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
关于模拟DateTime有一个特别的注意事项。现在使用TypeMock…
DateTime的值。现在必须放置到一个变量中,以便正确地模拟。例如:
这行不通:
if ((DateTime.Now - message.TimeOpened.Value) > new TimeSpan(1, 0, 0))
然而,这样做:
var currentDateTime = DateTime.Now;
if ((currentDateTime - message.TimeOpened.Value) > new TimeSpan(1, 0, 0))
其他回答
模拟对象。
一个模拟DateTime,返回适合您的测试的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);
}
最好的策略是将当前时间封装在一个抽象中,并将该抽象注入到使用者中。
或者,你也可以将时间抽象定义为环境上下文:
public abstract class TimeProvider
{
private static TimeProvider current =
DefaultTimeProvider.Instance;
public static TimeProvider Current
{
get { return TimeProvider.current; }
set
{
if (value == null)
{
throw new ArgumentNullException("value");
}
TimeProvider.current = value;
}
}
public abstract DateTime UtcNow { get; }
public static void ResetToDefault()
{
TimeProvider.current = DefaultTimeProvider.Instance;
}
}
这将使你能够像这样消费它:
var now = TimeProvider.Current.UtcNow;
在单元测试中,您可以替换TimeProvider。当前带有Test Double/Mock对象。使用Moq的例子:
var timeMock = new Mock<TimeProvider>();
timeMock.SetupGet(tp => tp.UtcNow).Returns(new DateTime(2010, 3, 11));
TimeProvider.Current = timeMock.Object;
但是,当使用静态状态进行单元测试时,始终记得通过调用TimeProvider.ResetToDefault()来删除fixture。
好的做法是,当DateTimeProvider实现IDisposable时。
public class DateTimeProvider : IDisposable
{
[ThreadStatic]
private static DateTime? _injectedDateTime;
private DateTimeProvider()
{
}
/// <summary>
/// Gets DateTime now.
/// </summary>
/// <value>
/// The DateTime now.
/// </value>
public static DateTime Now
{
get
{
return _injectedDateTime ?? DateTime.Now;
}
}
/// <summary>
/// Injects the actual date time.
/// </summary>
/// <param name="actualDateTime">The actual date time.</param>
public static IDisposable InjectActualDateTime(DateTime actualDateTime)
{
_injectedDateTime = actualDateTime;
return new DateTimeProvider();
}
public void Dispose()
{
_injectedDateTime = null;
}
}
接下来,您可以为单元测试注入假DateTime
using (var date = DateTimeProvider.InjectActualDateTime(expectedDateTime))
{
var bankAccount = new BankAccount();
bankAccount.DepositMoney(600);
var lastTransaction = bankAccount.Transactions.Last();
Assert.IsTrue(expectedDateTime.Equals(bankAccount.Transactions[0].TransactionDate));
}
参见示例DateTimeProvider示例
我也有同样的问题,但我在想我们不应该在同一类上使用设置日期时间的东西。因为有一天可能会导致滥用。我使用了提供商
public class DateTimeProvider
{
protected static DateTime? DateTimeNow;
protected static DateTime? DateTimeUtcNow;
public DateTime Now
{
get
{
return DateTimeNow ?? System.DateTime.Now;
}
}
public DateTime UtcNow
{
get
{
return DateTimeUtcNow ?? System.DateTime.UtcNow;
}
}
public static DateTimeProvider DateTime
{
get
{
return new DateTimeProvider();
}
}
protected DateTimeProvider()
{
}
}
对于测试,在测试项目中创建了一个助手来处理设置的事情,
public class MockDateTimeProvider : DateTimeProvider
{
public static void SetNow(DateTime now)
{
DateTimeNow = now;
}
public static void SetUtcNow(DateTime utc)
{
DateTimeUtcNow = utc;
}
public static void RestoreAsDefault()
{
DateTimeNow = null;
DateTimeUtcNow = null;
}
}
在代码
var dateTimeNow = DateTimeProvider.DateTime.Now //not DateTime.Now
var dateTimeUtcNow = DateTimeProvider.DateTime.UtcNow //not DateTime.UtcNow
在测试中
[Test]
public void Mocked_Now()
{
DateTime now = DateTime.Now;
MockDateTimeProvider.SetNow(now); //set to mock
Assert.AreEqual(now, DateTimeProvider.DateTime.Now);
Assert.AreNotEqual(now, DateTimeProvider.DateTime.UtcNow);
}
[Test]
public void Mocked_UtcNow()
{
DateTime utcNow = DateTime.UtcNow;
MockDateTimeProvider.SetUtcNow(utcNow); //set to mock
Assert.AreEqual(utcNow, DateTimeProvider.DateTime.UtcNow);
Assert.AreNotEqual(utcNow, DateTimeProvider.DateTime.Now);
}
但需要记住一件事,有时真正的DateTime和提供程序的DateTime并不相同
[Test]
public void Now()
{
Assert.AreEqual(DateTime.Now.Kind, DateTimeProvider.DateTime.Now.Kind);
Assert.LessOrEqual(DateTime.Now, DateTimeProvider.DateTime.Now);
Assert.LessOrEqual(DateTimeProvider.DateTime.Now - DateTime.Now, TimeSpan.FromMilliseconds(1));
}
我假设差异是maximum TimeSpan.FromMilliseconds(0.00002)。但大多数时候甚至更少
在MockSamples中找到样本