有是一堆选择取决于您的要求：

1） @JsonIgnore 可以用于避免字段的序列化。

  @OneToMany（mappedBy =manager）
 @JsonIgnore 
私人套餐<员工>雇员; 
 

2） @JsonView 可隐藏关系的一部分作为内部视图（但如果您将使用内部视图）编写JSON对象，则会出现：

  @OneToMany（mappedBy =manager）
 @JsonView（Views.Internal.class）
 private Set< Employee>雇员; 

 @ManyToOne（cascade = {CascadeType.ALL}）
 @JoinColumn（name =DepartmentID）
 @JsonView（Views.Public.class）
 private部门部门; 
 

3） 使用自定义serialiazer ，您可以确定自己构建JSON对象的规则。

4）在类上使用 @JsonIdentityInfo （以指示该类型的属性应该启用功能）以及单独的属性（以支持类型本身不能被注释的情况;或者使用不同的id生成序列）。

示例1 @JsonIdentityInfo

示例2 @JsonIdentityInfo

class employee:

@Entity
@Table(name = "Employee")
public class Employee {

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    @Column(name = "EmployeeID")
    private int EmployeeID;

    @Column(name = "ManagerID")
    private Integer ManagerID;

    @ManyToOne(cascade={CascadeType.ALL})
    @JoinColumn(name="ManagerID", insertable = false, updatable = false)
    @JsonBackReference
    private Employee manager;

    @OneToMany(mappedBy="manager")
    @JsonManagedReference
    private Set<Employee> employees;

    @ManyToOne(cascade={CascadeType.ALL})
    @JoinColumn(name = "DepartmentID")
    private Department department;

    @ManyToOne(cascade={CascadeType.ALL})
    @JoinColumn(name = "SalaryTypeID")
    private SalaryType salaryType;

    @Column(name = "Name")
    private String Name;
    //setters and getters here, wont be posting them
}

Whenever I create an instance of employee I get this infinite json error:

SEVERE: Servlet.service() for servlet [SpringMVC] in context with path 
[/SpringMVC] threw exception [Handler processing failed; nested exception is 
java.lang.***Error] with root cause
java.lang.***Error
at java.nio.CharBuffer.<init>(Unknown Source)
at java.nio.HeapCharBuffer.<init>(Unknown Source)
at java.nio.CharBuffer.wrap(Unknown Source)
at sun.nio.cs.StreamEncoder.implWrite(Unknown Source)
at sun.nio.cs.StreamEncoder.write(Unknown Source)
at sun.nio.cs.StreamEncoder.write(Unknown Source)
at java.io.OutputStreamWriter.write(Unknown Source)
at java.io.Writer.write(Unknown Source)
at com.google.gson.stream.JsonWriter.string(JsonWriter.java:534)
at com.google.gson.stream.JsonWriter.writeDeferredName(JsonWriter.java:402)
at com.google.gson.stream.JsonWriter.value(JsonWriter.java:495)

(and then it keeps going)

Since I am self referencing manager which is an employee in the employee class, how can I fix this?

There are bunch of options depends on your requirement:

1) @JsonIgnore can be used to avoid serialization of the field.

@OneToMany(mappedBy="manager")
@JsonIgnore
private Set<Employee> employees;

2) @JsonView can hide one part of the relationship as internal view (but will appear if you will write JSON object with Internal view):

@OneToMany(mappedBy="manager")
@JsonView(Views.Internal.class)
private Set<Employee> employees;

@ManyToOne(cascade={CascadeType.ALL})
@JoinColumn(name = "DepartmentID")
@JsonView(Views.Public.class)
private Department department;

3) Using custom serialiazer you can determine the rules of building your JSON object yourself.

4) Using @JsonIdentityInfo on classes (to indicate that properties of that type should have feature enabled) as well as on individual properties (to support cases where type itself can not be annotated; or to use different id generation sequence).

Example 1 @JsonIdentityInfo

Example 2 @JsonIdentityInfo