java – Jackson：反序列化递归对象

2023年5月29日 240次阅读

我正在尝试将KendoUI网格发送的过滤器参数解析到我的Web服务,并且有一些问题说服Jackson解析这个
JSON.据我所知,我可以控制剑道发送的参数格式,但我不知道如何将参数编组成更好的格式,以便它们现在保持不变.

我打算将这些参数转换为Oracle数据库的SQL查询.

示例JSON：

{
    "filters":
    [
        {
            "field": "Name",
            "operator": "contains",
            "value": "John"
        },
        {
            "filters": [
                {
                    "field": "Age",
                    "operator": "gt",
                    "value": 20
                },
                {
                    "field": "Age",
                    "operator": "lt",
                    "value": 85
                }
            ],
            "logic", "and"
        },
        {
            "field": "Address",
            "operator": "doesnotcontain",
            "value": "street"
        }
    ],
    "logic": "or"
}

过滤器. Java的

public class Filters {
    private List<Filter> filters;
    private String logic;
    // accessors/mutators/toString
}

Filter.java

public class Filter {
    private String field;
    private String operator;
    private String value;
    // accessors/mutators/toString
}

单元测试

public class KendoGridFilterTest {
    private ObjectMapper mapper;

    @Before
    public void before() {
        mapper = new ObjectMapper();
    }

    @Test
    public void jsonParseTest() {
        final String json = "{\"filters\":[{\"field\":\"Name\",\"operator\":\"contains\",\"value\":\"John\"},{filters: [{\"field\":\"Age\",\"operator\": \"eq\",\"value\": 85},{\"field\": \"Age\",\"operator\": \"eq\",\"value\": 85}]\"logic\", \"and\",},{\"field\": \"Address\",\"operator\": \"doesnotcontain\",\"value\": \"street\"}],\"logic\":\"or\"}";
        Filters filters = mapper.readValue(json, Filters.class);
        assertTrue(json.equals(filters.writeValueAsString(filters);
    }
}

错误

com.fasterxml.jackson.databind.UnrecognizedPropertyException: Unrecognized field 
    'logic'(com.example.Filter) not market as ignorable (3 known properties 
    "value", "field", "operator")
at [Source: java.io.StringReader@3bb2b8; line: 1, column 76] (through reference 
    chain: com.example.Filters["filters"]->com.example.Filter["logic"]

我也尝试将@JsonIdentityInfo(generator = ObjectIdGenerators.IntSequenceGenerator.class,property =“@ id”)添加到Filters类并获得相同的错误.

最佳答案 您的Filter类不正确.它应该扩展过滤器.

在更正单元测试(json不正确)后,它可以将您的json加载到Filters对象中.

public class Filter extends Filters {
    private String field;
    private String operator;
    private String value;
    // accessors/mutators/toString
}
点赞