Spring CrudRepository findBy *方法没有返回任何结果

Question

我已经实现了扩展CrudRepository的存储库。底层模型（Account）有一个plaidAccountId字段，相应的查询方法是findByPlaidAccountId。最好的我可以告诉我，我已正确连线，因为其他模型上的其他查询正常工作。但是，此查询不返回任何结果。我已手动验证数据库中存在的项目是否与提供的查询参数匹配，但仍返回null。

以下是相关配置。让我知道，如果我发布任何其他有用的内容。提前致谢。

我正在执行的查询：

accountService.findByPlaidAccountId(account.getAccountId()

我已经记录以验证accountService已初始化并且account.getAccount()提供了预期的字符串值。

// AccountRepository.java
@Repository
public interface AccountRepository extends CrudRepository<Account, Long> {

    Set<Account> findAllByUser(User user);
    Account findByPlaidAccountId(String plaidAccountId);
    Account findById(int id);
    Account findAccountByPlaidAccountId(String plaidAccountId);
}

--

// AccountService.java
@Service
public class AccountService {

    private AccountRepository accountRepository;

    @Autowired
    public AccountService(AccountRepository repository) {
        this.accountRepository = repository;
    }

    public Account findById(int id) {
        return accountRepository.findById(id);
    }

    public Account findByPlaidAccountId(String plaidAccountId) {
        return accountRepository.findByPlaidAccountId(plaidAccountId);
    }

    public Iterable<Account> findAll() {
        return accountRepository.findAll();
    }

    public Set<Account> findAllByUser(User user) {
        return accountRepository.findAllByUser(user);
    }

    public void saveAccount(Account account) {
        accountRepository.save(account);
    }

    public Account findAccountByPlaidAccountId(String plaidAccountId) {
        return accountRepository.findAccountByPlaidAccountId(plaidAccountId);
    }
}

--

// Account.java
@Entity
@Table(name = "accounts")
public class Account {

@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
@Column(name = "account_id")
private int id;

@ManyToOne(fetch = FetchType.EAGER)
@JoinColumn(name = "user_id", nullable = false)
private User user;

@Column(name = "plaid_account_id")
@NotEmpty(message = "Plaid account number is required")
private String plaidAccountId;

@Column(name = "account_type")
private String accountType;

@Column(name = "account_subtype")
private String accountSubtype;

@Column(name = "institution_id")
private String institutionId;

@Column(name = "current_balance")
private double currentBalance;

@Column(name = "available_balance")
private double availableBalance;

@Column(name = "account_limit")
private double accountLimit;

@Column(name = "name")
private String name;

@Column(name = "official_name")
private String officialName;

@Column(name = "mask")
private String mask;

@ManyToOne
@JoinColumn(name = "plaid_item_id", nullable = false)
private PlaidItem plaidItem;

public Account() {
}
// ... getters and setters

更新：JPA SQL日志记录

Hibernate: select account0_.account_id as account_1_0_, account0_.account_limit as account_2_0_, account0_.account_subtype as account_3_0_, account0_.account_type as account_4_0_, account0_.available_balance as availabl5_0_, account0_.current_balance as current_6_0_, account0_.institution_id as institut7_0_, account0_.mask as mask8_0_, account0_.name as name9_0_, account0_.official_name as officia10_0_, account0_.plaid_account_id as plaid_a11_0_, account0_.plaid_item_id as plaid_i12_0_, account0_.user_id as user_id13_0_ from accounts account0_ where account0_.plaid_account_id=?

Answer 1

另一答案

Answer 2

另一答案