问题
I am working on pagination in flask(Python framework) using flask-paginate (just for ref)
I am able to achieve pagination for just a find
query as below:
from flask_paginate import Pagination
from flask_paginate import get_page_args
def starting_with_letter(letter):
page, per_page, offset = get_page_args()
collection_name=letter.lower()+'_collection'
words=db[collection_name]
data_db=words.find()
data=data_db.limit(per_page).skip(offset) '''Here I have achieved the limit and skip'''
pagination = Pagination(page=page, total=data.count(),per_page=per_page,offset=offset,record_name='words')
return render_template('startingwords.html',data=data,pagination=pagination)
But I am not able to do the same for the aggregate here:
def test():
page, per_page, offset = get_page_args()
cursor_list=[] '''appending each cursor in iteration of for loop '''
collections=db.collection_names()
for collection in collections:
cursor_objects = db[collection].aggregate([
{
"$match": {
"$expr": {"$eq": [{"$strLenCP": "$word"}, 6]}
}
},
{"$skip": offset},
{"$limit": per_page}
])
for cursor in cursor_objects:
cursor_list.append(cursor)
pagination = Pagination(page=page, total=len(cursor_list),per_page=per_page,offset=offset,record_name='words')
return render_template('lettersearch.html',data=cursor_list,pagination=pagination)
The results are displayed as :
Here all the 39
results are shown at single page
On hitting page 2
it showed :
Note: By default flask-paginate sets initially per_page
as 10
and offset
as 0
after referring many links i have tried:
placing skip
and limit
above match
which is wrong any way
Also learnt that limit
is always followed by skip
I am stuck with this, Any help is appreciated
回答1:
Your issue is not with the skip()
and limit()
; that is working fine. The issue is with your overall logic; you are iterating all 39 collections in the first loop and then appending each result of the aggregation to cursor_list
.
I can't figure out the logic of what you are trying to do, as the first example is looking in a words collection and second is looking in all collections for a word field; with that said, you can likely simplify your approach to something like:
offset = 0
per_page = 10
collections = db.list_collection_names()
#
# Add some logic on the collections array to filter what is needed
#
print(collections[offset:offset+per_page])
EDIT to reflect comments. Full worked example of a function to perform this. No need for an aggregation query - this adds complexity.
from pymongo import MongoClient
from random import randint
db = MongoClient()['testdatabase1']
# Set up some data
for i in range(39):
coll_name = f'collection{i}'
db[coll_name].delete_many({}) # Be careful; testing only; this deletes your data
for k in range (randint(0, 2)):
db[coll_name].insert_one({'word': '123456'})
# Main function
def test(offset, per_page, word_to_find):
found = []
collections = db.list_collection_names()
for collection in sorted(collections):
if db[collection].find_one({word_to_find: { '$exists': True}}) is not None:
found.append(collection)
print(found[offset:offset+per_page])
test(offset=0, per_page=10, word_to_find='word')
来源:https://stackoverflow.com/questions/59573513/skip-and-limit-for-pagination-for-a-mongo-aggregate