根据建议here我尝试在@JsonIgnore
和@JsonProperty
字段上使用import com.fasterxml.jackson.annotation.*;
和password
(来自passwordSalt
)我的AppUser
。
@Entity
@Table(name = "app_user")
public class AppUser extends AbstractTimestampEntity {
// ..
@JsonIgnore
public String getPassword() {
return password;
}
@JsonProperty
public void setPassword(String password) {
this.password = password;
}
@JsonIgnore
public String getPasswordSalt() {
return passwordSalt;
}
@JsonProperty
public void setPasswordSalt(String passwordSalt) {
this.passwordSalt = passwordSalt;
}
}
但是,出于某种原因,这两个字段始终为null
。当我尝试“注册”用户时,我发送的JSON没有完全反序列化。 password
设置为null
:
我想要实现的是能够接收用户的密码以便存储它,但同时确保存储的(加密的)密码没有被序列化并且也被发送回客户端。
我在这里做错了什么?
答案 0 :(得分:0)
对于Jackson版本2.9.0,以下内容适用于我:
@NoArgsConstructor
@AllArgsConstructor
public class Credentials {
private String username;
private String password;
public String getUsername() {
return username;
}
public void setUsername(String username) {
this.username = username;
}
@JsonProperty(access = Access.WRITE_ONLY)
public String getPassword() {
return password;
}
public void setPassword(String password) {
this.password = password;
}
}
......还有正在运行的测试示例......
@Test
public void jsonIgnoreSerialization() throws Exception {
ObjectMapper objectMapper = new ObjectMapper();
String serialized = objectMapper.writeValueAsString(new Credentials("user", "pass"));
assertThat(serialized).isEqualTo("{\"username\":\"user\"}");
Credentials credentials = objectMapper.readValue("{\"username\":\"user\",\"password\":\"pass\"}", Credentials.class);
assertThat(credentials.getUsername()).isEqualTo("user");
assertThat(credentials.getPassword()).isEqualTo("pass");
}
为简单起见,构造函数是来自Lombok的构造函数,而断言来自AssertJ。
你可以尝试一下吗?