Jackson自定义反序列化器字符串到List，并删除\r\n

Question

我从一个服务获得以下JSON作为响应。我使用Spring RestTemplate调用服务，该服务还将JSON反序列化回一个对象。我试图将其反序列化为一个对象，该对象除其他字段外还有一个列表。反序列化时出现以下错误：

Can not instantiate value of type [simple type, class com.org.EmployeeInfo] from String value; no single-String constructor/factory method.

下面是我想要反序列化的JSON：

{
    "employees": {       
        "employeeInfo": [
            "{\r\n  \"id\": \"123\",\r\n  \"group\": \"MARKETING\",\r\n  \"role\": \"MANAGER\",\r\n}",
            "{\r\n  \"id\": \"256\",\r\n  \"group\": \"IT\",\r\n  \"role\": \"DIRECTOR\",\r\n}",
            "{\r\n  \"id\": \"789\",\r\n  \"group\": \"SALES\",\r\n  \"role\": \"CEO\",\r\n}"
        ]
    },
    "status": "EMPLOYED",
    "somethingElse": {
        "something": []
    }
}

如果我有列表，默认反序列化器失败

我计划编写一个自定义反序列化程序，将其反序列化为具有列表的对象，并删除响应的一部分。我该怎么做？感谢您的回复。

这是我的POJO：

public class Response {
    private Employees employees;
    private String status;
    private SomethingElse somethingElse; 

   // getters, setters
}

public class Employees {
    List<EmployeeInfo> employeeInfo;
    // getters, setters
}

public class EmployeeInfo
{
   private String id, group, role;
   // getters, setters 
}

谢谢




Answer 1

该JSON是无法解析的，因为它无效，即使您有自定义反序列化器，您也无法反序列化它。

无论如何，一个自定义反序列化器示例（不起作用，仅用于学习purpouse）

public class EmployeeInfoDeSerializer extends JsonDeserializer<List<EmployeeInfo>> {

    @Override
    public List<EmployeeInfo> deserialize(JsonParser jp, DeserializationContext ctxt) throws IOException, JsonProcessingException {

        // for learning purpouses only, does not work
        ObjectMapper mapper = new ObjectMapper();
        mapper.configure(Feature.ALLOW_UNQUOTED_FIELD_NAMES, true);

        List<EmployeeInfo> employeeInfos = new ArrayList<>();

        while (jp.nextToken() != JsonToken.END_OBJECT) {

            if (jp.getCurrentToken() != JsonToken.END_ARRAY) {

                JsonParser jParser = mapper.getJsonFactory().createJsonParser(jp.getText());
                employeeInfos.add(jParser.readValueAs(EmployeeInfo.class));
            }
        }
        return employeeInfos;
    }
}

在您的员工课程中

public class Employees {

    @JsonDeserialize(using=EmployeeInfoDeSerializer.class)
    private List<EmployeeInfo> employeeInfo;

    public List<EmployeeInfo> getEmployeeInfo() {
        return employeeInfo;
    }

    public void setEmployeeInfo(List<EmployeeInfo> employeeInfo) {
        this.employeeInfo = employeeInfo;
    }


}

Answer 2

我用以下方法解决了这个问题（一些逻辑需要优化）

public class EmployeeInfoJsonDeserializer extends JsonDeserializer<List<EmployeeInfo>> {

    @Override
    public List<EmployeeInfo> deserialize(final JsonParser jp, final DeserializationContext ctxt)
                    throws IOException, JsonProcessingException {

        final ObjectMapper mapper = (ObjectMapper) jp.getCodec();

        final JsonNode node = (JsonNode) mapper.readTree(jp);

        // TODO - Write the following logic in a better way
        String toStr = node.toString();
        toStr = StringUtils.replace(toStr, "\"{", "{");
        toStr = StringUtils.replace(toStr, "}\"", "}");
        toStr = StringUtils.remove(toStr, "\\r\\n");
        toStr = StringUtils.remove(toStr, "\\");

        final JsonNode newNode = mapper.readTree(toStr);

        final EmployeeInfo[] empInfo = mapper.convertValue(newNode, EmployeeInfo[].class);

        return Arrays.asList(empInfo);
    }
}

在Pojo中：

public class Employees {
       List<EmployeeInfo> employeeInfo;

       public List<EmployeeInfo> getEmployeeInfo() {
            return employeeInfo;
       }

       @JsonDeserialize(using = EmployeeInfoJsonDeserializer .class)
       public void setEmployeeInfo(final List<EmployeeInfo> employeeInfo) {
            this.employeeInfo= employeeInfo;
       }
}