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I have a parameter of integers. I need to write a custom Spring JPA query where the equivalent SQL would be . . . AND x.propertyStatusId IN (1,2) . . .

I see in the Spring Data JPA documentation the keyword is "IN" as in "findByAgeIn(Collection<Age> ages)" that translates to "where x.age in ?1" But I get an error. I am using an arrayList of integers. 

ArrayList<Integer> propertyStatusId = new ArrayList<Integer>();
propertyStatusId.add(1);
propertyStatusId.add(2);
findByPropertyOwnerIdAndIn(propertyStatusId)AndListingDateGreaterThanEqual(user.getId(),propertyStatusId, ListingDate, pageable);
Hien Nguyen
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STP
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    Possible duplicate of [Spring CrudRepository findByInventoryIds(List inventoryIdList) - equivalent to IN clause](https://stackoverflow.com/questions/18987292/spring-crudrepository-findbyinventoryidslistlong-inventoryidlist-equivalen) – Scrobot May 08 '19 at 04:22

1 Answers1

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Assuming that you have an entity called Product defined as below:

@Entity
public class Product {

    @Id
    @GeneratedValue
    private Integer id;
    private String title;
    private String description;
    private Date createDate;

// getters & setters 

}

If you would need to query based on title, List of ids and greater than or equal to createDate then you can define the method like below:

List<Product> findByTitleAndIdInAndCreateDateGreaterThanEqual(String title, List<Integer> ids, Date createDate);
snmaddula
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  • yes this compiled and helped me get past the compilation errors around the construction of this specific part. I am still having an issue with the error about it doesn't recognize the property name for some reason. But I will resolve that as I have had that before ( I am aware it is the entity name not the SQL name etc). Thanks – STP May 08 '19 at 17:16
  • I'm glad to hear this. Thank you @STP! – snmaddula May 08 '19 at 17:52