2014-11-14 48 views
1

我想查询存储在我的MongoDB中的java.util.UUID,使用与存储的UUID所基于的字符串相对应的字节[]。MongoDB [Java]:如何以字节[]格式查询存储的UUID?

我见过Mongo Driver自动将UUID转换为字节并将其存储为BinData(子类型)。下面是我用来保存UUID代码:

 UUID originalUUID= UUID.fromString("7ee973c0-54b5-11e4-aaed-0002a5d5c51b"); 
     MongoClient mongoClient = new MongoClient("localhost" , 27017); 
     DB db = mongoClient.getDB("test"); 
     DBCollection coll = db.getCollection("test"); 

     BasicDBObject query = new BasicDBObject("_id", originalUUID); 
     coll.save(query); 

我知道我们可以很容易地检索与UUID对象在BasicDBObject值这个文件,但是我想用一个byte []代替。毕竟,它在我的Mongo中必须是字节吗? 我知道Java在发送给Mongo之前搞乱了字节顺序,但是我一直无法使它工作。我的检索代码是在这里:

 String stringOne = "7ee973c0-54b5-11e4-aaed-0002a5d5c51b"; 
     // It's the same UUID String 

     BasicDBObject find = new BasicDBObject("_id", stringOne.getBytes()); 

     DBCursor cursor = null; 
     cursor = coll.find(find); 
     while(cursor.hasNext()){ 
      BasicDBObject dbObject = (BasicDBObject)cursor.next(); 
      byte[] received = (byte[])dbObject.get("_id"); 
      System.out.println(new String(received)); 
     } 

我也试过小端编码是这样的:

 byte[] bytes = new byte[16]; 
     ByteBuffer bb = ByteBuffer.wrap(bytes); 
     bb.order(ByteOrder.LITTLE_ENDIAN); 

     bb.putLong(originalUUID.getMostSignificantBits()); 
     bb.putLong(originalUUID.getLeastSignificantBits()); 

     BasicDBObject find = new BasicDBObject("_id", bb.array()); //Also tried flipping bb. 

这一切都不奏效。有人可以通过解决方案提供一些见解吗?

谢谢!

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你可以得到光标? – 2014-11-14 17:27:38

+0

我不明白。获得“光标”是什么意思? – 2014-11-14 17:56:42

回答

2

你非常接近。它不工作的唯一原因是查询匹配器也匹配二进制的子类型。试试这个:

UUID originalUUID = UUID.fromString("7ee973c0-54b5-11e4-aaed-0002a5d5c51b"); 
    MongoClient mongoClient = new MongoClient(); 
    DB db = mongoClient.getDB("test"); 
    DBCollection coll = db.getCollection("test"); 

    BasicDBObject query = new BasicDBObject("_id", originalUUID); 
    coll.save(query); 

    byte[] bytes = new byte[16]; 
    ByteBuffer bb = ByteBuffer.wrap(bytes); 
    bb.order(ByteOrder.LITTLE_ENDIAN); 

    bb.putLong(originalUUID.getMostSignificantBits()); 
    bb.putLong(originalUUID.getLeastSignificantBits()); 

    System.out.println(coll.findOne(originalUUID)); 
    System.out.println(coll.findOne(new Binary(Bytes.B_UUID, bb.array()))); // note use of the B_UUID subtype 

它应该打印:

{ "_id" : { "$uuid" : "7ee973c0-54b5-11e4-aaed-0002a5d5c51b"}} 
    { "_id" : { "$uuid" : "7ee973c0-54b5-11e4-aaed-0002a5d5c51b"}} 
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谢谢,这工作。但我只能使用一个字节数组,而没有别的。有没有办法做到这一点? (我正在使用vertx Mongo-Persistor。更多详细信息,请点击: https://groups.google.com/forum/#!topic/vertx/2uynNUXeoY8) – 2014-11-14 20:26:24

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而Mongo-Persistor不会让我把二进制或任何其他类型的字符串或字节[] – 2014-11-14 20:27:02