2015-09-01 106 views

回答

0

也許是這樣一個快速和骯髒的轉換:

import com.mongodb.DBObject; 
import com.mongodb.util.JSON; 
import java.io.StringReader; 
import java.io.StringWriter; 
import javax.json.Json; 
import javax.json.JsonObject; 
import javax.json.JsonWriter; 

public class MongoDBTranslator { 

    public static JsonObject convertDocumentToJson(org.bson.Document bson) { 
     JsonObject obj = null; 
     try (StringReader sReader = new StringReader(bson.toJson()); 
       javax.json.JsonReader reader = Json.createReader(sReader)) { 
      obj = reader.readObject(); 
      reader.close(); 
     } 
     return obj; 
    } 

    public static DBObject convertJsonToDocument(JsonObject json) { 
     StringWriter sw = new StringWriter(); 
     try (JsonWriter writer = Json.createWriter(sw)) { 
      writer.writeObject(json); 
      writer.close(); 
     } catch (Exception ex) { 
      //error 
      return null; 
     }  

     return (DBObject) JSON.parse(sw.toString()); 
    } 

}