如何将 java.lang.String 的空白 JSON 字符串值反序列化为 null?

问题描述

我正在尝试使用简单的 JSON 反序列化为 java 对象.但是，我为 java.lang.String 属性值获取了空的 String 值.在其余属性中，空白值正在转换为 null 值(这是我想要的).

I am trying a simple JSON to de-serialize in to java object. I am however, getting empty String values for java.lang.String property values. In rest of the properties, blank values are converting to null values(which is what I want).

下面列出了我的 JSON 和相关的 Java 类.

My JSON and related Java class are listed below.

JSON 字符串:

{
  "eventId" : 1,
  "title" : "sample event",
  "location" : ""
}

EventBean 类 POJO:

public class EventBean {

    public Long eventId;
    public String title;
    public String location;

}

我的主类代码:

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

try {
    File file = new   File(JsonTest.class.getClassLoader().getResource("event.txt").getFile());

    JsonNode root = mapper.readTree(file);
    // find out the applicationId

    EventBean e = mapper.treeToValue(root, EventBean.class);
    System.out.println("It is " + e.location);
}

我期待打印它是空的".相反，我得到它是".显然，在转换为我的 String 对象类型时，Jackson 不会将空白字符串值视为 NULL.

I was expecting print "It is null". Instead, I am getting "It is ". Obviously, Jackson is not treating blank String values as NULL while converting to my String object type.

我在某处读到这是预期的.但是，对于 java.lang.String 我也想避免这种情况.有没有简单的方法?

I read somewhere that it is expected. However, this is something I want to avoid for java.lang.String too. Is there a simple way?

推荐答案

对于其他对象，Jackson 会给你 null，但对于 String 它会给空 String.

Jackson will give you null for other objects, but for String it will give empty String.

但是您可以使用自定义 JsonDeserializer 来执行此操作:

But you can use a Custom JsonDeserializer to do this:

class CustomDeserializer extends JsonDeserializer<String> {

    @Override
    public String deserialize(JsonParser jsonParser, DeserializationContext context) throws IOException, JsonProcessingException {
        JsonNode node = jsonParser.readValueAsTree();
        if (node.asText().isEmpty()) {
            return null;
        }
        return node.toString();
    }

}

在课堂上，您必须将其用于位置字段:

In class you have to use it for location field:

class EventBean {
    public Long eventId;
    public String title;

    @JsonDeserialize(using = CustomDeserializer.class)
    public String location;
}

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