我想使用com.fasterxml.jackson.databind.ObjectMapper对不可变对象进行序列化和反序列化。

不可变类看起来像这样(只有3个内部属性，getter和constructor)：

public final class ImportResultItemImpl implements ImportResultItem {

private final ImportResultItemType resultType;

private final String message;

private final String name;

public ImportResultItemImpl(String name, ImportResultItemType resultType, String message) {

super();

this.resultType = resultType;

this.message = message;

this.name = name;

}

public ImportResultItemImpl(String name, ImportResultItemType resultType) {

super();

this.resultType = resultType;

this.name = name;

this.message = null;

}

@Override

public ImportResultItemType getResultType() {

return this.resultType;

}

@Override

public String getMessage() {

return this.message;

}

@Override

public String getName() {

return this.name;

}

}

但是当我运行这个单元测试时：

@Test

public void testObjectMapper() throws Exception {

ImportResultItemImpl originalItem = new ImportResultItemImpl("Name1", ImportResultItemType.SUCCESS);

String serialized = new ObjectMapper().writeValueAsString((ImportResultItemImpl) originalItem);

System.out.println("serialized: " + serialized);

//this line will throw exception

ImportResultItemImpl deserialized = new ObjectMapper().readValue(serialized, ImportResultItemImpl.class);

}

我得到这个例外：

com.fasterxml.jackson.databind.JsonMappingException: No suitable constructor found for type [simple type, class eu.ibacz.pdkd.core.service.importcommon.ImportResultItemImpl]: can not instantiate from JSON object (missing default constructor or creator, or perhaps need to add/enable type information?)

at [Source: {"resultType":"SUCCESS","message":null,"name":"Name1"}; line: 1, column: 2]

at

... nothing interesting here

这个异常要求我创建一个默认构造函数，但这是一个不可变的对象，所以我不想拥有它。它将如何设置内部属性？这将完全混淆API的用户。

所以我的问题是：我可以以某种方式去/不序列化不变的对象没有默认的构造函数？