我有以下实体,我将其用作对 Controller 的请求之一的目标 POJO:
Entity
@Table(name="user_account_entity")
@JsonIgnoreProperties(ignoreUnknown = true)
@JsonSerialize(using = UserAccountSerializer.class)
public class UserAccountEntity implements UserDetails {
//...
private String username;
private String password;
@PrimaryKeyJoinColumn
@OneToOne(mappedBy= "userAccount", cascade = CascadeType.ALL, fetch = FetchType.LAZY)
private UserEntity user;
@PrimaryKeyJoinColumn
@OneToOne(mappedBy= "userAccount", cascade = CascadeType.ALL, fetch = FetchType.LAZY)
private UserAccountActivationCodeEntity activationCode;
@JsonCreator
public UserAccountEntity(@JsonProperty(value="username", required=true) final String username, @JsonProperty(value="password", required=true) final String password) {
//....
}
public UserAccountEntity() {}
//.....
}
当我在请求中放入意外字段时,它会抛出 MismatchedInputException
并失败并显示以下消息:
com.fasterxml.jackson.databind.exc.MismatchedInputException: Cannot construct instance of `com.myproject.project.core.entity.userAccountActivationCode.UserAccountActivationCodeEntity` (although at least one Creator exists): no String-argument constructor/factory method to deserialize from String value ('9WL4J')
at [Source: (PushbackInputStream); line: 4, column: 20] (through reference chain: com.myproject.project.core.entity.userAccount.UserAccountEntity["activationCode"])
在我的 Controller 中:
@InitBinder
public void binder(WebDataBinder binder) {
binder.addValidators(new CompoundValidator(new Validator[] {
new UserAccountValidator(),
new UserAccountActivationCodeDTOValidator() }));
}
我向其发出请求的端点是:
@Override
public UserAccountEntity login(@Valid @RequestBody UserAccountEntity account,
HttpServletResponse response) throws MyBadCredentialsException, InactiveAccountException {
return userAccountService.authenticateUserAndSetResponsenHeader(
account.getUsername(), account.getPassword(), response);
}
更新 1
UserAccountSerializer
的代码:
public class UserAccountSerializer extends StdSerializer<UserAccountEntity> {
public UserAccountSerializer() {
this(null);
}
protected UserAccountSerializer(Class<UserAccountEntity> t) {
super(t);
}
@Override
public void serialize(UserAccountEntity value, JsonGenerator gen,
SerializerProvider provider) throws IOException {
gen.writeStartObject();
gen.writeStringField("id", value.getId());
gen.writeStringField("username", value.getUsername());
gen.writeEndObject();
}
}
最佳答案
错误被触发是因为你的 json 中有:
"activationCode" : "9WL4J"
但是 Jackson 不知道如何将字符串“9WL4J”映射到对象 UserAccountActivationCodeEntity
我猜字符串“9WL4J”是 UserAccountActivationCodeEntity 的主键 ID 的值,在这种情况下,您应该在 json 中包含:
"activationCode" : {"id" : "9WL4J"}
如果不是这种情况,请使用自定义反序列化器告诉 Jackson 如何将字符串映射到对象。您可以在实体上使用 @JsonDeserialize。
关于java - JsonIgnoreProperties 不适用于 JsonCreator 构造函数,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/52661057/