无法在杰克逊中构造“javax.naming.Name”的实例

Question

我们使用 Spring LDAP 模块开发 Spring 引导和 oracle LDAP (Oid)。使用 spring 连接到 LDAP 是可以的，但是当我们想使用 ldapTemplate.create() 方法创建 LDAP 组时，我们从 Jackson 那里得到了这个错误：请帮我们看看这里出了什么问题！谢谢

Type definition error: [simple type, class javax.naming.Name]; nested exception is com.fasterxml.jackson.databind.exc.InvalidDefinitionException: Cannot construct instance of javax.naming.Name (no Creators, like default construct, exist)

//Group Entry
    @Getter
    @Setter
    @Entry(objectClasses = {"top", "groupOfUniqueNames"}, base = "cn=Groups")
    public final class Group {

        private static final String BASE_DN = "dc=eis,dc=msc,dc=ir";

        @Id
        private Name dn;

        @Attribute(name="cn")
        @DnAttribute("cn")
        private String name;

        @Attribute(name = "displayName")
        private String description;

        @Attribute(name = "owner")
        private String owner;

        @Attribute(name="uniqueMember")
        private Set members;

        public Group() {
        }

        public Group(String name, Set members) {
                Name dn =  LdapNameBuilder.newInstance(BASE_DN)
                        .add("ou", "groups")
                        .add("cn", name)
                        .build();
                this.dn = dn;
                this.name = name;
                this.members = members;
        }

        public Group(Name dn, String name, String description, String owner, Set members) {
            this.dn = dn;
            this.name = name;
            this.description = description;
            this.owner = owner;
            this.members = members;
        }

        public void addMember(Name member) {
            if (this.members == null){
                this.members = new HashSet<>();
            }
            members.add(member);
        }

        public void removeMember(Name member) {
            members.remove(member);
        }

        @Override
        public String toString() {
            return "Group{" +
                    "dn=" + dn +
                    ", name='" + name + '\'' +
                    ", members=" + members +
                    '}';
        }

    //Service

    @Override
        public int createOIDGroupByJob(Group group) {
            try {
                ldapTemplate.create(group);
                return 1;
            }catch (Exception e){
                return 0;
            }
        }

    @PostMapping("/api/ldap/group")
            public ResponseEntity<?> add(@RequestBody Group group) {
                int retVal = ldapService.createOIDGroupByJob(group);
                if (retVal==1) {
                    return new ResponseEntity<>(HttpStatus.OK);
                } else {
                    return new ResponseEntity<>(HttpStatus.EXPECTATION_FAILED);
                }
            }
//in postman

    {
    "dn": {
        "rdns": [
            {
                "value": "Groups",
                "type": "cn"
            },
            {
                "value": "some data",
                "type": "cn"
            }
        ]
    },
    "fullName": "some data",
    "lastName": "some data"

}

Answer 1

终于！！我解决了问题！问题是杰克逊想要反序列化一个接口！ (javax.naming.Name) 它没有醒来！

我发现解决方案是在构造函数上使用 @JsonCreator，如下所示：

//User Ldap Entry
 public User(Name dn, String name, String lastName, String group) {
        this.dn = dn;
        this.name = name;
        this.lastName = lastName;
        this.group = group;
    }

    @JsonCreator
    public User(@JsonProperty("dn") @JsonDeserialize(as=LdapName.class) final Name dn) {
        this.dn = dn;
    }

通过上面的示例，在 @JsonDeserialize(as=LdapName.class) 中，我传递了 LdapName.class ，这是 Name 实现之一！它工作正常:)

无法在杰克逊中构造“javax.naming.Name”的实例

Cannot construct instance of `javax.naming.Name` in jackson

spring

ldap

spring-ldap

jackson

spring-boot