如何断言日期时间戳在 2 分钟内？

Question

我想要断言以确保检索到的数据在

1. 打印数据1 2020-03-16 09:08:49

2. 打印数据2 2020-03-16 09:09:15

3. 断言 data1 和 data2 之间的时间间隔不超过 2 分钟，然后通过。

我有示例代码，这是最好的方法吗？ 任何建议，请发表评论。

 //data1 
 Date data1 = new Date();
 //data2 
 Date data2 = new Date();
 //assert
 assertThat(data1, DateMatchers.within(2, ChronoUnit.MINUTES, data2 ));

最新脚本

 import static org.assertj.core.api.Assertions.* import java.sql.* import java.text.SimpleDateFormat import com.kms.katalon.core.webui.keyword.WebUiBuiltInKeywords as WebUI import internal.GlobalVariable as GlobalVariable import java.text.ParseException import java.text.SimpleDateFormat import java.util.Date import com.kms.katalon.core.configuration.RunConfiguration GlobalVariable.TestIssueKey = null WebUI.delay(1) //SQL statement dbQuery2 = /SELECT * FROM drugs.sync/ //Connect to PostgresSQL, global variable is stored at profile List results = CustomKeywords.'test.database.getPostgresSQLResults'(GlobalVariable.dbConnString2 , GlobalVariable.dbUsername2 , GlobalVariable.dbPassword2 ,GlobalVariable.dbDriver2 ,dbQuery2 ) //SimpleDateFormat sdf = new SimpleDateFormat("yyyy-MM-dd'T'HH:mm:ss.SSS'Z'") SimpleDateFormat sdf = new SimpleDateFormat("yyyy-mm-dd hh:mm:ss", Locale.ENGLISH) String date = sdf.format(new Date()) //print the "lastupdatedwm6" column for PULL String lastupdatedwm6 = results.get(0).get('lastupdatedwm6') //store the lastupdatedwm6 to file def lastupdatedwm6aft = new File(RunConfiguration.getProjectDir() + "/Data Files/lastupdatedwm6aft.txt") lastupdatedwm6aft.newWriter().withWriter { it << lastupdatedwm6 } println lastupdatedwm6aft.text WebUI.delay(2) //Read data before drugsync Pull def lastupdatedwm6bef = new File(RunConfiguration.getProjectDir() + "/Data Files/lastupdatedwm6bef.txt") Date data1 = sdf.parse(lastupdatedwm6bef.text); Date data2 = sdf.parse(lastupdatedwm6aft.text); long diffInMillies = Math.abs(data2.getTime() - data1.getTime()); long minute_millis_2 = 2 * 60 * 1000; long diffTime = minute_millis_2 - diffInMillies; assertTrue(diffTime > 0);

Answer 1

@Test
public void test() throws ParseException {
    SimpleDateFormat sdf = new SimpleDateFormat("yyyy-mm-dd hh:mm:ss", Locale.ENGLISH);

    Date data1 = sdf.parse("2020-03-16 09:08:49");
    Date data2 = sdf.parse("2020-03-16 09:09:49");

    long diffInMillies = Math.abs(data2.getTime() - data1.getTime());
    long minute_millis_2 = 2 * 60 * 1000;
    long diffTime = minute_millis_2 - diffInMillies;
    assertTrue(diffTime > 0);
}

Answer 2

另一种方法是使用超时。

// This will fail if the test takes more than 1000 milliseconds
@Test(timeout=1000)
public void test() {
  objectToTest.methodThatTakesALongTime();
}