比较具有不同精度级别的日期对象

还有一个解决办法,我会这样做:

assertTrue("Dates aren't close enough to each other!", (date2.getTime() - date1.getTime()) < 1000);

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Apache commons-lang 在类路径上,可以使用 DateUtils.truncate 将日期截断为某个字段。

assertEquals(DateUtils.truncate(date1,Calendar.SECOND),
             DateUtils.truncate(date2,Calendar.SECOND));

assertTrue(DateUtils.truncatedEquals(date1,date2,Calendar.SECOND));

请注意,12:00:00.001和11:59:00.999将截断为不同的值,因此这可能并不理想。为此,有以下几点:

assertEquals(DateUtils.round(date1,Calendar.SECOND),
             DateUtils.round(date2,Calendar.SECOND));

从3.7.0版开始, AssertJ isCloseTo 断言,如果您使用的是Java8日期/时间API。

LocalTime _07_10 = LocalTime.of(7, 10);
LocalTime _07_42 = LocalTime.of(7, 42);
assertThat(_07_10).isCloseTo(_07_42, within(1, ChronoUnit.HOURS));
assertThat(_07_10).isCloseTo(_07_42, within(32, ChronoUnit.MINUTES));

它也适用于传统的java日期:

Date d1 = new Date();
Date d2 = new Date();
assertThat(d1).isCloseTo(d2, within(100, ChronoUnit.MILLIS).getValue());

使用 DateFormat 对象,其格式仅显示要匹配的部分,并执行 assertEquals() 在生成的字符串上。您也可以轻松地将其包装在自己的文件中 assertDatesAlmostEqual() 方法。

你可以这样做:

assertTrue((date1.getTime()/1000) == (date2.getTime()/1000));

在JUnit中,您可以编写两个断言方法,如下所示:

public class MyTest {
  @Test
  public void test() {
    ...
    assertEqualDates(expectedDateObject, resultDate);

    // somewhat more confortable:
    assertEqualDates("01/01/2012", anotherResultDate);
  }

  private static final String DATE_PATTERN = "dd/MM/yyyy";

  private static void assertEqualDates(String expected, Date value) {
      DateFormat formatter = new SimpleDateFormat(DATE_PATTERN);
      String strValue = formatter.format(value);
      assertEquals(expected, strValue);
  }

  private static void assertEqualDates(Date expected, Date value) {
    DateFormat formatter = new SimpleDateFormat(DATE_PATTERN);
    String strExpected = formatter.format(expected);
    String strValue = formatter.format(value);
    assertEquals(strExpected, strValue);
  }
}

LocalDateTime now = LocalDateTime.now().truncatedTo(ChronoUnit.SECONDS);
// e.g. in MySQL db "timestamp" is without fractional seconds precision (just up to seconds precision)
assertEquals(myTimestamp, now);

我不知道JUnit中是否有支持,但有一种方法可以做到:

import java.text.SimpleDateFormat;
import java.util.Date;

public class Example {

    private static SimpleDateFormat formatter = new SimpleDateFormat("dd MMM yyyy HH:mm:ss");

    private static boolean assertEqualDates(Date date1, Date date2) {
        String d1 = formatter.format(date1);            
        String d2 = formatter.format(date2);            
        return d1.equals(d2);
    }    

    public static void main(String[] args) {
        Date date1 = new Date();
        Date date2 = new Date();

        if (assertEqualDates(date1,date2)) { System.out.println("true!"); }
    }
}

这实际上是一个比看起来更难的问题,因为在边界情况下,您不关心的方差超过了您正在检查的值的阈值。e、 g.毫秒差小于一秒,但两个时间戳跨过第二阈值、分钟阈值或小时阈值。这使得任何DateFormat方法都天生容易出错。

相反,我建议比较实际的毫秒时间戳,并提供一个方差增量,指示您认为两个日期对象之间的可接受差异。下面是一个过于冗长的示例:

public static void assertDateSimilar(Date expected, Date actual, long allowableVariance)
{
    long variance = Math.abs(allowableVariance);

    long millis = expected.getTime();
    long lowerBound = millis - allowableVariance;
    long upperBound = millis + allowableVariance;

    DateFormat df = DateFormat.getDateTimeInstance();

    boolean within = lowerBound <= actual.getTime() && actual.getTime() <= upperBound;
    assertTrue(MessageFormat.format("Expected {0} with variance of {1} but received {2}", df.format(expected), allowableVariance, df.format(actual)), within);
}

匹配器 用于根据您选择的精度确定测试日期。在本例中,匹配器将字符串格式表达式作为参数。对于这个例子,代码没有更短。但是,matcher类可以重用;如果你给它一个描述的名字,你可以用一种优雅的方式记录测试的意图。

import static org.junit.Assert.assertThat;
// further imports from org.junit. and org.hamcrest.

