此错误的可能性非常多，其他一些可能性也出现在添加页或编辑页上。在我的案例中，我试图保存一个对象AdvanceSalary。问题是，在编辑AdvanceSalary employee.employee_id时为空，因为在编辑时我没有设置employee。

    @Entity(name = "ic_advance_salary")
    @Table(name = "ic_advance_salary")
    public class AdvanceSalary extends BaseDO{

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

        @ManyToOne(fetch = FetchType.EAGER)
        @JoinColumn(name = "employee_id", nullable = false)
        private Employee employee;

        @Column(name = "employee_id", insertable=false, updatable=false)
        @NotNull(message="Please enter employee Id")
        private Long employee_id;

        @Column(name = "advance_date")
        @DateTimeFormat(pattern = "dd-MMM-yyyy")
        @NotNull(message="Please enter advance date")
        private Date advance_date;

        @Column(name = "amount")
        @NotNull(message="Please enter Paid Amount")
        private Double amount;

        @Column(name = "cheque_date")
        @DateTimeFormat(pattern = "dd-MMM-yyyy")
        private Date cheque_date;

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

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

        public AdvanceSalary() {
        }

        public AdvanceSalary(Integer advance_salary_id) {
            this.id = advance_salary_id;
        }

        public Integer getId() {
            return id;
        }

        public void setId(Integer id) {
            this.id = id;
        }

        public Employee getEmployee() {
            return employee;
        }

        public void setEmployee(Employee employee) {
            this.employee = employee;
        }


        public Long getEmployee_id() {
            return employee_id;
        }

        public void setEmployee_id(Long employee_id) {
            this.employee_id = employee_id;
        }

    }

除了所有其他好的答案之外，如果您使用merge来持久化一个对象，并且意外地忘记在父类中使用该对象的合并引用，那么可能会发生这种情况。考虑以下示例

merge(A);
B.setA(A);
persist(B);

在这种情况下，您合并了A，但忘记了使用A的合并对象。为了解决这个问题，您必须像这样重写代码。

A=merge(A);//difference is here
B.setA(A);
persist(B);

我刚刚收到这个错误，因为我在保存之前设置了一个未代理的实体而不是另一个实体。

我应该链接一个代理实体实例。

参见以下说明：

Child saved = childRepository.save(child);

// INCORRECT
parent.setChild(child);  // <-- 'child' is NOT managed (not proxied)

// CORRECT
parent.setChild(saved);  // <-- 'saved' is managed (proxied)

这不是错误的唯一原因。我刚才遇到了它，因为我的代码中有一个错别字，我相信它设置了一个已经保存的实体的值。

X x2 = new X();
x.setXid(memberid); // Error happened here - x was a previous global entity I created earlier
Y.setX(x2);

我通过准确查找导致错误的变量（在本例中为Stringxid）发现了错误。我在保存实体并打印痕迹的整个代码块周围使用了捕获。

{
   code block that performed the operation
} catch (Exception e) {
   e.printStackTrace(); // put a break-point here and inspect the 'e'
   return ERROR;
}

此错误的可能性非常多，其他一些可能性也出现在添加页或编辑页上。在我的案例中，我试图保存一个对象AdvanceSalary。问题是，在编辑AdvanceSalary employee.employee_id时为空，因为在编辑时我没有设置employee。

    @Entity(name = "ic_advance_salary")
    @Table(name = "ic_advance_salary")
    public class AdvanceSalary extends BaseDO{

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

        @ManyToOne(fetch = FetchType.EAGER)
        @JoinColumn(name = "employee_id", nullable = false)
        private Employee employee;

        @Column(name = "employee_id", insertable=false, updatable=false)
        @NotNull(message="Please enter employee Id")
        private Long employee_id;

        @Column(name = "advance_date")
        @DateTimeFormat(pattern = "dd-MMM-yyyy")
        @NotNull(message="Please enter advance date")
        private Date advance_date;

        @Column(name = "amount")
        @NotNull(message="Please enter Paid Amount")
        private Double amount;

        @Column(name = "cheque_date")
        @DateTimeFormat(pattern = "dd-MMM-yyyy")
        private Date cheque_date;

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

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

        public AdvanceSalary() {
        }

        public AdvanceSalary(Integer advance_salary_id) {
            this.id = advance_salary_id;
        }

        public Integer getId() {
            return id;
        }

        public void setId(Integer id) {
            this.id = id;
        }

        public Employee getEmployee() {
            return employee;
        }

        public void setEmployee(Employee employee) {
            this.employee = employee;
        }


        public Long getEmployee_id() {
            return employee_id;
        }

        public void setEmployee_id(Long employee_id) {
            this.employee_id = employee_id;
        }

    }

不要使用Cascade.All，除非你真的需要。角色和权限具有双向manyToMany关系。那么下面的代码就可以正常工作了

    Permission p = new Permission();
    p.setName("help");
    Permission p2 = new Permission();
    p2.setName("self_info");
    p = (Permission)crudRepository.save(p); // returned p has id filled in.
    p2 = (Permission)crudRepository.save(p2); // so does p2.
    Role role = new Role();
    role.setAvailable(true);
    role.setDescription("a test role");
    role.setRole("admin");
    List<Permission> pList = new ArrayList<Permission>();
    pList.add(p);
    pList.add(p2);
    role.setPermissions(pList);
    crudRepository.save(role);

而如果对象只是一个“新”对象，那么它将抛出相同的错误。