英文:
Parse JSON in Java with intermediate element ({"realKey": {"badKey": "goodValue"}}))
问题
public class MyClass {private String name;private Double number;}
英文:
I have a problem parsing a JSON to a Java class. I have read about Elasticsearch Search App APIs, and I have found responses are like this one:
{"name": {"raw": "Hello world"},"number": {"raw": 7.5}}
That is, every field is an object with a key called raw and the real value I want. I would like to have a Java class like this one:
public class MyClass {private String name;private Double number;}
I haven't seen in Jackson or any other mapping library something in order to map this JSON easily.
I would expect something like that if we use Jackson (is not a requisite, we could use any other parsing library):
ObjectMapper objectMapper = new ObjectMapper();MyClass jsonParsed = objectMapper.readValue(json, MyClass.class);
答案1
得分: 1
你可以将 JSON 属性视为 Map 并解压所需的属性并赋值给相应的属性。你应该能够做类似以下的事情：
class MyClass {private String name;private Double number;@JsonProperty("name")private void unpackNameFromNestedRawObject(Map<String, String> name) {this.name = name.get("raw");}@JsonProperty("number")private void unpackNumberFromNestedRawObject(Map<String, Double> number) {this.number = number.get("raw");}public String getName() {return name;}public void setName(final String name) {this.name = name;}public Double getNumber() {return number;}public void setNumber(final Double number) {this.number = number;}@Overridepublic String toString() {return "MyClass{" + "name='" + name + '\'' + ", number=" + number + '}';}}
英文:
You can read the JSON Property as Map and unpack the needed attribute and assign to the property. You should be able to do something like:
class MyClass {private String name;private Double number;@JsonProperty("name")private void unpackNameFromNestedRawObject(Map<String, String> name) {this.name = name.get("raw");}@JsonProperty("number")private void unpackNumberFromNestedRawObject(Map<String, Double> number) {this.number = number.get("raw");}public String getName() {return name;}public void setName(final String name) {this.name = name;}public Double getNumber() {return number;}public void setNumber(final Double number) {this.number = number;}@Overridepublic String toString() {return "MyClass{" + "name='" + name + '\'' + ", number=" + number + '}';}}
答案2
得分: 0
对于有相同疑问的任何人，我最终创建了一个类似这样的类：
public class Wrapper<T> {T value;// getters/setters}
然后我创建了我的类，类似这样：
public class MyClass {Wrapper<String> aStringField;Wrapper<Integer> aIntegerField;...}
有了这个，JSON 被正确解析。如果你想将其转换为另一个没有 RawWrappers 但具有相同字段的类，我在这种情况下使用了 Mapstruct 来指示如何从 RawWrapper<String> 转换为 String，RawWrapper<Integer> 转换为 Integer，等等，就是这样！
英文:
For anyone who have the same doubt, I finally created a class like this one:
public class Wrapper<T> {T value;// getters/setters}
And then I created my class like this one:
public class MyClass {RawWrapper<String> aStringField;RawWrapper<Integer> aIntegerField;...}
With that, JSON is parsed correctly. If then you want to transform it to another class with the same fields but without RawWrappers, in my case I have used Mapstruct to indicate how to pass from RawWrapper<String> to String, RawWrapper<Integer> to Integer, etc., and that's all!
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