Question

我已阅读到，当我使用pymongo的upsert且不提供“ _id”时，upsert将尝试生成新的ID，这将导致操作失败。这是真的？以及如何在不使用“ _id”的情况下向上插入？

这是我使用pymongo的replace_one：

db['dataitemdetails'].replace_one({'asset_Id':tdata['asset_id'],'period_type':tdata['period_type'],'detail_id':tdata['detail_id'], 'currencycode':tdata['currencycode'],'dataitem_Id':tdata['dataitem_id'],'period_end':tdata['period_end'], 'scenario_id':tdata['scenario_id'],}, tdata, upsert=True)

我已经创建了一个复合索引，并使用搜索条件将其设置为唯一，这样，在集合中，如果搜索到的内容，我搜索的内容将始终唯一。

，并与replace_one我收到此错误：

pymongo.errors.DuplicateKeyError: E11000 duplicate key error collection: finance.dataitemdetails index: asset_id_1_dataitem_id_1_detail_id_1_period_type_1_scenario_id_1_currencycode_1_period_end_1 dup key: { : 19, : 1211, : 0, : "Month", : 1, : "RC", : new Date(949276800000) }

这是我用相同过滤器查询的查询，它返回1个文档。

> db.dataitemdetails.find({'asset_id':19,'dataitem_id':1211,'detail_id':0,'period_type':'Month','currencycode':'RC','period_end':new Date(949276800000)})
{ "_id" : ObjectId("5c7721c17314e53a85be7e89"), "Value" : "USD", "period_end" : ISODate("2000-01-31T00:00:00Z"), "currencycode" : "RC", "scenario_id" : 1, "dataitem_id" : 1211, "period_type" : "Month", "detail_id" : 0, "asset_id" : 19 }

我坚持不知道要尝试什么以及如何解决它。

完整的错误消息：

Traceback (most recent call last):
  File "./periodic_update.sh", line 307, in <module>
    db['dataitemdetails'].replace_one({'asset_Id':tdata['asset_id'],'period_type':tdata['period_type'],'detail_id':tdata['detail_id'], 'currencycode':tdata['currencycode'],'dataitem_Id':tdata['dataitem_id'],'period_end':tdata['period_end'], 'scenario_id':tdata['scenario_id'],}, tdata, upsert=True)
  File "/usr/local/lib64/python3.7/site-packages/pymongo/collection.py", line 925, in replace_one
    collation=collation, session=session),
  File "/usr/local/lib64/python3.7/site-packages/pymongo/collection.py", line 851, in _update_retryable
    _update, session)
  File "/usr/local/lib64/python3.7/site-packages/pymongo/mongo_client.py", line 1248, in _retryable_write
    return self._retry_with_session(retryable, func, s, None)
  File "/usr/local/lib64/python3.7/site-packages/pymongo/mongo_client.py", line 1201, in _retry_with_session
    return func(session, sock_info, retryable)
  File "/usr/local/lib64/python3.7/site-packages/pymongo/collection.py", line 847, in _update
    retryable_write=retryable_write)
  File "/usr/local/lib64/python3.7/site-packages/pymongo/collection.py", line 818, in _update
    _check_write_command_response(result)
  File "/usr/local/lib64/python3.7/site-packages/pymongo/helpers.py", line 217, in _check_write_command_response
    _raise_last_write_error(write_errors)
  File "/usr/local/lib64/python3.7/site-packages/pymongo/helpers.py", line 198, in _raise_last_write_error
    raise DuplicateKeyError(error.get("errmsg"), 11000, error)
pymongo.errors.DuplicateKeyError: E11000 duplicate key error collection: finance.dataitemdetails index: asset_id_1_dataitem_id_1_detail_id_1_period_type_1_scenario_id_1_currencycode_1_period_end_1 dup key: { : 19, : 1211, : 0, : "Month", : 1, : "RC", : new Date(949276800000) }

```


Find in Mongodb console returned only 1 document


```


>db.dataitemdetails.find({'asset_id':19,'dataitem_id':1211,'detail_id':0,'period_type':'Month','currencycode':'RC','period_end':new Date(949276800000)})
{ "_id" : ObjectId("5c7721c17314e53a85be7e89"), "Value" : "USD", "period_end" : ISODate("2000-01-31T00:00:00Z"), "currencycode" : "RC", "scenario_id" : 1, "dataitem_id" : 1211, "period_type" : "Month", "detail_id" : 0, "asset_id" : 19 }
>

```

Here is my replace one query in the mongo console instead of pymongo:

```
> db.dataitemdetails.replaceOne({'asset_id':19,'dataitem_id':1211,'detail_id':0,'period_type':'Month','currencycode':'RC','period_end':new Date(949276800000)} ,{'asset_id':19,'dataitem_id':1211,'detail_id':0,'period_type':'Month','currencycode':'RC','period_end':new Date(949276800000), 'Value':'USD'}, upsert = true)
{ "acknowledged" : true, "matchedCount" : 1, "modifiedCount" : 1 }
>
```

some other notes:

If I remove the compound index, the error message would go away, but I would end up with duplicate documents that are the same.

I tried to drop the collection, database and it would end up the same.

Answer 1

我发现了问题：

在这行代码中：

db['dataitemdetails'].replace_one({'asset_Id':tdata['asset_id'],'period_type':tdata['period_type'],'detail_id':tdata['detail_id'], 'currencycode':tdata['currencycode'],'dataitem_Id':tdata['dataitem_id'],'period_end':tdata['period_end'], 'scenario_id':tdata['scenario_id'],}, tdata, upsert=True)

我的过滤器有误。

其中asset_Id应该为asset_id，而dataitem_Id应该为dataitem_id。

因此找不到匹配项，但是在插入重复的键错误时出现。

Answer 2

您正在告诉mongodb替换一个文档。您的过滤器很可能会返回多个文档。

使用带有相同过滤器的find并找出一个将始终返回一个文档的过滤器。在大多数情况下，只要正确布置架构，仅使用id就足以返回一个文档。

Answer 3

要回答第一个问题，

如果您不提供upsert = true的_id字段值，MongoDB将生成一个新的ID

出现重复键错误的原因是，您没有在过滤条件中使用period字段。

带有upsert = true的Mongodb replace_one（）引发重复键错误

3 个答案: