Jackson JSON:未识别字段，未标记为可忽略

我需要将某个JSON字符串转换为Java对象。我正在使用Jackson进行JSON处理。我无法控制输入JSON(我从web服务读取)。这是我的输入JSON:

{"wrapper":[{"id":"13","name":"Fred"}]}

下面是一个简化的用例:

private void tryReading() {
    String jsonStr = "{\"wrapper\"\:[{\"id\":\"13\",\"name\":\"Fred\"}]}";
    ObjectMapper mapper = new ObjectMapper();  
    Wrapper wrapper = null;
    try {
        wrapper = mapper.readValue(jsonStr , Wrapper.class);
    } catch (Exception e) {
        e.printStackTrace();
    }
    System.out.println("wrapper = " + wrapper);
}

我的实体类是:

public Class Student { 
    private String name;
    private String id;
    //getters & setters for name & id here
}

我的Wrapper类基本上是一个容器对象来获取我的学生列表:

public Class Wrapper {
    private List<Student> students;
    //getters & setters here
}

我一直得到这个错误和“包装器”返回null。我不知道少了什么。有人能帮帮我吗?

org.codehaus.jackson.map.exc.UnrecognizedPropertyException: 
    Unrecognized field "wrapper" (Class Wrapper), not marked as ignorable
 at [Source: java.io.StringReader@1198891; line: 1, column: 13] 
    (through reference chain: Wrapper["wrapper"])
 at org.codehaus.jackson.map.exc.UnrecognizedPropertyException
    .from(UnrecognizedPropertyException.java:53)

当前回答

Java 11及更新版本的简单解决方案:

var mapper = new ObjectMapper()
        .registerModule(new JavaTimeModule())
        .disable(FAIL_ON_UNKNOWN_PROPERTIES)
        .disable(WRITE_DATES_AS_TIMESTAMPS)
        .enable(ACCEPT_EMPTY_ARRAY_AS_NULL_OBJECT);

重要的忽略是禁用'FAIL_ON_UNKNOWN_PROPERTIES'

2022-09-03 18:16:36

其他回答

你可以使用

ObjectMapper objectMapper = getObjectMapper();
objectMapper.configure(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES, false);

它将忽略所有未声明的属性。

2012-10-04 15:27:48

如果由于某种原因，你不能将@JsonIgnoreProperties注释添加到你的类中，并且你是在一个web服务器/容器中，比如Jetty。您可以在自定义提供程序中创建和定制ObjectMapper

import javax.ws.rs.ext.ContextResolver;
import javax.ws.rs.ext.Provider;

import com.fasterxml.jackson.annotation.JsonInclude.Include;
import com.fasterxml.jackson.databind.DeserializationFeature;
import com.fasterxml.jackson.databind.ObjectMapper;

@Provider
public class CustomObjectMapperProvider implements ContextResolver<ObjectMapper> {

    private ObjectMapper objectMapper;

    @Override
    public ObjectMapper getContext(final Class<?> cls) {
        return getObjectMapper();
    }

    private synchronized ObjectMapper getObjectMapper() {
        if(objectMapper == null) {
            objectMapper = new ObjectMapper();
            objectMapper.configure(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES, false);
        }
        return objectMapper;
    }
}

2021-02-25 15:12:38

如果您正在使用Jackson 2.0

ObjectMapper mapper = new ObjectMapper();
mapper.disable(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES);

2014-01-07 19:07:12

在我的例子中，它很简单:更新了rest -服务JSON对象(添加了一个属性)，但没有更新rest -客户端JSON对象。只要我已经更新了JSON客户端对象的“不可识别的字段…’这个例外已经消失了。

2016-06-13 10:20:49

根据文档，您可以忽略选定的字段或所有uknown字段:

 // to prevent specified fields from being serialized or deserialized
 // (i.e. not include in JSON output; or being set even if they were included)
 @JsonIgnoreProperties({ "internalId", "secretKey" })

 // To ignore any unknown properties in JSON input without exception:
 @JsonIgnoreProperties(ignoreUnknown=true)

2016-02-01 17:03:48

Jackson JSON:未识别字段，未标记为可忽略

推荐文章

最新文章

标签