英文:
How to deserialize complex JSON to java object
问题
我需要将JSON反序列化为Java类。
我有以下JSON：
{"data": {"text": "John"},"fields":[{"id": "testId","name": "fieldName","options": {"color": "#000000","required": true}},{"id": "testId","name": "fieldName1","options": {"color": "#000000","required": false}}]}
我需要将这个JSON（仅限于"fields"部分）反序列化为以下Java类：
public class Field {public final String id;public final String name;public final String color;public final boolean required;}
并且我需要获得以下类似的结果：
// 键是字段对象中的id（它可以在多个对象中相同）Map<String, List<Field>> fields = objectMapper.readValue(json, new TypeReference<Map<String, List<Field>>>() {});
如何使用Jackson做到这一点？
英文:
I need to deserialize JSON to java class.<br/>
I have JSON like the following:
{"data": {"text": "John"},"fields":[{"id": "testId","name": "fieldName","options": {"color": "#000000","required": true}},{"id": "testId","name": "fieldName1","options": {"color": "#000000","required": false}}]}
and I need to deserialize this JSON (only "fields" section) to java class like the following:
public class Field {public final String id;public final String name;public final String color;public final boolean required;}
and I need to get something like the following:
// The key is the id from field object (it can be the same in the multiple objects.)Map<String, List<Field>> fields = objectMapper.readValue(json, Map<String, List<Field>>);
How can I do it using Jackson?
答案1
得分: 1
> 只要jackson不支持@JsonWrapped，您就必须使用以下解决方法。
首先，您需要创建一个包含fields的自定义类：
public class Fields {public List<Field> fields;}
根据您的ObjectMapper配置，您需要向Fields类添加@JsonIgnoreProperties(ignoreUnknown = true)，以忽略任何其他属性。
接下来，您必须定义仅用于临时使用的嵌套Options类：
public class Options {public String color;public boolean required;}
最后，将此构造函数添加到您的Field类中：
@JsonCreatorpublic Field(@JsonProperty("id") String id, @JsonProperty("name") String name, @JsonProperty("options") Options options){this.id = id;this.name = name;this.color = options.color;this.required = options.required;}
@JsonCreator注解告诉jackson需要使用此构造函数进行反序列化。此外，@JsonProperty注解是必需的，因为构造函数和方法的参数不会保留在字节码中。
然后，您可以像这样反序列化您的json数据：
List<Field> fields = objectMapper.readValue(json, Fields.class).fields;
英文:
> As long as jackson doesn't support @JsonWrapped, you have to use the following work around.
First you need to create a custom class which contains the fields:
public class Fields {public List<Field> fields;}
Depending on your ObjectMapper configuration you have to add @JsonIgnoreProperties(ignoreUnknown = true) to the Fields class, to ignore any other properties.
Next is that you have to define the nested Options class which is solely used temporarily:
public class Options {public String color;public boolean required;}
And at last add this constructor to your Field class:
@JsonCreatorpublic Field(@JsonProperty("id") String id, @JsonProperty("name") String name, @JsonProperty("options") Options options){this.id = id;this.name = name;this.color = options.color;this.required = options.required;}
The @JsonCreator annotation indicates to jackson that this constructor needs to be used for the deserialization. Also the @JsonProperty annotations are required as arguments to constructors and methods are not preserved in the bytecode
Then you can deserialize your json just like this:
List<Field> fields = objectMapper.readValue(json, Fields.class).fields;
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