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

我在你的代码中看到一些可能会产生未来问题的东西，我不知道它们是否会让你头疼，但将来会给你一个。

// AccountRepository.java 
 @Repository 
 public interface AccountRepository extends CrudRepository<Account, Long> {        
    Set<Account> findAllByUser(User user);        
    Account findByPlaidAccountId(String plaidAccountId);      
    Account findById(int id);        
    Account findByPlaidAccountId(String plaidAccountId); 
 }

第一件事是“CrudRepository<Account, Long>”，你在这里很长，但是你的班级的ID是“int”，它们应该是相同的。

Account findByPlaidAccountId(String plaidAccountId);
Account findAccountByPlaidAccountId(String plaidAccountId);

这两种方法期望返回一个Account对象，但是plaidAccountId不是唯一的或pk，所以你必须要一个列表。如果您确定只有一个元素而没有一个元素来处理您的应用程序中的列表，只需将存储库更改为Set/List<Account>，服务保持如下：

public Account findByPlaidAccountId(String plaidAccountId) {
        return accountRepository.findByPlaidAccountId(plaidAccountId).get(0);
}

要么

findFirstByPlaidAccountId

祝零回报好运;）。

Answer 2

出于某种原因，我的值已保存到数据库中，包括它们的引号，因此正确的查询在技术上是qazxsw poi。

Answer 3

如果你通过"<value>"搜索（如何调用你的方法）我想你想传递那个id而不是plaidAccountId

account.getAccountId()