@Test
public void testAddEventsToBaby() {
    Date referenceDate = new Date();
    // Do something..
    Date testDate = new Date();

    //assertThat(referenceDate, equalTo(testDate)); // Test on equal could fail; it is a race condition
    assertThat(referenceDate, sameCalendarDay(testDate, "yyyy MM dd"));
}

public static Matcher<Date> sameCalendarDay(final Object testValue, final String dateFormat){

    final SimpleDateFormat formatter = new SimpleDateFormat(dateFormat);

    return new BaseMatcher<Date>() {

        protected Object theTestValue = testValue;


        public boolean matches(Object theExpected) {
            return formatter.format(theExpected).equals(formatter.format(theTestValue));
        }

        public void describeTo(Description description) {
            description.appendText(theTestValue.toString());
        }
    };
}

对Joda Time使用AssertJ断言( http://joel-costigliola.github.io/assertj/assertj-joda-time.html

import static org.assertj.jodatime.api.Assertions.assertThat;
import org.joda.time.DateTime;

assertThat(new DateTime(dateOne.getTime())).isEqualToIgnoringMillis(new DateTime(dateTwo.getTime()));

测试失败消息更具可读性

java.lang.AssertionError: 
Expecting:
  <2014-07-28T08:00:00.000+08:00>
to have same year, month, day, hour, minute and second as:
  <2014-07-28T08:10:00.000+08:00>
but had not.

如果你用的是乔达,你可以用 Fest Joda Time .

只需比较您感兴趣的日期部分:

Date dateOne = new Date();
dateOne.setTime(61202516585000L);
Date dateTwo = new Date();
dateTwo.setTime(61202516585123L);

assertEquals(dateOne.getMonth(), dateTwo.getMonth());
assertEquals(dateOne.getDate(), dateTwo.getDate());
assertEquals(dateOne.getYear(), dateTwo.getYear());

// alternative to testing with deprecated methods in Date class
Calendar calOne = Calendar.getInstance();
Calendar calTwo = Calendar.getInstance();
calOne.setTime(dateOne);
calTwo.setTime(dateTwo);

assertEquals(calOne.get(Calendar.MONTH), calTwo.get(Calendar.MONTH));
assertEquals(calOne.get(Calendar.DATE), calTwo.get(Calendar.DATE));
assertEquals(calOne.get(Calendar.YEAR), calTwo.get(Calendar.YEAR));

JUnit有一个内置的断言,用于比较double,并指定它们需要接近的程度。在本例中,增量是指您认为日期相等的毫秒数。此解决方案没有边界条件,测量绝对方差,可以轻松指定精度,并且不需要编写额外的库或代码。

    Date dateOne = new Date();
    dateOne.setTime(61202516585000L);
    Date dateTwo = new Date();
    dateTwo.setTime(61202516585123L);
    // this line passes correctly 
    Assert.assertEquals(dateOne.getTime(), dateTwo.getTime(), 500.0);
    // this line fails correctly
    Assert.assertEquals(dateOne.getTime(), dateTwo.getTime(), 100.0);

注 它必须是100.0而不是100(或者需要转换为double),才能强制将它们作为double进行比较。

isEqualToIgnoringMillis 或 isCloseTo

assertThat(thing)
  .usingRecursiveComparison()
  .withComparatorForType(
      (a, b) -> a.truncatedTo(ChronoUnit.MILLIS).compareTo(b.truncatedTo(ChronoUnit.MILLIS)),
      OffsetDateTime.class
  )

类似的方法可能会奏效:

assertEquals(new SimpleDateFormat("dd MMM yyyy").format(dateOne),
                   new SimpleDateFormat("dd MMM yyyy").format(dateTwo));

new Date 您可以直接创建一个小型协作者,您可以在测试中模拟它:

public class DateBuilder {
    public java.util.Date now() {
        return new java.util.Date();
    }
}

新日期 到 dateBuilder.now()

import java.util.Date;

public class Demo {

    DateBuilder dateBuilder = new DateBuilder();

    public void run() throws InterruptedException {
        Date dateOne = dateBuilder.now();
        Thread.sleep(10);
        Date dateTwo = dateBuilder.now();
        System.out.println("Dates are the same: " + dateOne.equals(dateTwo));
    }

    public static void main(String[] args) throws InterruptedException {
        new Demo().run();
    }
}

Dates are the same: false

在测试中,您可以注入一个 DateBuilder 并让它返回您喜欢的任何值。例如,使用Mockito或覆盖 now() :

public class DemoTest {

    @org.junit.Test
    public void testMockito() throws Exception {
        DateBuilder stub = org.mockito.Mockito.mock(DateBuilder.class);
        org.mockito.Mockito.when(stub.now()).thenReturn(new java.util.Date(42));

        Demo demo = new Demo();
        demo.dateBuilder = stub;
        demo.run();
    }

    @org.junit.Test
    public void testAnonymousClass() throws Exception {
        Demo demo = new Demo();
        demo.dateBuilder = new DateBuilder() {
            @Override
            public Date now() {
                return new Date(42);
            }
        };
        demo.run();
    }
}

SimpleDateFormat formatter = new SimpleDateFormat("yyyy-MM-dd HH:mm:ss");
String expectedDate = formatter.format(dateOne));
String dateToTest = formatter.format(dateTwo);
assertEquals(expectedDate, dateToTest);

https://stackoverflow.com/a/37168645/5930242

这是一个实用函数,它为我完成了这项工作。

    private boolean isEqual(Date d1, Date d2){
        return d1.toLocalDate().equals(d2.toLocalDate());
    }

isEqualToIgnoringSeconds 方法忽略秒并仅按分钟进行比较:

Date d1 = new Date();
Thread.sleep(10000);
Date d2 = new Date();
assertThat(d1).isEqualToIgnoringSeconds(d2); // true

我将对象强制转换为java.util.Date并进行比较

assertEquals((Date)timestamp1,(Date)timestamp2);