我在这里找到了一个很好的分页模板。不过,这个示例是使用SQLlite编写的,但我已经重写了它以便能够使用PyMongo。
这个脚本使用PyMongo创建用户并将其保存到MongoDB中:
import sys
from pymongo import MongoClient
def fill_data(users_no):
for i in range(users_no):
doc = {
'_id': str(i),
'uname': "name_" + str(i),
}
sDB.insert(doc)
if __name__ == '__main__':
db = MongoClient().test
sDB = db.users
fill_data(1000)
我修改了原始的app.py脚本,以便能够使用PyMongo。这是修改后的脚本:
#!/usr/bin/env python
#-*- coding: utf-8 -*-
from __future__ import unicode_literals
from flask import Flask, render_template, g, current_app, request
from flask.ext.paginate import Pagination
from pymongo import MongoClient
app = Flask(__name__)
app.config.from_pyfile('app.cfg')
@app.route('/')
def index():
db_name = "test"
col_name = "users"
db = MongoClient()
sDB = db[db_name][col_name]
total = sDB.find().count()
print total
page, per_page, offset = get_page_items()
users = sDB.find().skip(offset).limit(per_page)
for u in users:
print u
pagination = get_pagination(page=page,
per_page=per_page,
total=total,
record_name=users,
)
return render_template('index.html', users=users,
page=page,
per_page=per_page,
pagination=pagination,
)
def get_css_framework():
return current_app.config.get('CSS_FRAMEWORK', 'bootstrap3')
def get_link_size():
return current_app.config.get('LINK_SIZE', 'sm')
def show_single_page_or_not():
return current_app.config.get('SHOW_SINGLE_PAGE', False)
def get_page_items():
page = int(request.args.get('page', 1))
per_page = request.args.get('per_page')
if not per_page:
per_page = current_app.config.get('PER_PAGE', 10)
else:
per_page = int(per_page)
offset = (page - 1) * per_page
return page, per_page, offset
def get_pagination(**kwargs):
kwargs.setdefault('record_name', 'records')
return Pagination(css_framework=get_css_framework(),
link_size=get_link_size(),
show_single_page=show_single_page_or_not(),
**kwargs
)
if __name__ == '__main__':
app.run(debug=True)
有没有一种方法可以通过点击分页菜单中的下一页或上一页来避免使用
count()
?我忘记改变什么,以便在浏览器中显示实际用户,因为目前我只能看到分页菜单而没有用户?
from flask.ext.mongoengine.pagination import Pagination
。 - hamidfzm