使用Jackson将JSON字符串或对象反序列化为String字段 [英] Use Jackson to deserialize JSON string or object into a String field

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本文介绍了使用Jackson将JSON字符串或对象反序列化为String字段的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我正在使用Jackson 2库,我正在尝试阅读JSON响应,如下所示:

I am using Jackson 2 library and I am trying to read a JSON response, which looks like:

{ "value":"Hello" }

当值为空时,JSON响应如下:

When value is empty, JSON response looks like:

{ "value":{} }

我的模型POJO类看起来像这样

My model POJO class looks like this

public class Hello {

    private String value;

    public String getValue() {
        return value;
    }

    public void setValue(String value) {
        this.value = value;
    }

}

问题是当响应看起来像{value:{}},杰克逊试图读取一个对象,但我的模型类字段是一个字符串,所以它抛出一个异常:

The problem is that when response looks like {value:{}}, Jackson is trying to read an Object, but my model class field is a string, so it throws an Exception:

JsonMappingException: Can not deserialize instance of java.lang.String out of START_OBJECT token. 

我的问题是杰克逊如何成功阅读看起来像的JSON:

My question is how Jackson can successfully read JSONs who look like:

 {"value":"something"} 

同时如果响应看起来像这个{value:{}}(对我来说是空响应),则将null传递给我的Hello模型类的value字段。

and at the same time if response looks like this {"value":{}} (empty response for me), pass null to value field of my Hello model class.

我使用下面的代码来读取JSON字符串:

I am using the code below in order to read JSON string:

String myJsonAsString = "{...}";
ObjectMapper mapper = new ObjectMapper();
mapper.readValue(myJsonAsString , Hello.class);


推荐答案

您可以为此feld使用自定义反序列化器。如果它在那里则返回一个字符串,或者在任何其他情况下返回null:

You can use a custom deserializer for this feld. Here is one that returns the string if it's there, or null in any other case:

public class Hello {

    @JsonDeserialize(using = StupidValueDeserializer.class)
    private String value;

    public String getValue() {
        return value;
    }

    public void setValue(String value) {
        this.value = value;
    }
}

public class StupidValueDeserializer extends JsonDeserializer<String> {
    @Override
    public String deserialize(JsonParser p, DeserializationContext ctxt) throws IOException, JsonProcessingException {
        JsonToken jsonToken = p.getCurrentToken();
        if (jsonToken == JsonToken.VALUE_STRING) {
            return p.getValueAsString();
        }
        return null;
    }
}

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