Spring Jpa按登录用户及其ID选择 [英] Spring Jpa Choose By Logged User and his Id

问题描述

我试图发送一个由id为的稀疏用户创建的收入列表，例如，我得到了company和customer，在表收入中，它们都具有相同的clinet ID，因为它们是由用户ID定义的，所以如何我拉正确的ID，因为如果我尝试发送get ID 1，例如，它将拉我表中的所有1.ty

i tried to send an list of income that created by the spacific user that have id , i got company for example and customer , in the table income they both have the same clinet id becouse they id defined by the user id so how i pull the right id becouse if i will try to send get id 1 for example it will pull me all the 1's in the table . ty

我现在构建的一些代码可能不正确:

some code that i build for now that probably not right :

@SuppressWarnings("static-access")
    public List<Income> viewIncomeByCompany(int clientid) throws Exception {
        LoginUser loginUser = new LoginUser();
        if (loginUser.getClientType().COMPANY != null) {
            List<Income> allIncomesByCompany = incomeRepo.findAllByClientId(clientid);
            return allIncomesByCompany;
        } else {

            throw new Exception();
        }

    }

在公司财务负责人中:

    @GetMapping("/allincomecompany/{companyid}")
    public List<Income> viewIncomeByCompanyId(HttpServletRequest req) throws Exception {
        List<Income> allcompanyincome = incomeService.viewIncomeByCompany(getLoggedUser(req).getUserId());
        return allcompanyincome;

    }

收入表:

@Entity
@Table(name = "income")
public class Income {

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

    @Basic(optional = false)
    @Column(nullable = false)
    private long clientId;

    @Column(nullable = false)
    @Basic(optional = false)
    private String name;

    @Column(nullable = false)
    @Basic(optional = false)
    @JsonFormat(pattern="dd-MM-yyyy")
//  @DateTimeFormat(pattern = "dd-MM-yyyy")
    private Date date;

    @Column(nullable = false)
    @Enumerated(EnumType.STRING)
    private IncomeType description;

    @Column(nullable = false)
    @Basic(optional = false)
    private double price;

    public int getId() {
        return id;
    }

收入库:

@Repository
public interface IncomeRepo extends JpaRepository<Income, Long> {

    List<Income> findAllByClientId(int clientid);

}

推荐答案

a)

尝试:

findByClientId(Long clientId)

或:

List< T>findAll(Specification< T> spec);

b)

您正在使用

int客户ID

它在哪里:

长

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