我有一些单元测试,期望“当前时间”与DateTime不同。显然,我不想改变电脑的时间。
实现这一目标的最佳策略是什么?
我有一些单元测试,期望“当前时间”与DateTime不同。显然,我不想改变电脑的时间。
实现这一目标的最佳策略是什么?
当前回答
以下是我对这个问题的回答。我将“环境上下文”模式与IDisposable结合起来。你可以使用DateTimeProvider。当前在您的正常程序代码和测试中,您使用using语句覆盖范围。
using System;
using System.Collections.Immutable;
namespace ambientcontext {
public abstract class DateTimeProvider : IDisposable
{
private static ImmutableStack<DateTimeProvider> stack = ImmutableStack<DateTimeProvider>.Empty.Push(new DefaultDateTimeProvider());
protected DateTimeProvider()
{
if (this.GetType() != typeof(DefaultDateTimeProvider))
stack = stack.Push(this);
}
public static DateTimeProvider Current => stack.Peek();
public abstract DateTime Today { get; }
public abstract DateTime Now {get; }
public void Dispose()
{
if (this.GetType() != typeof(DefaultDateTimeProvider))
stack = stack.Pop();
}
// Not visible Default Implementation
private class DefaultDateTimeProvider : DateTimeProvider {
public override DateTime Today => DateTime.Today;
public override DateTime Now => DateTime.Now;
}
}
}
下面是如何在单元测试中使用上述DateTimeProvider
using System;
using Xunit;
namespace ambientcontext
{
public class TestDateTimeProvider
{
[Fact]
public void TestDateTime()
{
var actual = DateTimeProvider.Current.Today;
var expected = DateTime.Today;
Assert.Equal<DateTime>(expected, actual);
using (new MyDateTimeProvider(new DateTime(2012,12,21)))
{
Assert.Equal(2012, DateTimeProvider.Current.Today.Year);
using (new MyDateTimeProvider(new DateTime(1984,4,4)))
{
Assert.Equal(1984, DateTimeProvider.Current.Today.Year);
}
Assert.Equal(2012, DateTimeProvider.Current.Today.Year);
}
// Fall-Back to Default DateTimeProvider
Assert.Equal<int>(expected.Year, DateTimeProvider.Current.Today.Year);
}
private class MyDateTimeProvider : DateTimeProvider
{
private readonly DateTime dateTime;
public MyDateTimeProvider(DateTime dateTime):base()
{
this.dateTime = dateTime;
}
public override DateTime Today => this.dateTime.Date;
public override DateTime Now => this.dateTime;
}
}
}
其他回答
这些都是很好的答案,这是我在另一个项目中所做的:
用法:
获取今天的真实日期时间
var today = SystemTime.Now().Date;
而不是使用DateTime。现在,你需要使用SystemTime.Now()…这并不难改变,但这种解决方案可能并不适合所有项目。
时间旅行(让我们去5年后)
SystemTime.SetDateTime(today.AddYears(5));
获得我们虚假的“今天”(从“今天”开始算起5年)
var fakeToday = SystemTime.Now().Date;
重置日期
SystemTime.ResetDateTime();
/// <summary>
/// Used for getting DateTime.Now(), time is changeable for unit testing
/// </summary>
public static class SystemTime
{
/// <summary> Normally this is a pass-through to DateTime.Now, but it can be overridden with SetDateTime( .. ) for testing or debugging.
/// </summary>
public static Func<DateTime> Now = () => DateTime.Now;
/// <summary> Set time to return when SystemTime.Now() is called.
/// </summary>
public static void SetDateTime(DateTime dateTimeNow)
{
Now = () => dateTimeNow;
}
/// <summary> Resets SystemTime.Now() to return DateTime.Now.
/// </summary>
public static void ResetDateTime()
{
Now = () => DateTime.Now;
}
}
一种简单的方法是注入VirtualTime。它可以让你控制时间。 首先安装VirtualTime
Install-Package VirtualTime
例如,这允许在所有对DateTime的调用上使时间加快5倍。Now或UtcNow
var DateTime = DateTime.Now.ToVirtualTime(5);
放慢时间(如放慢5倍)
var DateTime = DateTime.Now.ToVirtualTime(0.5);
让时间静止
var DateTime = DateTime.Now.ToVirtualTime(0);
回到过去还没有测试过
下面是一个测试示例:
[TestMethod]
public void it_should_make_time_move_faster()
{
int speedOfTimePerMs = 1000;
int timeToPassMs = 3000;
int expectedElapsedVirtualTime = speedOfTimePerMs * timeToPassMs;
DateTime whenTimeStarts = DateTime.Now;
ITime time = whenTimeStarts.ToVirtualTime(speedOfTimePerMs);
Thread.Sleep(timeToPassMs);
DateTime expectedTime = DateTime.Now.AddMilliseconds(expectedElapsedVirtualTime - timeToPassMs);
DateTime virtualTime = time.Now;
Assert.IsTrue(TestHelper.AreEqualWithinMarginOfError(expectedTime, virtualTime, MarginOfErrorMs));
}
你可以在这里查看更多测试:
https://github.com/VirtualTime/VirtualTime/blob/master/VirtualTimeLib.Tests/when_virtual_time_is_used.cs
什么DateTime.Now.ToVirtualTime扩展给你是一个ITime的实例,你传递给一个依赖于ITime的方法/类。一些DateTime.Now.ToVirtualTime被设置在你选择的DI容器中
下面是注入类构造函数的另一个例子
public class AlarmClock
{
private ITime DateTime;
public AlarmClock(ITime dateTime, int numberOfHours)
{
DateTime = dateTime;
SetTime = DateTime.UtcNow.AddHours(numberOfHours);
Task.Run(() =>
{
while (!IsAlarmOn)
{
IsAlarmOn = (SetTime - DateTime.UtcNow).TotalMilliseconds < 0;
}
});
}
public DateTime SetTime { get; set; }
public bool IsAlarmOn { get; set; }
}
[TestMethod]
public void it_can_be_injected_as_a_dependency()
{
//virtual time has to be 1000*3.75 faster to get to an hour
//in 1000 ms real time
var dateTime = DateTime.Now.ToVirtualTime(1000 * 3.75);
var numberOfHoursBeforeAlarmSounds = 1;
var alarmClock = new AlarmClock(dateTime, numberOfHoursBeforeAlarmSounds);
Assert.IsFalse(alarmClock.IsAlarmOn);
System.Threading.Thread.Sleep(1000);
Assert.IsTrue(alarmClock.IsAlarmOn);
}
模拟对象。
一个模拟DateTime,返回适合您的测试的Now。
摩尔数:
[Test]
public void TestOfDateTime()
{
var firstValue = DateTime.Now;
MDateTime.NowGet = () => new DateTime(2000,1,1);
var secondValue = DateTime.Now;
Assert(firstValue > secondValue); // would be false if 'moleing' failed
}
免责声明-我工作的鼹鼠
我们使用的是静态SystemTime对象,但是在运行并行单元测试时遇到了问题。我尝试使用Henk van Boeijen的解决方案,但在派生异步线程上有问题,最终以类似于下面的方式使用AsyncLocal:
public static class Clock
{
private static Func<DateTime> _utcNow = () => DateTime.UtcNow;
static AsyncLocal<Func<DateTime>> _override = new AsyncLocal<Func<DateTime>>();
public static DateTime UtcNow => (_override.Value ?? _utcNow)();
public static void Set(Func<DateTime> func)
{
_override.Value = func;
}
public static void Reset()
{
_override.Value = null;
}
}
来源:https://gist.github.com/CraftyFella/42f459f7687b0b8b268fc311e6b4af08