$匹配（聚合）失败，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") } 

我想经营一个简单聚合管道（大大简化了解释）

我运行这个蒙戈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()); } 

这个结果（这是绝对正确）

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") } 

我想将其转换为蟒蛇如下：

... 
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) 

这个结果：

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：

的$匹配操作从未采取的ObjectId语法考虑。

如何正确写入？

感谢您的任何提示

基督教

Answer 1

你要使用的ObjectId类标准库

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

Possibly Related

眼下的比赛，并找到字段问_id是一个名为“ObjectId（”5908e63cd15fa104356eaf64）“的字符串，但是在Python中，您必须首先分配一个唯一的标识符

你的最后一场比赛是实际打印一切，因为你是问，如果所有_ids不等于字符串“的ObjectId（” 5908e63cd15fa104356eaf64）”，而不是的ObjectId