Question

我需要将我拥有的一些十六进制字符串转换为UUID，并将它们存储在MongoDB中。

我尝试了以下内容：

import pymongo
import uuid
[...]
document = { '_id': 123, 'my_uuid': uuid.UUID('b7aef1d4830843750f3846b34606528f') }
my_collection.save(document)

但是这会插入一个LUUID而不是UUID。

我错过了什么吗？

Answer 1

但是这会插入一个LUUID而不是UUID。

LUUID不是BSON或PyMongo类型。它只是一个名称，Robomongo uses表示存储为BSON二进制文件的UUID，子类型为3而不是子类型4.

 For example the source code of UuidRepresentation class in mongodb java library:
 /**
  * The representation to use when converting a UUID to a BSON binary value.
  * This class is necessary because the different drivers used to have different
  * ways of encoding UUID, with the BSON subtype: \x03 UUID old.
  *
  * @since 3.0
  */
 public enum UuidRepresentation {
     /**
      * The canonical representation of UUID
      *
      * BSON binary subtype 4
      */
     STANDARD,

     /**
      * The legacy representation of UUID used by the C# driver
      *
      * BSON binary subtype 3
      */
     C_SHARP_LEGACY,

     /**
      * The legacy representation of UUID used by the Java driver
      *
      * BSON binary subtype 3
      */
     JAVA_LEGACY,

     /**
      * The legacy representation of UUID used by the Python driver, which is the same
      * format as STANDARD, but has the UUID old BSON subtype (\x03)
      *
      * BSON binary subtype 3
      */
     PYTHON_LEGACY
 }

有关二进制子类型的更多信息，请参阅BSON spec。实际上，您正在存储UUID。如果你决定在Robomongo中看到UUID而不是LUUID，那么你可以像这样在PyMongo中切换UUID表示：

client = MongoClient(uuidRepresentation='standard')

我最近写了历史in this ticket。

如何从已知的十六进制字符串创建UUID并使用pymongo将其存储在MongDB中？

1 个答案: