'How to set current date in yaml file

I am creating a dataset in yml format to initialize a table in a postgresql database

- id : 5
    description : "test5"
    is_deleted : false
    last_redaction_date : 2021-12-13 15:09:52.715570
    persist_date : 2022-03-02 15:09:52.715570
    title : "test5"
    user_id : 15

How can I set the current date in the required format to the persist_date variable so as not to manually enter the date each time?

@AutoConfigureMockMvc
@SpringBootTest(classes = JmApplication.class)
public class TestQuestionResourceController extends AbstractClassForDRRiderMockMVCTests {

    @Autowired
    private MockMvc mvc;

      @Test
      @DataSet(cleanBefore = true,
                value = {
                        "dataset/testQuestionResourceController/question_different_date.yml",
                   
                },
                strategy = SeedStrategy.CLEAN_INSERT
        )
        public void getQuestionSortedByWeightForTheWeek() throws Exception {
javapostgresqlyamldbunitspring-test-dbunit


Solution 1:[1]

You can get the current date and time from ZonedDateTime:

ZonedDateTime now = ZonedDateTime.now();

And then you can format the time using DateTimeFormatter's ISO_LOCAL_DATE and ISO_LOCAL_TIME:

String currentDateTime = DateTimeFormatter.ISO_LOCAL_DATE.format(now) +
    DateTimeFormatter.ISO_LOCAL_TIME.format(now);

Which produces "2022-03-14 12:40:32.8071372".

Sources

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Source: Stack Overflow

Solution Source
Solution 1 David Conrad

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