Criteria API in Spring Boot

Criteria API in Spring Boot

By : Nomar Oscar Mora Tov
Date : November 18 2020, 07:00 PM
I wish this help you You should have a look at Spring Data JPA Specifications:
code :
public interface CustomerRepository 
       extends CrudRepository<Customer, Long>, JpaSpecificationExecutor {
List<T> findAll(Specification<T> spec);
public static Specification<Animal> bySizeAndType(String size, String type) {

    return new Specification<Animal>() {

      public Predicate toPredicate(Root<Animal> root, CriteriaQuery<?> query,
            CriteriaBuilder builder) {

         // JOIN Attributes
         // JOIN Detail

         if (size != null) {
           // add condition
         if (type != null) {
           // add condition

         return builder.where(...);

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How to use Criteria Queries in Spring Boot Data Jpa Application

How to use Criteria Queries in Spring Boot Data Jpa Application

By : Developer Vishakha
Date : March 29 2020, 07:55 AM
To fix the issue you can do From the docs
code :
public interface StudentRepositoryCustom {

    List<String> nameByCourse(String coursename);

class StudentRepositoryImpl implements StudentRepositoryCustom {

    private EntityManager em;

    public List<String> nameByCourse(String coursename) {            
        CriteriaBuilder cb = em.getCriteriaBuilder();
        //Using criteria builder you can build your criteria queries.

public interface StudentRepository extends CrudRepository<StudentEntity, Integer>, StudentRepositoryCustom {

Spring Boot JPA Select by criteria getting NullPointerException

Spring Boot JPA Select by criteria getting NullPointerException

By : user2153690
Date : March 29 2020, 07:55 AM
fixed the issue. Will look into that further When you use nativeQuery = true it will run your query natively so you should write query in this way:
code :
@Query(value="SELECT * FROM User u WHERE u.is_user_logged_in = 1", nativeQuery = true)
List<User> findUserLastLoggedIn();
@Query(value="SELECT u FROM User u WHERE u.isUserLoggedIn = 1")
List<User> findUserLastLoggedIn();
userController = new UserController();
 public class WalletController {

WalletRepository walletRepository;
UserRepository userRepository;

@RequestMapping(value = "/createWallet")
public ResponseEntity<String> createWallet() {
    Wallet wallet = new Wallet();
    List<User> userList = userRepository.findUserLastLoggedIn();
    if (userList == null) {
        return new ResponseEntity<>("Active user not found. You must login 
 to the system.", HttpStatus.OK);
    if (userList.size() == 1) {
        User user = null;//userList.get(0);
        WalletKey walletKey = WalletKey.getWalletKeys();
        return new ResponseEntity<>("Wallet created.", HttpStatus.OK);
    return new ResponseEntity<String>("Wallet creation failed", 
Spring Boot + Mongo - com.mongodb.BasicDocument, you can't add a second 'id' criteria

Spring Boot + Mongo - com.mongodb.BasicDocument, you can't add a second 'id' criteria

By : user3336831
Date : March 29 2020, 07:55 AM
should help you out Any ideas why I get this error when making a query: , Based on your code, you need to use either
code :
var subGenreName = mongoTemplate.find(query, SubGenre::class.java)
var subGenreName = mongoTemplate.findById(it, SubGenre::class.java)
How to use Criteria in JPA in Spring boot?

How to use Criteria in JPA in Spring boot?

By : user3477938
Date : March 29 2020, 07:55 AM
To fix this issue I am using JPA and spring boot.I am new to JPA.I want to retireve a object by passing the value of the function.But in example,it is done through hibernate config.I have not config sessionFactory bean in my class path.I want to use JPA to retrieve the object.Here in example it is: , You can use JPQL like this
code :
public Account findAccount(String userName) {
  return em.createQuery("select account from Account account where account.username = :username", Account.class)
        .setParameter("username", userName).getSingleResult();
Implement complex search feature using Spring BOOT REST and Spring Data JPA using Criteria API

Implement complex search feature using Spring BOOT REST and Spring Data JPA using Criteria API

By : Danielle
Date : March 29 2020, 07:55 AM
help you fix your problem I think the best choice, in your case, is the specification-arg-resolver lib that provide convenient way to build specification declaratively. For example, this code:
code :
public Object findByName(
            @Spec(path="registrationDate", params="registeredBefore", spec=DateBefore.class),
            @Spec(path="lastName", spec=Like.class)}) Specification<Customer> customerSpec) {

    return customerRepo.findAll(customerSpec);
GET http://myhost/customers?registeredBefore=2015-01-18&lastName=Simpson
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