我有一个小函数,它接受输入的JSON字符串,使用boon解析为Map,替换特定键的值,返回修改后的Map的JSON字符串。

代码如下:

// inputJson = {"key3":"A","key2":"B","key1":null,"keyX":[{"x":2019,"y":123,"z":456},{"x":2017,"y":234,"z":345},{"x":2018,"y":456,"z":567}]}

private static String sorter(String inputJson) {
    JsonParserAndMapper mapper = new JsonParserFactory().strict().create();

    Map<String, Object> map = mapper.parseMap(inputJson);

    List<?> l1 = (List<?>) map.get("keyX");

    sort(l1, Sort.sortBy("x"));

    map.replace("keyX", l1);

    for (String x: map.keySet())
        System.out.println(map.get(x));

    String outputJson = toJson(map);  // problem seems to be here

    return outputJson

// outputJson = {"key2":"B","key3":"A","keyX":[{"x":2017,"y":234,"z":345},{"x":2018,"y":456,"z":567},{"x":2019,"y":123,"z":456}]}


问题是,当我执行toJson(map)时,它将删除带有null值的键。因此,如果inputJson包含具有空值的键,则它不会出现在输出中。 (注意:输出中缺少key1

我该如何解析而不丢失空字段?

最佳答案

使用toJson,您正在使用默认的序列化器工厂。从boon源代码:

public class JsonFactory {


    private static ObjectMapper json = JsonFactory.create();

    public static ObjectMapper create () {
        JsonParserFactory jsonParserFactory = new JsonParserFactory();
        jsonParserFactory.lax();

        return new ObjectMapperImpl(jsonParserFactory,  new JsonSerializerFactory());
    }
....
)


代替使用toJson,尝试使用带有includeNulls()的序列化器工厂

JsonSerializer factory = new JsonSerializerFactory().includeNulls().create();

关于java - Java boon JSON解析器从输出中删除空值,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/42125603/

10-09 04:33