10

我一直在尝试升级 JSON 模块以使用 Jackson 的 FasterXML (2.6.3) 版本而不是旧的 Codehaus 模块。在升级过程中,我注意到使用 FasterXML 而不是 Codehaus 时的命名策略有所不同。

Codehaus 在命名策略方面更加灵活。下面的测试突出了我在使用 FasterXML 时面临的问题。如何配置ObjectMapper它,使其遵循与 Codehaus 相同的策略?

我无法更改JSONProperty注释,因为它们有数百个。我希望升级在命名策略方面向后兼容。

import java.io.IOException;

import com.fasterxml.jackson.annotation.JsonIgnoreProperties;
import com.fasterxml.jackson.annotation.JsonProperty;
import com.fasterxml.jackson.databind.ObjectMapper;
import com.fasterxml.jackson.databind.PropertyNamingStrategy;
/*import org.codehaus.jackson.annotate.JsonIgnoreProperties;
import org.codehaus.jackson.annotate.JsonProperty;
import org.codehaus.jackson.map.ObjectMapper;
import org.codehaus.jackson.map.PropertyNamingStrategy;*/
import org.junit.Assert;
import org.junit.Test;

public class JSONTest extends Assert {

    @JsonIgnoreProperties(ignoreUnknown = true)
    public static class Product {

        @JsonProperty(value = "variationId")
        private String variantId;

        @JsonProperty(value = "price_text")
        private String priceText;

        @JsonProperty(value = "listPrice")
        public String listPrice;

        @JsonProperty(value = "PRODUCT_NAME")
        public String name;

        @JsonProperty(value = "Product_Desc")
        public String description;
    }

    private static final String VALID_PRODUCT_JSON =
            "{ \"list_price\": 289," +
             " \"price_text\": \"269.00\"," +
             " \"variation_id\": \"EUR\"," +
             " \"product_name\": \"Product\"," +
             " \"product_desc\": \"Test\"" +
            "}";

    @Test
    public void testDeserialization() throws IOException {
        ObjectMapper mapper = new ObjectMapper();
        mapper.setPropertyNamingStrategy(PropertyNamingStrategy.CAMEL_CASE_TO_LOWER_CASE_WITH_UNDERSCORES);

        Product product = mapper.readValue(VALID_PRODUCT_JSON, Product.class);
        System.out.println(mapper.writerWithDefaultPrettyPrinter().writeValueAsString(product));
        assertNotNull(product.listPrice);
        assertNotNull(product.variantId);
        assertNotNull(product.priceText);
        assertNotNull(product.name);
        assertNotNull(product.description);
    }
}
4

2 回答 2

10

@JsonProperty覆盖自2.4.0版本以来PropertyNamingStrategy 的 fasterxml 中的任何内容。但是,尚未发布的2.7.0版将提供一项功能,允许您选择恢复旧行为。还有一个未实现的建议可以在每个注释级别切换它,但这对您没有帮助。

Codehaus 似乎在映射时确实应用了PropertyNamingStrategy这些@JsonProperty值,尽管我找不到任何明确的文档。这似乎也是 2.4.0 之前的 fastxml 中的行为。是另一个注意到行为相同差异的人的例子。

于 2015-11-16T21:46:42.197 回答
2

虽然 SkinnyJ 提供的解决方案非常适合您的问题,但是如果您不能等到 2.7 发布,您可以应用以下 hack 来解决问题。

这个想法是转换传入的 JSON 以匹配 bean 定义中的属性。下面的代码就是这样做的。应注意以下几点:

  1. 如果您正在处理嵌套结构,则必须实现一个递归函数来实现这种转换。
  2. 进行转换会涉及一些开销。

代码:

public class JSONTest extends Assert {

    @JsonIgnoreProperties(ignoreUnknown = true)
    public static class Product {

        @JsonProperty(value = "variationId")
        private String variantId;

        @JsonProperty(value = "price_text")
        private String priceText;

        @JsonProperty(value = "listPrice")
        public String listPrice;

        @JsonProperty(value = "PRODUCT_NAME")
        public String name;

        @JsonProperty(value = "Product_Desc")
        public String description;
    }

    private static final String VALID_PRODUCT_JSON =
            "{ \"list_price\": 289," +
             " \"price_text\": \"269.00\"," +
             " \"variation_id\": \"EUR\"," +
             " \"product_name\": \"Product\"," +
             " \"product_desc\": \"Test\"" +
            "}";

    @Test
    public void testDeserialization() throws IOException {
        ObjectMapper mapper = new ObjectMapper();
        mapper.setPropertyNamingStrategy(PropertyNamingStrategy.CAMEL_CASE_TO_LOWER_CASE_WITH_UNDERSCORES);

        //Capture the original JSON in org.json.JSONObject
        JSONObject obj = new JSONObject(VALID_PRODUCT_JSON);
        JSONArray keys = obj.names();

        //New json object to be created using property names defined in bean
        JSONObject matchingJson = new JSONObject();

        //Map of lowercased key to original keys in incoming json. eg: Prod_id > prodid
        Map<String, String> jsonMappings = new LinkedHashMap<String, String>();
        for (int i = 0; i < keys.length(); i++) {
            String key = lowerCaseWithoutUnderScore(keys.getString(i));
            String value = keys.getString(i);
            jsonMappings.put(key, value);
        }

        /*
         * Iternate all jsonproperty beans and create new json
         * such that keys in json map to that defined in bean
         */
        Field[] fields = Product.class.getDeclaredFields();
        for (Field field : fields) {
            JsonProperty prop = field.getAnnotation(JsonProperty.class);
            String propNameInBean = prop.value();
            String keyToLook = lowerCaseWithoutUnderScore(propNameInBean);
            String keyInJson = jsonMappings.get(keyToLook);
            matchingJson.put(propNameInBean, obj.get(keyInJson));
        }

        String json = matchingJson.toString();
        System.out.println(json);

        //Pass the matching json to Object mapper
        Product product = mapper.readValue(json, Product.class);
        System.out.println(mapper.writerWithDefaultPrettyPrinter().writeValueAsString(product));
        assertNotNull(product.listPrice);
        assertNotNull(product.variantId);
        assertNotNull(product.priceText);
        assertNotNull(product.name);
        assertNotNull(product.description);
    }

    private String lowerCaseWithoutUnderScore(String key){
        return key.replaceAll("_", "").toLowerCase();
    }

}
于 2015-11-23T04:09:45.007 回答