Question

我想忽略多少属性（我从UI获得），如

filtervalue0，filtervalue1，filtervalue2等

然而，我从UI

获得此计数

即。 if filterscount = 3 我知道，会有 filtervalue0，filtervalue1，filtervalue2

是否可以动态地进行此工作（即@JsonIgnoreProperties）

Answer 1

您可以使用@JsonFilter动态过滤字段Serialization或Deserialization。

例如：

@JsonFilter("myFilter")
static class Bean {
    public String filtervalue0 = "a";
    public String filtervalue1 = "b";
    public int filtercount = 1;
    public int totalFilters = 2;
}

static void test() throws JsonProcessingException{
    Bean value = new Bean();
    ObjectMapper mapper = new ObjectMapper();
    Set<String> filtersToExclude = new HashSet<>();

    //logic for applying filter
    for(int i=value.totalFilters-1; i>value.totalFilters-value.filtercount-1; i--){
        filtersToExclude.add("filtervalue"+i);
    }

   //dynamic field filtering is provided by FilterProvider
    FilterProvider filters = new SimpleFilterProvider().addFilter("myFilter",
        SimpleBeanPropertyFilter.filterOutAllExcept(filtersToExclude));
    String json = mapper.setFilterProvider(filters).writeValueAsString(value);
    System.out.println(json); // prints {"filtervalue1":"b"}
}

public static void main(String[] args) throws JsonProcessingException {
    test();
}

@JsonIgnoreProperties用于动态参数数量

1 个答案: