249

我一直在看杰克逊，但似乎我必须将地图转换为 JSON，然后将生成的 JSON 转换为 POJO。

有没有办法将 Map 直接转换为 POJO？

4

9 回答 9

470

好吧，你也可以通过杰克逊实现这一目标。（自从您考虑使用杰克逊以来，它似乎更舒服）。

使用方法ObjectMapper：convertValue

final ObjectMapper mapper = new ObjectMapper(); // jackson's objectmapper
final MyPojo pojo = mapper.convertValue(map, MyPojo.class);

无需转换成 JSON 字符串或其他东西；直接转换的速度要快得多。

于 2013-05-08T00:18:59.637 回答
73

Gson的解决方案：

Gson gson = new Gson();
JsonElement jsonElement = gson.toJsonTree(map);
MyPojo pojo = gson.fromJson(jsonElement, MyPojo.class);
于 2016-05-09T13:14:40.617 回答
7

如果你的类中有泛型类型，你应该使用TypeReferencewith convertValue()。

final ObjectMapper mapper = new ObjectMapper();
final MyPojo<MyGenericType> pojo = mapper.convertValue(map, new TypeReference<MyPojo<MyGenericType>>() {});

您也可以使用它将 pojo 转换为java.util.Mapback。

final ObjectMapper mapper = new ObjectMapper();
final Map<String, Object> map = mapper.convertValue(pojo, new TypeReference<Map<String, Object>>() {});
于 2019-04-05T12:49:48.600 回答
5 
 ObjectMapper objectMapper = new ObjectMapper();
 //if all properties are not in class use this
 objectMapper.configure(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES, false);
 final MyPojo pojo =     objectMapper.convertValue(map, MyPojo.class);

与第一个答案相同，但我使用它时出错，因为我不希望地图的所有属性都转换为 calss。我 objectMapper.configure(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES, false);发现这是解决方案

于 2021-05-12T09:20:39.457 回答
4

是的，绝对可以避免中间转换为 JSON。使用像Dozer这样的深拷贝工具，您可以将地图直接转换为 POJO。这是一个简单的例子：

示例 POJO：

public class MyPojo implements Serializable {
    private static final long serialVersionUID = 1L;

    private String id;
    private String name;
    private Integer age;
    private Double savings;

    public MyPojo() {
        super();
    }

    // Getters/setters

    @Override
    public String toString() {
        return String.format(
                "MyPojo[id = %s, name = %s, age = %s, savings = %s]", getId(),
                getName(), getAge(), getSavings());
    }
}

示例转换代码：

public class CopyTest {
    @Test
    public void testCopyMapToPOJO() throws Exception {
        final Map<String, String> map = new HashMap<String, String>(4);
        map.put("id", "5");
        map.put("name", "Bob");
        map.put("age", "23");
        map.put("savings", "2500.39");
        map.put("extra", "foo");

        final DozerBeanMapper mapper = new DozerBeanMapper();
        final MyPojo pojo = mapper.map(map, MyPojo.class);
        System.out.println(pojo);
    }
}

输出：

MyPojo[id = 5, name = Bob, age = 23, Savings = 2500.39]

注意：如果您将源映射更改为 a，Map<String, Object>那么您可以复制任意深度的嵌套属性（Map<String, String>您只能获得一个级别）。

于 2013-05-07T22:00:18.167 回答
3

我测试了 Jackson 和 BeanUtils，发现 BeanUtils 更快。
在我的机器（Windows8.1，JDK1.7）中，我得到了这个结果。

BeanUtils t2-t1 = 286
Jackson t2-t1 = 2203



public class MainMapToPOJO {

public static final int LOOP_MAX_COUNT = 1000;

public static void main(String[] args) {
    Map<String, Object> map = new HashMap<>();
    map.put("success", true);
    map.put("data", "testString");

    runBeanUtilsPopulate(map);

    runJacksonMapper(map);
}

private static void runBeanUtilsPopulate(Map<String, Object> map) {
    long t1 = System.currentTimeMillis();
    for (int i = 0; i < LOOP_MAX_COUNT; i++) {
        try {
            TestClass bean = new TestClass();
            BeanUtils.populate(bean, map);
        } catch (IllegalAccessException e) {
            e.printStackTrace();
        } catch (InvocationTargetException e) {
            e.printStackTrace();
        }
    }
    long t2 = System.currentTimeMillis();
    System.out.println("BeanUtils t2-t1 = " + String.valueOf(t2 - t1));
}

private static void runJacksonMapper(Map<String, Object> map) {
    long t1 = System.currentTimeMillis();
    for (int i = 0; i < LOOP_MAX_COUNT; i++) {
        ObjectMapper mapper = new ObjectMapper();
        TestClass testClass = mapper.convertValue(map, TestClass.class);
    }
    long t2 = System.currentTimeMillis();
    System.out.println("Jackson t2-t1 = " + String.valueOf(t2 - t1));
}}
于 2016-06-14T21:05:38.953 回答
2

