16

我有一个 JSON 字符串，如：

"shipping_profiles": {
  "563": {
    "name": "name",
    "value": "value"            
  },
  "564": {
    "name": "name",
    "value": "value"            
  },
  "565": {
    "name": "name",
    "value": "value"            
  },
  "566": {
    "name": "name",
    "value": "value"            
  }
}

现在我用Jackson 2.0解析它。我正在尝试List<shipping_profiles>从 JSON 字符串中获取一个。

可能吗？

4

2 回答 2

20

您的shipping_profiles财产看起来不像数组。它表示具有动态属性的对象，因此我们应该将其视为对象。如果我们对属性一无所知，我们可以使用@JsonAnySetter注解。算法可能如下所示：

  1. 将 JSON 反序列化为 JSON 模型类。
  2. 使用 ObjectMapper 将动态对象（地图）转换为应用程序的 POJO 类
  3. 随时使用应用程序的 POJO。

请参阅我的示例实现。我希望它可以帮助您解决问题。输入 JSON：

{
   "shipping_profiles":{
      "563":{
         "name":"name563",
         "value":"value563"
      },
      "564":{
         "name":"name564",
         "value":"value564"
      },
      "565":{
         "name":"name565",
         "value":"value565"
      },
      "566":{
         "name":"name566",
         "value":"value566"
      }
   }
}

示例程序：

import java.io.File;
import java.io.IOException;
import java.util.ArrayList;
import java.util.List;
import java.util.Map;

import com.fasterxml.jackson.annotation.JsonAnySetter;
import com.fasterxml.jackson.annotation.JsonProperty;
import com.fasterxml.jackson.databind.ObjectMapper;

public class JacksonProgram {

    public static void main(String[] args) throws IOException {
        ObjectMapper mapper = new ObjectMapper();

        File source = new File("X:/test.json");
        Entity entity = mapper.readValue(source, Entity.class);
        ShippingProfiles shippingProfiles = entity.getShippingProfiles();
        List<Map<String, String>> profileMaps = shippingProfiles.getProfiles();

        List<Profile> profiles = new ArrayList<Profile>(profileMaps.size());
        for (Map<String, String> item : profileMaps) {
            profiles.add(mapper.convertValue(item, Profile.class));
        }
        System.out.println(profiles);
    }
}

class Entity {

    @JsonProperty("shipping_profiles")
    private ShippingProfiles shippingProfiles;

    public ShippingProfiles getShippingProfiles() {
        return shippingProfiles;
    }

    public void setShippingProfiles(ShippingProfiles shippingProfiles) {
        this.shippingProfiles = shippingProfiles;
    }
}

class ShippingProfiles {

    private List<Map<String, String>> profiles = new ArrayList<Map<String, String>>();

    @JsonAnySetter
    public void setDynamicProperty(String name, Map<String, String> map) {
        profiles.add(map);
    }

    public List<Map<String, String>> getProfiles() {
        return profiles;
    }

    public void setProfiles(List<Map<String, String>> profiles) {
        this.profiles = profiles;
    }
}

class Profile {

    private String name;
    private String value;

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }

    public String getValue() {
        return value;
    }

    public void setValue(String value) {
        this.value = value;
    }

    @Override
    public String toString() {
        return "Profile [name=" + name + ", value=" + value + "]";
    }
}

上面的应用程序打印：

[Profile [name=name563, value=value563], Profile [name=name564, value=value564], Profile [name=name565, value=value565], Profile [name=name566, value=value566]]
于 2013-07-17T11:16:55.170 回答
0

我用@michalziober 提供的方式解析了具有动态属性的json。

"commandClasses": {
        "32": {
          "name": "Basic",
          "data": {
          "name": "devices.1.instances.1.commandClasses.32.data",
          "value": null,
          "type": "NoneType"
         },
         "38": {
          "name": "SwitchMultilevel",
          "data": {
          "name": "devices.1.instances.1.commandClasses.38.data",
          "value": null,
          "type": "NoneType"
         },
         "43": {
          "name": "SceneActivation",
          "data": {
          "name": "devices.1.instances.1.commandClasses.43.data",
          "value": null,
          "type": "NoneType"
         }

有了这个 json，我还需要保存那个动态属性，所以我添加了另一个 List 来存储它。

public class CommandClasses {

    private List<String> nameList = new ArrayList<String>();
    private List<CommandClass> commmandClasses = new ArrayList<CommandClass>();
    private Logger logger = Logger.getInstance(CommandClasses.class);

    @JsonAnySetter
    public void setDynamicCommandClass(String name, CommandClass cc) {
       logger.d("@ adding new CC : " + name);
       nameList.add(name);
       commmandClasses.add(cc);
    }

    public List<CommandClass> getCommmandClasses() {
        return commmandClasses;
    }

    public void setCommmandClasses(List<CommandClass> commmandClasses) {
        this.commmandClasses = commmandClasses;
    }
}

现在我也可以作为 id 访问该字段，以便稍后发送请求。

于 2013-11-15T03:43:36.330 回答