法中取出并全部放入类中。加上现在时间也是随机的。这是结果。
class RandomDateTime
{
DateTime start;
Random gen;
int range;
public RandomDateTime()
{
start = new DateTime(1995, 1, 1);
gen = new Random();
range = (DateTime.Today - start).Days;
}
public DateTime Next()
{
return start.AddDays(gen.Next(range)).AddHours(gen.Next(0,24)).AddMinutes(gen.Next(0,60)).AddSeconds(gen.Next(0,60));
}
}
And example how to use to write 100 random DateTimes to console:
以及如何使用将 100 个随机日期时间写入控制台的示例：
RandomDateTime date = new RandomDateTime();
for (int i = 0; i < 100; i++)
{
Console.WriteLine(date.Next());
}
回答by Hamit Gündogdu
I am a bit late in to the game, but here is one solution which works fine:
我进入游戏有点晚了，但这里有一个工作正常的解决方案：
void Main()
{
var dateResult = GetRandomDates(new DateTime(1995, 1, 1), DateTime.UtcNow, 100);
foreach (var r in dateResult)
Console.WriteLine(r);
}
public static IList<DateTime> GetRandomDates(DateTime startDate, DateTime maxDate, int range)
{
var randomResult = GetRandomNumbers(range).ToArray();
var calculationValue = maxDate.Subtract(startDate).TotalMinutes / int.MaxValue;
var dateResults = randomResult.Select(s => startDate.AddMinutes(s * calculationValue)).ToList();
return dateResults;
}
public static IEnumerable<int> GetRandomNumbers(int size)
{
var data = new byte[4];
using (var rng = new System.Security.Cryptography.RNGCryptoServiceProvider(data))
{
for (int i = 0; i < size; i++)
{
rng.GetBytes(data);
var value = BitConverter.ToInt32(data, 0);
yield return value < 0 ? value * -1 : value;
}
}
}
回答by BernardV
Small method that returns a random date as string, based on some simple input parameters. Built based on variations from the above answers:
基于一些简单的输入参数，以字符串形式返回随机日期的小方法。基于上述答案的变化而构建：
public string RandomDate(int startYear = 1960, string outputDateFormat = "yyyy-MM-dd")
{
DateTime start = new DateTime(startYear, 1, 1);
Random gen = new Random(Guid.NewGuid().GetHashCode());
int range = (DateTime.Today - start).Days;
return start.AddDays(gen.Next(range)).ToString(outputDateFormat);
}