到目前为止使用 Jackson 提供的答案非常好，但您仍然可以使用util函数来帮助您转换不同POJO的 s，如下所示：

    public static <T> T convert(Map<String, Object> aMap, Class<T> t) {
        try {
            return objectMapper
                    .convertValue(aMap, objectMapper.getTypeFactory().constructType(t));
        } catch (Exception e) {
            log.error("converting failed! aMap: {}, class: {}", getJsonString(aMap), t.getClass().getSimpleName(), e);
        }
        return null;
    }
于 2020-03-29T07:43:06.047 回答
1

将 Map 转换为 POJO 示例。注意 Map 键包含下划线，字段变量是驼峰。

用户类 POJO

import com.fasterxml.jackson.annotation.JsonProperty;
import lombok.Data;

@Data
public class User {
    @JsonProperty("user_name")
    private String userName;
    @JsonProperty("pass_word")
    private String passWord;
}

App.class 测试示例

import java.util.HashMap;
import java.util.Map;

import com.fasterxml.jackson.databind.ObjectMapper;

public class App {
    public static void main(String[] args) {
        Map<String, String> info = new HashMap<>();
        info.put("user_name", "Q10Viking");
        info.put("pass_word", "123456");

        ObjectMapper mapper = new ObjectMapper();
        User user = mapper.convertValue(info, User.class);

        System.out.println("-------------------------------");
        System.out.println(user);
    }
}
/**output
-------------------------------
User(userName=Q10Viking, passWord=123456)
 */
于 2019-07-17T05:02:13.567 回答
0

@Hamedz 如果使用很多数据，请使用 Jackson 转换光数据，使用 apache ... 测试用例： 



import java.lang.reflect.InvocationTargetException;
import java.util.HashMap;
import java.util.Map;

import org.apache.commons.beanutils.BeanUtils;

import com.fasterxml.jackson.databind.ObjectMapper;

import lombok.AllArgsConstructor;
import lombok.Data;
import lombok.NoArgsConstructor;

public class TestPerf {

    public static final int LOOP_MAX_COUNT = 1000;

    public static void main(String[] args) {
        Map<String, Object> map = new HashMap<>();
        map.put("success", true);
        map.put("number", 1000);
        map.put("longer", 1000L);
        map.put("doubler", 1000D);
        map.put("data1", "testString");
        map.put("data2", "testString");
        map.put("data3", "testString");
        map.put("data4", "testString");
        map.put("data5", "testString");
        map.put("data6", "testString");
        map.put("data7", "testString");
        map.put("data8", "testString");
        map.put("data9", "testString");
        map.put("data10", "testString");

        runBeanUtilsPopulate(map);

        runJacksonMapper(map);
    }

    private static void runBeanUtilsPopulate(Map<String, Object> map) {
        long t1 = System.currentTimeMillis();
        for (int i = 0; i < LOOP_MAX_COUNT; i++) {
            try {
                TestClass bean = new TestClass();
                BeanUtils.populate(bean, map);
            } catch (IllegalAccessException e) {
                e.printStackTrace();
            } catch (InvocationTargetException e) {
                e.printStackTrace();
            }
        }
        long t2 = System.currentTimeMillis();
        System.out.println("BeanUtils t2-t1 = " + String.valueOf(t2 - t1));
    }

    private static void runJacksonMapper(Map<String, Object> map) {
        long t1 = System.currentTimeMillis();
        for (int i = 0; i < LOOP_MAX_COUNT; i++) {
            ObjectMapper mapper = new ObjectMapper();
            TestClass testClass = mapper.convertValue(map, TestClass.class);
        }
        long t2 = System.currentTimeMillis();
        System.out.println("Jackson t2-t1 = " + String.valueOf(t2 - t1));
    }

    @Data
    @AllArgsConstructor
    @NoArgsConstructor
    public static class TestClass {
        private Boolean success;
        private Integer number;
        private Long longer;
        private Double doubler;
        private String data1;
        private String data2;
        private String data3;
        private String data4;
        private String data5;
        private String data6;
        private String data7;
        private String data8;
        private String data9;
        private String data10;

    }
}
于 2019-12-27T18:49:07.137 回答