In my reservation-entity i have a column "bookingDate" --> example: "2021-05-10 12:00:00".
So in this object the date and starttime of an user-booking gets displayed.
If a user wants to book a timeslot, i want to check first if the selected timeslot is empty. So i want to query the database by date&startTime.
I tried it with https://www.baeldung.com/spring-data-jpa-query-by-date , but it didnt work. I got the errors: "The annotation #Temporal is disallowed for this location" & "#Temporal cant be used for variables"
these are the relevant classes:
Reservation.java
#Entity
public class Reservation {
#Id #GeneratedValue
private int reservationId;
#DateTimeFormat(iso = DateTimeFormat.ISO.DATE_TIME)
private LocalDateTime bookingDate;
private int court = 1;
private String playerNames;
private int userIdReservation;
//getter and setters
With the method "findByBookingDate()" i want to query the database, if the selected timeslot is empty...
VerificationClass.java
public boolean validateReservation(Reservation r) {
LocalDateTime tempDate = r.getBookingDate();
if(reservationRepository.findByBookingDate(tempDate)){ // todo: + and Court
logger.getLogger().info(this.getClass().getName() + "||Booking Slot is empty -- Reservation created||");
return true;
}
logger.getLogger().info(this.getClass().getName() + "||Booking Slot is full -- Reservation failed||");
return false;
}
ReservationRepository.java
#Repository
#Repository
public interface ReservationRepository extends JpaRepository<Reservation, Integer>{
#Query("Select r from reservation r where r.booking_date = :tempDate")
boolean findByBookingDate(#Param("tempDate") LocalDateTime tempDate);
}
If I run it like this i always get an "org.springframework.beans.factory.UnsatisfiedDependencyException: Error creating bean with name 'backyardcodersSpringReactApplication'" --> so the application does not successfully start up.
Im very thankful for every tip and critique!
cheers!
Not understood completely. this is just a lead maybe not a perfect solution.
You can use java.time.LocalDateTime . and annotation be like #DateTimeFormat(iso = DateTimeFormat.ISO.DATE_TIME).
And the query should be like. [the query will check all the reservation for that day]
Select
from reservation_table
Where
timeSlot between ‘2021-05-10 00:00:00’ and ‘2021-05-10 23:59:59’
I copied your code in my local file. Instead of
import org.springframework.data.jpa.repository.Temporal;
I used
import javax.persistence.Temporal;
in your Reservation.java file.
Also this is my very first answer on Stackoverflow.
First of all #Temporal have three different arguments and it can be applied to the variable of type date , time and time stamp.
Usage
#Temporal(Temporal type.DATE)
private Date date;
#Temporal(Temporal type.TIMESTAMP) // INSERT BOTH DATE WITH TIME TILL MILLISECONDS
private Calendar date;
Why not just just extract the localdate from your LocalDateTime and pass it on? and extract the hour and pass it on and query 2 different columns with it.
LocalDateTime.toLocalDate()
LocalDateTime.getHour()
Related
I'm trying to select all entities where the value of the date field is below a given argument. However, my code doesn't seem to work.
SomeEntity.java
#Entity
public class SomeEntity {
#Id #GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
private ZonedDateTime date = ZonedDateTime.now(ZoneId.of("UTC"));
}
SomeEntityRepository.java
public interface SomeEntityRepositoryextends CrudRepository<SomeEntity, Long> {
/**
* Get all SomeEntity created before :until
*/
#Query("Select e from SomeEntity e where e.date < :until")
List<SomeEntity> findAllUntil(#Param("until") ZonedDateTime until);
}
My problem is this:
I will have a SomeEntity object where the date is 2019-08-31T10:00:00.000+02:00 and the value of until is 2019-08-24T10:00:00.000+02:00 (seven days earlier)
However, the object is still returned from the query.
I am using PostgreSQL 9.6
Not every JDBC driver supports it. You can use one of following solutions:
Use TimeStamp instead of ZonedDateTime.
Use proper driver and follow its configuration requirements. For instance, for Oracle you may need to specify converter:
#Convert(converter=OracleZonedDateTimeSeriliazer.class)
private ZonedDateTime date = ZonedDateTime.now(ZoneId.of("UTC"));
For other databases you may need to implement such a convertor and specify it via #Convert.
In my Spring application I need to design a data model which is having date range. Is there any data type which can be used?
Currently I did something like this.
#Entity
#Table(name = "scheduled_period")
public class ScheduledPeriod {
private Date oneDay;
private Date fromDate;
private Date toDate;
}
I want to put fromDate and toDate together in to something like dateRange
Best solution is yours, by using LocalDate.
