springboot使用JpaRepository实现简单的条件查询

分3个步骤，如下：
1 编写实体

@Entity
public class user {
    
    
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private  Integer userid;
    @Column(length = 16)
    private  String pwd;
    @Column(length = 30)
    private  String username;
    @Column(length = 11)
    private  String phone;
    private  Integer stutas;

    public Integer getUserid() {
    
    
        return userid;
    }

    public String getPwd() {
    
    
        return pwd;
    }

    public String getUsername() {
    
    
        return username;
    }

    public String getPhone() {
    
    
        return phone;
    }

    public Integer getStutas() {
    
    
        return stutas;
    }

    public void setUserid(Integer userid) {
    
    
        this.userid = userid;
    }

    public void setPwd(String pwd) {
    
    
        this.pwd = pwd;
    }

    public void setUsername(String username) {
    
    
        this.username = username;
    }

    public void setPhone(String phone) {
    
    
        this.phone = phone;
    }

    public void setStutas(Integer stutas) {
    
    
        this.stutas = stutas;
    }
}

2 继承JpaRepository（根据属性名查询，在继承JpaRepository中写findBy+属性,如数据库表里有一个属性为phone重写方法为findByphone即可，等价于sql语句的select * from 表(对象实体）where phone=参数(findByphone的参数确定)）

public interface userDao extends JpaRepository<user,Integer> {
    
    
    user findByphone(String phone);
}

3操作数据库`

@Controller
public class loginAction {
    
    
    @Autowired
    userDao userdao;
    @GetMapping("/login")
    @ResponseBody
    public String loogin(String phone,String password ,user User){
    
    
        User=userdao.findByphone(phone);
      System.out.println(User.getUserid());
        return User.toString();
    }
}

需要注意的一点就是，实体的属性名必须是小写字母（大写或者下划线都会报错）