如何反序列化json对象数组为json字符串数组? [英] How deserialize json object array as array of json strings?

问题描述

考虑json输入:

{
    companies: [
        {
            "id": 1,
            "name": "name1"
        },
        {
            "id": 1,
            "name": "name1"
        }
    ],
    nextPage: 2
}

如何将其反序列化为类:

How deserialize this into class:

public class MyClass {
    List<String> companies;
    Integer nextPage;
}

List<String> companies;由字符串组成:

{"id": 1,"name": "name1"}
{"id": 1,"name": "name1"}

@JsonRawValue不适用于List<String> companies;

是否可以配置 Jackson 序列化，以使公司仅使用带有注释的原始json字符串进行排列? (例如，无需编写自定义反序列化器)

Is there a way to configure Jackson serialization to keep companies array with raw json string with annotations only? (E.g. without writing custom deserializator)

推荐答案

没有针对您问题的仅注释解决方案.您必须以某种方式将JSON Object转换为java.lang.String，并且需要指定该转换.

There is no annotation-only solution for your problem. Somehow you have to convert JSON Object to java.lang.String and you need to specify that conversion.

您可以:

1. 编写自定义反序列化器，这可能是最明显的解决方案，但被禁止使用.
2. 注册自定义com.fasterxml.jackson.databind.deser.DeserializationProblemHandler并以更复杂的方式处理com.fasterxml.jackson.databind.exc.MismatchedInputException情况.
3. 实现com.fasterxml.jackson.databind.util.Converter接口并将JsonNode转换为String.这是解决问题的半注释方式，但我们不会执行最糟糕的部分-反序列化.

1. Write custom deserializer which is probably most obvious solution but forbidden in question.
2. Register custom com.fasterxml.jackson.databind.deser.DeserializationProblemHandler and handle com.fasterxml.jackson.databind.exc.MismatchedInputException situation in more sophisticated way.
3. Implement com.fasterxml.jackson.databind.util.Converter interface and convert JsonNode to String. It is semi-annotational way to solve a problem but we do not implement the worst part - deserialisation.

让我们马上转到第2点.

Let's go to point 2. right away.

解决方案非常简单:

ObjectMapper mapper = new ObjectMapper();
mapper.addHandler(new DeserializationProblemHandler() {
    @Override
    public Object handleUnexpectedToken(DeserializationContext ctxt, JavaType targetType, JsonToken t, JsonParser p, String failureMsg) throws IOException {
        if (targetType.getRawClass() == String.class) {
            // read as tree and convert to String
            return p.readValueAsTree().toString();
        }
        return super.handleUnexpectedToken(ctxt, targetType, t, p, failureMsg);
    }
});

将整个JSON读取为TreeNode，然后使用toString方法将其转换为String.有用的是，toString会生成有效的JSON.缺点是，此解决方案具有给定ObjectMapper实例的全局范围.

Read a whole piece of JSON as TreeNode and convert it to String using toString method. Helpfully, toString generates valid JSON. Downside, this solution has a global scope for given ObjectMapper instance.

此解决方案需要实现com.fasterxml.jackson.databind.util.Converter接口，该接口将com.fasterxml.jackson.databind.JsonNode转换为String:

This solution requires to implement com.fasterxml.jackson.databind.util.Converter interface which converts com.fasterxml.jackson.databind.JsonNode to String:

class JsonNode2StringConverter implements Converter<JsonNode, String> {
    @Override
    public String convert(JsonNode value) {
        return value.toString();
    }

    @Override
    public JavaType getInputType(TypeFactory typeFactory) {
        return typeFactory.constructType(new TypeReference<JsonNode>() {
        });
    }

    @Override
    public JavaType getOutputType(TypeFactory typeFactory) {
        return typeFactory.constructType(new TypeReference<String>() {
        });
    }
}

现在，您可以使用如下注释:

and now, you can use annotation like below:

@JsonDeserialize(contentConverter = JsonNode2StringConverter.class)
private List<String> companies;

解决方案2和3.几乎以相同的方式解决此问题-读取节点并将其转换回JSON，但是使用不同的方法.

Solutions 2. and 3. solve this problem almost in the same way - read node and convert it back to JSON, but uses different approaches.

如果要避免反序列化和序列化过程，可以看一下本文提供的解决方案:

If, you want to avoid deserialising and serialising process you can take a look on solution provided in this article: Deserializing JSON property as String with Jackson and take a look at:

• 如何使用数组字段到带有字符串字段的对象?
• 如何获取杰克逊使用纯文本将JSON的一部分作为纯文本
• 如何提取部分内容杰克逊(Jackson)的JSON中的原始文本?

• How to serialize JSON with array field to object with String field?
• How to get a part of JSON as a plain text using Jackson
• How to extract part of the original text from JSON with Jackson?

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