将Java对象到JSON

Question

在Java中我有这种Map<Member, Map<CustomerVO, Set<Vehicle>>>的对象和另一List<DeviceOrder>我想它转换为JSON，我试图与Gson但它抛出错误为“忘记注册一个类型的适配器？”

有人可以请帮我怎么做，如果不是gson什么是其他方式转换。

上述错误是，当我试图这样

List<DeviceOrder> devLst; 
Gson gson = new Gson(); 
String jsonStr = gson.toJson(devLst); 

为另一种为

请让我知道我做错了。后来我想通过这个jsp来显示。

更新： 甚至尝试过这种方式，但没有用。

Gson gson = new Gson(); 
Type type = new TypeToken<List<DeviceOrder>>(){}.getType(); 
String jsonstr = gson.toJson(devLst, type); 

Answer 1

您可以尝试走出标准执行Java API for JSON processing这是J2EE一部分。

为了您List<DeviceOrder> devLst，定义bean：

public class DeviceOrder implements Serializable { 
    private static final long serialVersionUID = 893438341L; 

    public DeviceOrder() { 
    } 

    public DeviceOrder(int id, String desc, Date date) { 
     this.id = id; 
     this.desc = desc; 
     this.date = date; 
    } 

    public int getId() { 
     return id; 
    } 

    public void setId(int id) { 
     this.id = id; 
    } 

    public String getDesc() { 
     return desc; 
    } 

    public void setDesc(String desc) { 
     this.desc = desc; 
    } 

    public Date getDate() { 
     return date; 
    } 

    public void setDate(Date date) { 
     this.date = date; 
    } 

    @Override 
    public int hashCode() { 
     final int prime = 31; 
     int result = 1; 
     result = prime * result + ((date == null) ? 0 : date.hashCode()); 
     result = prime * result + ((desc == null) ? 0 : desc.hashCode()); 
     result = prime * result + id; 
     return result; 
    } 

    private int id; 
    private String desc; 
    private Date date; 
} 

然后用：

final SimpleDateFormat sdf = new SimpleDateFormat("yyyy-MM-dd"); 

List<DeviceOrder> devLst = new ArrayList<DeviceOrder>() { 
    { 
     add(new DeviceOrder(1, "order 1", sdf.parse("2010-05-01"))); 
     add(new DeviceOrder(2, "order 2", sdf.parse("2010-06-01"))); 
     add(new DeviceOrder(3, "order 3", sdf.parse("2010-07-01"))); 
    } 
}; 
DeviceOrder[] devArr = devLst.toArray(new DeviceOrder[devLst.size()]); 

JsonArrayBuilder devArrBuilder = Json.createArrayBuilder(); 
for (DeviceOrder devOrder : devArr) { 
    JsonObjectBuilder jsonObject = Json.createObjectBuilder() 
      .add("id", devOrder.getId()) 
      .add("desc", devOrder.getDesc()) 
      .add("date", sdf.format(devOrder.getDate())); 
    devArrBuilder.add(jsonObject); 
} 
JsonArray jsonArray = devArrBuilder.build(); 

Map<String, Object> prop = new HashMap<String, Object>() { 
    { 
     put(JsonGenerator.PRETTY_PRINTING, true); 
    } 
}; 
JsonWriter jsonWriter = Json.createWriterFactory(prop).createWriter(System.out); 
jsonWriter.writeArray(jsonArray); 
jsonWriter.close(); 

输出应该是：

[ 
    { 
     "id":1, 
     "desc":"order 1", 
     "date":"2010-05-01" 
    }, 
    { 
     "id":2, 
     "desc":"order 2", 
     "date":"2010-06-01" 
    }, 
    { 
     "id":3, 
     "desc":"order 3", 
     "date":"2010-07-01" 
    } 
] 

你Map<Member, Map<CustomerVO, Set<Vehicle>>> map可以降低到List并以类似的方式JSON。