By using two LocalDate to Start Date and End Date.
and working with them, see:
HQL / JPA find available items between date range
some DBMS's like PostgreSQL has datatype named: daterange
for example:
CREATE TABLE reservation (room int, during daterange);
INSERT INTO reservation VALUES
(1108, '[2010-01-01 14:30, 2010-01-01 15:30)');
but Java (JPA) does not support datarange data type.
If you generate that table into Entity, it generated to Object.
private Object during;
public Object getDuring() {
return this.during;
}
public void setDuring(Object during) {
this.during = during;
}
finally, if you persist to use rangedate, you can define a new class (Entity) named RangeDate with two LocalDates or use array of LocalDate (if your DMBS support it).
I am storing objects with timestamp in the database(Realm).
public class Met extends RealmObject{
private String name;
private int met;
private long timestamp;
}
I want to show them date wise, like grouping them by date.
Since it is a timestamp and will be different for rows of the same day, I am not able to get it to work.
This comes from the backend and I cannot change it to date.
The only idea I have is to add an extra date field, so that it would be easy to query.
Is there a way to achieve this at the query level without any extra fields?
To query by the same day, you could easily set up a query that queries between the start of the day and the start of the next day.
public class Met extends RealmObject {
private String name;
private int met;
#Index
private long timestamp;
}
And
Date startOfDay = //...get start of day
Date startOfNextDay = //... get start of next day;
RealmResults<Met> mets = realm.where(Met.class)
.greaterThanOrEqualTo(MetFields.TIMESTAMP, startOfDay.getTime())
.lessThan(MetFields.TIMESTAMP, startOfNextDay.getTime())
.findAllSorted(MetFields.TIMESTAMP, Sort.ASCENDING);
Using Hibernate 3.5.3:
In the database is a function (that uses Oracle's pipelined table functions) which returns a number of rows and values when supplying the ID of the entity. (This function is part of legacy code which can't be changed).
Result example:
select * from table(DateOptions_Function('ID_OF_ENTITY'));
OPTION DATE
-----------------------
startDate 2012/09/01
endDate 2013/04/01
otherDate 2011/01/01
I want to map the result of a #Formula (containing the above SQL) to an object on the entity.
public class DateOptions {
private LocalDate startDate;
private LocalDate endDate;
private LocalDate otherDate;
// getters and setters omitted
}
I want to map it like so in the entity:
#Formula("(select * from table(DateOptions_Function(ENTITY_ID)))")
#Type(type = "mypackage.DateOptionsUserType")
public DateOptions getDateOptions() {
return dateOptions;
}
I have tried creating a Hibernate UserType with the hope of creating a DateOptions object using the ResultSet in nullSafeGet(...).
I specify the sql types in my DateOptionsUserType
public int[] sqlTypes() {
return new int[] {Types.VARCHAR, Types.DATE};
}
I keep getting the following exception on startup though:
org.hibernate.MappingException: property mapping has wrong number of columns: mypackage.MyEnity.dateOptions type: mypackage.DateOptionsUserType (I have also tried CompositeUserType with the same result).
Any idea what might be causing the issue? Is it even possible (mapping a #Formula to custom non-entity object)?
since you basicly have 3 fields with a formula i would see if executing the function 3 times is still fast enough
#Formula("(select DATE from table(DateOptions_Function(ENTITY_ID))) WHERE option='startDate'")
private LocalDate startDate;
#Formula("(select DATE from table(DateOptions_Function(ENTITY_ID))) WHERE option='endDate'")
private LocalDate endDate;
#Formula("(select DATE from table(DateOptions_Function(ENTITY_ID))) WHERE option='otherDate'")
private LocalDate otherDate;
I have a table with users and are trying to get a list with the people who have birthday today so the app can send an email.
The User is defined as
#Entity
public class User {
#Size(max = 30)
#NotNull
private String name;
[...]
#Temporal(TemporalType.DATE)
#DateTimeFormat(style = "S-")
protected Date birthday;
}
and I've got a method which returns the people which were born today like so
#SuppressWarnings("unchecked")
public static List<User> findUsersWithBirthday() {
List<User> users =
entityManager().createQuery("select u from User u where u.birthday = :date")
.setParameter("date", new Date(), TemporalType.DATE)
.getResultList();
return users;
}
This is fine and all for finding people which were born today, however tha's not really that useful, so I've been struggling for a way to find all the users that were born today, independent of the year.
Using MySQL there's functions I can use like
select month(curdate()) as month, dayofmonth(curdate()) as day
However I've been struggling to find a JPA equivalent to that.
I'm using Spring 3.0.1 and Hibernate 3.3.2
I don't know how to do that in JPQL but HQL has day() and month() functions so you could run:
from User u
where day(u.birthday) = day(CURRENT_DATE)
and month(u.birthday) = month(CURRENT_DATE)
If using HQL is not an option, I would go for a native query.