$匹配(聚合)失败，而objectID值

Question

我有一个简单的集合。

> db.y.find({}, {'_id': 1})
{ "_id" : ObjectId("5908e63cd15fa104356eaf64") }
{ "_id" : ObjectId("5908e63cd15fa104356eaf65") }
{ "_id" : ObjectId("5908e63cd15fa104356eaf66") }
{ "_id" : ObjectId("5908e63cd15fa104356eaf67") }
{ "_id" : ObjectId("5908e63cd15fa104356eaf68") }
{ "_id" : ObjectId("5908e63cd15fa104356eaf69") }
{ "_id" : ObjectId("5908e63cd15fa104356eaf6a") }
{ "_id" : ObjectId("5908e63cd15fa104356eaf6b") }

我想操作一个简单的聚合管道(在很大程度上简化了解释(

我运行此mongo shell脚本：

print('find')
result = db.y.find({ '_id': ObjectId("5908e63cd15fa104356eaf64") }, {'_id':1})
while ( result.hasNext() ) { printjson( result.next() ); }
print('aggregate match direct')
result = db.y.aggregate( [ {'$match': {'_id': ObjectId("5908e63cd15fa104356eaf64") } }, {'$project': {'_id': 1}} ] )
while ( result.hasNext() ) { printjson( result.next() ); }
print('aggregate match with $eq')
result = db.y.aggregate( [ {'$match': {'_id': {'$eq': ObjectId("5908e63cd15fa104356eaf64") } } }, {'$project': {'_id': 1}} ] )
while ( result.hasNext() ) { printjson( result.next() ); }
print('aggregate match with $ne')
result = db.y.aggregate( [ {'$match': {'_id': {'$ne': ObjectId("5908e63cd15fa104356eaf64") } } }, {'$project': {'_id': 1}}, {'$limit': 5} ] )
while ( result.hasNext() ) { printjson( result.next() ); }

使用此结果(这是绝对正确的(

find
{ "_id" : ObjectId("5908e63cd15fa104356eaf64") }
aggregate match direct
{ "_id" : ObjectId("5908e63cd15fa104356eaf64") }
aggregate match with $eq
{ "_id" : ObjectId("5908e63cd15fa104356eaf64") }
aggregate match with $ne
{ "_id" : ObjectId("5908e63cd15fa104356eaf65") }
{ "_id" : ObjectId("5908e63cd15fa104356eaf66") }
{ "_id" : ObjectId("5908e63cd15fa104356eaf67") }
{ "_id" : ObjectId("5908e63cd15fa104356eaf68") }

然后，我想将其转换为Python，如下所示：

...
print('find')
result = y.find({ '_id': 'ObjectId("5908e63cd15fa104356eaf64")' }, {'_id':1})
for i, o in enumerate(result):
    print(i, o)
print('aggregate match direct')
result = y.aggregate( [ {'$match': {'_id': 'ObjectId("5908e63cd15fa104356eaf64")' } }, {'$project': {'_id': 1} } ] )
for i, o in enumerate(result):
    print(i, o)
print('aggregate match with $eq')
result = y.aggregate( [ {'$match': {'_id': {'$eq': 'ObjectId("5908e63cd15fa104356eaf64")' } } }, {'$project': {'_id': 1} } ] )
for i, o in enumerate(result):
    print(i, o)
print('aggregate match with $ne')
result = y.aggregate( [ {'$match': {'_id': {'$ne': 'ObjectId("5908e63cd15fa104356eaf64")' } } }, {'$project': {'_id': 1} }, {'$limit': 5} ] )
for i, o in enumerate(result):
    print(i, o)

与此结果：

find
aggregate match direct
aggregate match with $eq
aggregate match with $ne
0 {'_id': ObjectId('5908e63cd15fa104356eaf64')}
1 {'_id': ObjectId('5908e63cd15fa104356eaf65')}
2 {'_id': ObjectId('5908e63cd15fa104356eaf66')}
3 {'_id': ObjectId('5908e63cd15fa104356eaf67')}
4 {'_id': ObjectId('5908e63cd15fa104356eaf68')}

结论：

$匹配操作永远不会考虑objectid语法。

如何正确编写？

感谢任何提示

基督徒

Answer 1

您必须在标准库中使用objectid类

from bson.objectid import ObjectId
result = y.find({ '_id': ObjectId("5908e63cd15fa104356eaf64") }, {'_id':1})

可能相关

现在，比赛并找到字段正在询问_id是否是一个名为" ObjectID(" 5908E63CD15FA104356(的字符串"。但是，在Python中，您必须首先与ObjectID类别分配一个唯一的标识符

您的最后一场比赛实际上是在打印所有内容，因为您询问所有_ids是否不等于字符串" ObjectId(" 5908E63CD15FA10435644(("，而不是objectid