英文:
How to deserialize generic List<T> with Jackson?
问题
我多年来一直在使用Jackson对对象进行序列化/反序列化,但我始终觉得使用TypeReference<T>
来反序列化List
等内容非常复杂。我创建了一个简单的辅助函数:
public static <T> TypeReference<List<T>> list() {
return new TypeReference<List<T>>(){}
}
预期的用法如下:
List<Foo> foos = objectMapper.readValue(json, list());
它可以工作!在调试器中检查时,与其是Foo
的列表,它实际上是LinkedHashMap
的列表。我理解ObjectMapper
在类型为Object
时反序列化为LinkedHashMap
,我在这里阅读了关于这一点的解释:
https://stackoverflow.com/questions/6846244/jackson-and-generic-type-reference
然而,为什么它能够将List<LinkedHashMap>
赋值给List<Foo>
呢?至少这不应该引发某种形式的ClassCastException
吗?
另外,是否有任何方法可以在Java的类型系统中实现这一点?
注意:下面的方法声明存在相同的问题,因为额外的参数对于确定T
并不是必需的:
public static <T> TypeReference<List<T>> listOf(Class<T> ignored) {
return new TypeReference<List<T>>(){}
}
英文:
I've been using Jackson to serialize/deserialize objects for years and have always found it needlessly complicated to use TypeReference<T>
to deserialize List
etc. I created a simple helper function:
public static <T> TypeReference<List<T>> list() {
return new TypeReference<List<T>>(){}
}
With intended use:
List<Foo> foos = objectMapper.readValue(json, list());
And it works! Kind of. When inspecting through the debugger, rather than a list of Foo
, it is rather a list of LinkedHashMap
. I understand that ObjectMapper
deserializes into LinkedHashMap
for type Object
and I read the explanation for that here:
https://stackoverflow.com/questions/6846244/jackson-and-generic-type-reference
However, why is it able to assign List<LinkedHasMap>
to a List<Foo>
? At the very least shouldn't that be some sort of ClassCastException
?
Also, is there anyway to do this with Java's type system?
NOTE: the following method declaration has the same issue, which makes sense because the additional argument is not needed for T
to be determined:
public static <T> TypeReference<List<T>> listOf(Class<T> ignored) {
return new TypeReference<List<T>>(){}
}
答案1
得分: 12
这是由于Java
中的类型擦除(type erasure)机制导致的。在阅读本答案的后续部分之前,请先阅读以下内容:
根据你现在可能已经了解的情况,阅读上述文章后,你的方法在编译后会变成这样:
static <T> TypeReference<List> listOf(Class<T> ignored) {
return new TypeReference<List>(){};
}
Jackson
将尝试找到最适合的类型,对于JSON
对象来说,这个类型将是java.util.LinkedHashMap
。要创建无法否定的类型,你需要使用com.fasterxml.jackson.databind.type.TypeFactory
类。请参考以下示例:
import com.fasterxml.jackson.core.type.TypeReference;
import com.fasterxml.jackson.databind.JavaType;
import com.fasterxml.jackson.databind.ObjectMapper;
import com.fasterxml.jackson.databind.type.TypeFactory;
import java.io.File;
import java.util.List;
public class JsonTypeApp {
public static void main(String[] args) throws Exception {
File jsonFile = new File("./resource/test.json").getAbsoluteFile();
ObjectMapper mapper = new ObjectMapper();
System.out.println("使用'TypeFactory'尝试");
List<Id> ids = mapper.readValue(jsonFile, CollectionsTypeFactory.listOf(Id.class));
System.out.println(ids);
Id id1 = ids.get(0);
System.out.println(id1);
System.out.println("使用'TypeReference<List<T>>'尝试");
List<Id> maps = mapper.readValue(jsonFile, CollectionsTypeFactory.erasedListOf(Id.class));
System.out.println(maps);
Id maps1 = maps.get(0);
System.out.println(maps1);
}
}
class CollectionsTypeFactory {
static JavaType listOf(Class clazz) {
return TypeFactory.defaultInstance().constructCollectionType(List.class, clazz);
}
static <T> TypeReference<List> erasedListOf(Class<T> ignored) {
return new TypeReference<List>(){};
}
}
class Id {
private int id;
// getters, setters, toString
}
对于下面的JSON
数据:
[
{
"id": 1
},
{
"id": 22
},
{
"id": 333
}
]
以上示例将输出:
使用'TypeFactory'尝试
[{1}, {22}, {333}]
{1}
使用'TypeReference<List<T>>'尝试
[{id=1}, {id=22}, {id=333}]
在线程"main"中出现异常:java.lang.ClassCastException: 无法将java.util.LinkedHashMap强制转换为com.example.Id
at com.example.JsonTypeApp.main(JsonTypeApp.java:27)
另请参考:
英文:
It works like this because of type erasure in Java
. Please, read about it before you start reading next part of this answer:
As you probably know right now, after reading above articles, your method after compilation looks like this:
static <T> TypeReference<List> listOf(Class<T> ignored) {
return new TypeReference<List>(){};
}
Jackson
will try to find out the most appropriate type for it which will be java.util.LinkedHashMap
for a JSON Object
. To create irrefutable type
you need to use com.fasterxml.jackson.databind.type.TypeFactory
class. See below example:
import com.fasterxml.jackson.core.type.TypeReference;
import com.fasterxml.jackson.databind.JavaType;
import com.fasterxml.jackson.databind.ObjectMapper;
import com.fasterxml.jackson.databind.type.TypeFactory;
import java.io.File;
import java.util.List;
public class JsonTypeApp {
public static void main(String[] args) throws Exception {
File jsonFile = new File("./resource/test.json").getAbsoluteFile();
ObjectMapper mapper = new ObjectMapper();
System.out.println("Try with 'TypeFactory'");
List<Id> ids = mapper.readValue(jsonFile, CollectionsTypeFactory.listOf(Id.class));
System.out.println(ids);
Id id1 = ids.get(0);
System.out.println(id1);
System.out.println("Try with 'TypeReference<List<T>>'");
List<Id> maps = mapper.readValue(jsonFile, CollectionsTypeFactory.erasedListOf(Id.class));
System.out.println(maps);
Id maps1 = maps.get(0);
System.out.println(maps1);
}
}
class CollectionsTypeFactory {
static JavaType listOf(Class clazz) {
return TypeFactory.defaultInstance().constructCollectionType(List.class, clazz);
}
static <T> TypeReference<List> erasedListOf(Class<T> ignored) {
return new TypeReference<List>(){};
}
}
class Id {
private int id;
// getters, setters, toString
}
Above example, for below JSON
payload:
[
{
"id": 1
},
{
"id": 22
},
{
"id": 333
}
]
prints:
Try with 'TypeFactory'
[{1}, {22}, {333}]
{1}
Try with 'TypeReference<List<T>>'
[{id=1}, {id=22}, {id=333}]
Exception in thread "main" java.lang.ClassCastException: java.util.LinkedHashMap cannot be cast to com.example.Id
at com.example.JsonTypeApp.main(JsonTypeApp.java:27)
See also:
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